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this is a lecture No. 21 of a B 2 3 2

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Today we will study how to determine the resistance

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and capacitys of different types of resistors

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and capacitys using boundary value problem

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this will cover your section 6.5 of textbook

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this is Academic Week No. 13 of this semester

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the learning objectives of today's lecture would be

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to determine the resistance

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and capacity my solving be boundary value problem

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you remember that in previous lecture

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we have studied

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how to solve these boundary value problems

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using the Liplas and poisons equation

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first of all the resistance

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for conductors with uniform cross sectional area

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the resistance it is related by this relationship

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where to the

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resistance is

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directly proportional to the length of this conductor

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and it is inversely proportional to the surface area

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of that conductor

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if the cross section of conductor is not uniform

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then the resistance can

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can still be obtained using this Homs law

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but in that case your voltage is and current

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they will be of some integral form

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and here we also studied that

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how to find out the voltages

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using this integration of electrical

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and how to find out this current

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using the integration of this current density

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okay surface current density here

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so finding the resistance of a conductor with non

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uniform cross sectionalia

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it is also known as the boundary value problem

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so we will involve this boundary value problem

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to determine these two

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these voltage and grant for the non uniform

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cross sectionality of the resistors

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these are the steps that we will follow

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to find out the resistance of uh

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any conducting media material

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which is having non uniform surface area

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first of all we will uh

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choose a suitable coordinate system

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that can be your cartision

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slendicalant cervical coordinate system

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thereafter we will assume uh

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v not as the potential difference between the uh

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conductor terminals

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then will solve depending upon the situation

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either it is a lap last equation case

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or poison sequation case

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so if it is a lap last equation case

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then will involve this question

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and uh we will uh

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double integrate this equation to find out the v

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in terms of here v

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and then we will determine the E

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electrically using this relationship between the uh

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potential and the electrically

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and then we will uh

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use this integral equation to find out the current

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in terms of in terms of we not right

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so we will find out this in terms of we not

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and then using this Onstar relationship

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will find out the resistance

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by simply dividing this assumed constant

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potential

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difference between the terminals of the conductor

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divided by this current that we have determined

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for the non uniform surface surface conductor

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so it is also possible to operate in other ways

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that will first resume the current is constant

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okay and then we find out the potential

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a corresponding potential defence in terms of I

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not in terms of I not remember

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so that this in the previous case

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this we not and we not would be cancelled

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in this case this I not and I not would be cancelled

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and we are left with only be resistance

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to have contestants we must have two conductors

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and which carry that equal and opposite charges

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the conductors may be separated by free space

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or some insulating material

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that is known as dialect material

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so here's uh it is an example of your capactor

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which is having two plates

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and uh you can see over here

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the charges are of uh same magnitude

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but opposite polarity

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and they are separated by some distance and then they

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they can be this separation

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it can be filled either by this free space

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if this is free space

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then your option is simply abslan not

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and if it is other material

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then you will have abslan is equal to abslan

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not times D

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relative permittivity

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so the conductors are maintained at a potential

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that is v in this case

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and it is also

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the potential difference between these two conductors

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and it is uh related uh with this electric V E

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using this relationship

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we have already derived this relationship

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that how

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to find out

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the potential difference between two conductors

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by simply taking the line

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integral of this electric feed

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so you notice that this electric grid

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it is always normal this electric cross lines

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they are always normal to the surface

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of these conductors so all flatlines

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leaving one conductor

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must necessarily terminate at the surface of

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other conductor and also by a radical idea

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at the normal orientation

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to the surface of that originating conductor

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and terminating conductor

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so the capacitor saw

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of the capacitor

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is defined as the ratio of magnitude of the charge

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on one of the plate

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to the potential difference between them

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so this is the basic relationship

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that we have already utilized

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in our early classes as well

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in our early lectures as well

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the only difference over here is

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it is one non uniform surface area

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so if it is the in non uniform surface areas

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then obviously we are going to involve the integrals

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and here uh from glazes law

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we know that uh

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this charge enclosed is equal to the uh

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this uh surface integral of here

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uh flux uh this is electric flux density right

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this one

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okay and uh

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and this is the same relationship between the uh

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that uh

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voltage potential is equal to the integral of here

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align integral of here electric feed

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so there again

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there are two methods of finding all the capacitys

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for any two conductor capacitors

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so

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it is either by using the gauzesla or the liplactions

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by solving the liplactions equation

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so if we assume q the charge as a constant

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and then we determine this V in terms of this q

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then this will enroll your causes drop

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and the vice versa if we assume V as a constant

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and determine this q in terms of your V

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then we will we need to solve this

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the class integration we'll do these things

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uh one by one in our uh

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in our examples

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examples as well and in our innovations as well

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so magnet 1 involves the falling steps again

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you need to choose the suitable corner system

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according to your problem

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and then let the conducting plates uh

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let the let the two conducting plates carry a charges

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uh equal

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uh equal uh

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magnitude charges but with

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but with this opposite polarity vitamin E electrical

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either using your coolants law or Gaza's Law

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for symmetrical surface densities charges

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charge distributions

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and then if we know the electric field

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then obviously

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we can find out this voltage

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using this relationship of uh

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between this align integral of your electrically

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I know yes

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you know since you have found out the B in terms of Q

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then you can uh

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involve the specific relationship to find out

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the capacitors of these two

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uh conductor

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play conductor uh

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plate capacitors and hey remember

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that negative sign may be ignored

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as absolute value of V is required because

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because we know that capacity

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capacity is a skill of quantity

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uh these are the uh

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some of the examples of your capacitors as you can see

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as you can see as you can see

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that these are electrolytic types of the capacitors

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right these

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if you are you're seeing that these are

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these are ceramic types of the conductors

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so electrolytic types of the conductor

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they involve the polarity

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okay positive and negative like

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whereas the ceramic types of the conductors they are

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they are in the in nanos and fatter

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nanoferribs and PICO ferbs of the capacitors

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whereas these are you can as you can see here

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these are in microfranger milli fat

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whereas these are injured nano feathers and

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and PICO feathers

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so what is the purpose of your chair or this

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this capacitor okay

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a capacitor consists of two conductors of any shape

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placed near one another without touching each other

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without touching

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remember this without touching each other

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it is common to fill up

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the region between these two conductors

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with an insulating material

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known as the dialectric

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and we have seen that the capacity

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the storage

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is directly proportional to the dialectric value

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of the medium

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that is present in between these two conductor plates

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right this this is the dialect

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and uh we charge these plates with uh

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with opposing charges to set up an electric field

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so and uh so if we uh

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if we uh connect these two these two plates

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the better metallic connector plates with some uh

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external potential then

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then electric field is going to build up in between

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these two

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private plates

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and then you're going to store the charge between

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charge on this to pasture plates

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uh this is very important

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this slide is uh you can uh

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you can uh

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take this slide as a complete summary of your uh

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these two lectures that

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how you need to proceed to find out the capacity

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and the resistance of conductors

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so the first step if you keep this Q as a constant okay

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q constant

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and then you need to determine this V in terms of Q

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and for this purpose

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you will take help of this cooling stop

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okay the second

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the second one is your uh if

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if you keep this V as a constant

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v not as a constant between two plates

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and then you can find out this Q using this um

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definite definition of this uh flux

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flux uh

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this uh

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that charge is equal to the uh

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surface in general of your electric

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uh electric uh flux 10 3

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and then from here you can

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you know that

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this is the constitutive relationship between this

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Indian electrical

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and then if you if you have found out this electrical

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from this equation

265

00:11:41,166  -->  00:11:44,600
then you can use this relationship to find out

266

00:11:44,600  -->  00:11:46,200
the electrical and hey

267

00:11:46,200  -->  00:11:47,733
remember that we will

268

00:11:47,733  -->  00:11:50,166
we will take help of the Prussians

269

00:11:50,166  -->  00:11:52,500
or the poisons equation

270

00:11:52,500  -->  00:11:54,900
according to this case to find out this

271

00:11:55,466  -->  00:11:56,700
this v

272

00:11:59,000  -->  00:12:00,400
like here okay

273

00:12:01,733  -->  00:12:04,366
and the third thing is your resistance

274

00:12:04,700  -->  00:12:07,366
so for the resistance if you again

275

00:12:07,366  -->  00:12:10,866
keep this voltage across that resistance

276

00:12:10,866  -->  00:12:12,000
at that constant

277

00:12:12,000  -->  00:12:14,366
then you need to determine this current

278

00:12:14,366  -->  00:12:16,266
flowing to that conductor

279

00:12:16,800  -->  00:12:21,200
and that is your surface integral of your condensity

280

00:12:21,366  -->  00:12:24,666
and where this conduct is equal to using the ONS

281

00:12:24,666  -->  00:12:27,933
like it is equal to conductivity times e electrical

282

00:12:27,933  -->  00:12:29,966
and again you can find out this electrical age

283

00:12:29,966  -->  00:12:32,933
using this relationship and how to find out this V

284

00:12:32,933  -->  00:12:35,800
again you need to take the help of this uh

285

00:12:35,800  -->  00:12:39,566
solving this boundary problem or uh equently the uh

286

00:12:39,566  -->  00:12:42,666
solution of finding out the solution of this leclasin

287

00:12:42,666  -->  00:12:43,966
poison secretion

288

00:12:47,466  -->  00:12:50,700
let's first find out the capacitors of our power plate

289

00:12:50,700  -->  00:12:51,566
capacitor

290

00:12:52,166  -->  00:12:54,766
in an ideal power plate capacitors

291

00:12:54,900  -->  00:12:58,300
the plate separation this means the both conductors

292

00:12:58,300  -->  00:13:03,066
they are separated by a distance or separation of D

293

00:13:03,766  -->  00:13:05,166
it is very small

294

00:13:05,166  -->  00:13:07,666
as compared to the dimensions of the plate

295

00:13:07,666  -->  00:13:09,900
so this is the first Assumption

296

00:13:10,133  -->  00:13:13,500
if this is an Assumption that the distance is very

297

00:13:13,500  -->  00:13:15,600
very small as compared to the

298

00:13:16,500  -->  00:13:18,766
the size of this place of the capacitor

299

00:13:19,066  -->  00:13:22,466
then we can ignore the fringing fields

300

00:13:22,466  -->  00:13:25,933
at the edge of place so this can be ignored

301

00:13:25,933  -->  00:13:28,166
okay this can be ignored

302

00:13:29,166  -->  00:13:30,700
okay the first thing

303

00:13:31,166  -->  00:13:33,866
and then uh the second thing is the field

304

00:13:33,866  -->  00:13:36,933
uh the field between them is considered to be constant

305

00:13:36,933  -->  00:13:39,066
so they will be considered as constant

306

00:13:39,066  -->  00:13:40,666
you can see over here they're

307

00:13:40,666  -->  00:13:43,166
they're they're showing just like equal

308

00:13:43,500  -->  00:13:48,900
equal magnitude lines so the space between the plates

309

00:13:48,966  -->  00:13:51,400
it is filled up with a homogeneous

310

00:13:51,400  -->  00:13:52,333
another resumption

311

00:13:52,333  -->  00:13:55,666
homogeneous dialectic with permetuity absline

312

00:13:55,966  -->  00:13:58,866
and we and this flux

313

00:13:58,866  -->  00:14:01,166
fringing can be ignored by

314

00:14:01,166  -->  00:14:02,000
using this

315

00:14:02,000  -->  00:14:05,200
smaller separation between these two conductor plates

316

00:14:05,200  -->  00:14:08,800
then what we can we can we can into that

317

00:14:10,133  -->  00:14:13,466
the uh this electric trucks electric cross density

318

00:14:13,466  -->  00:14:16,733
it is uh using either using this boundary conditions

319

00:14:16,733  -->  00:14:18,333
it is equal to uh

320

00:14:18,333  -->  00:14:22,766
this surface surface chart density time

321

00:14:22,766  -->  00:14:27,166
this your vector country is your basically the normal

322

00:14:27,166  -->  00:14:29,366
the unit normal vector to the surface of this

323

00:14:29,366  -->  00:14:32,266
these two plates so we can uh uh

324

00:14:32,266  -->  00:14:34,500
we can uh we can proceed in another

325

00:14:34,500  -->  00:14:38,766
another way that this will derive in our chapter No. 4

326

00:14:38,766  -->  00:14:40,666
that this is the uh

327

00:14:40,666  -->  00:14:43,200
your electric field on the surface

328

00:14:43,200  -->  00:14:44,866
on one conductor surface

329

00:14:45,066  -->  00:14:48,666
uh which is having the surface on distribution of raw s

330

00:14:49,400  -->  00:14:51,466
and if it is of negative here

331

00:14:51,466  -->  00:14:53,766
if it is of negative charge

332

00:14:53,766  -->  00:14:56,100
then you can you can see over here

333

00:14:57,100  -->  00:14:59,033
you can see over here that uh

334

00:14:59,100  -->  00:15:01,533
for this uh for these two things

335

00:15:01,533  -->  00:15:03,800
uh if for these two two things

336

00:15:03,800  -->  00:15:07,566
the uh the normal it is in the opposite direction right

337

00:15:07,566  -->  00:15:08,566
so this is uh

338

00:15:08,566  -->  00:15:13,566
this this was your end and equation that was uh uh

339

00:15:13,566  -->  00:15:16,833
determine the electric field between two power plate

340

00:15:17,400  -->  00:15:19,766
uh capactor and uh

341

00:15:19,766  -->  00:15:22,633
it is in terms of the surface charge of uh

342

00:15:23,266  -->  00:15:24,966
both capactors uh

343

00:15:24,966  -->  00:15:27,666
and the unit normal between them

344

00:15:27,666  -->  00:15:29,800
in the direction of your electric feed

345

00:15:30,300  -->  00:15:34,866
so if you if we apply this equation on this case right

346

00:15:35,200  -->  00:15:38,800
we have this negative q negative rules

347

00:15:38,800  -->  00:15:40,233
you can also say okay

348

00:15:40,466  -->  00:15:44,466
and this positive rules you can also say right

349

00:15:44,933  -->  00:15:48,133
they are separated by a distance d and since uh

350

00:15:48,133  -->  00:15:51,333
this positive positive rate is at top

351

00:15:51,333  -->  00:15:54,766
so you can assume that this normal

352

00:15:54,866  -->  00:15:58,033
normal direction is in minus x

353

00:15:58,266  -->  00:15:59,400
right so

354

00:15:59,800  -->  00:16:02,700
the electric field originates from the positive plate

355

00:16:02,700  -->  00:16:06,166
and the terminates on the negative charge plate

356

00:16:06,800  -->  00:16:09,100
so put in this this normal

357

00:16:09,100  -->  00:16:11,466
uh you know normal factor

358

00:16:11,566  -->  00:16:13,566
uh in the direction of your electric field

359

00:16:13,566  -->  00:16:15,433
so this is your electric field

360

00:16:16,000  -->  00:16:17,166
uh for this uh

361

00:16:17,166  -->  00:16:18,900
particular uh example

362

00:16:18,900  -->  00:16:19,700
okay

363

00:16:20,600  -->  00:16:21,300
and uh

364

00:16:21,300  -->  00:16:24,300
it is just the definition of your surface identity

365

00:16:24,300  -->  00:16:25,900
that it is equal to your

366

00:16:26,800  -->  00:16:29,566
what is you remember that it is your uh

367

00:16:29,566  -->  00:16:32,333
that charge budget per unit alia

368

00:16:32,333  -->  00:16:36,333
right so first charge then is your charge per unit alia

369

00:16:36,333  -->  00:16:38,166
that is your Judi right

370

00:16:38,333  -->  00:16:41,900
yes so that is your question

371

00:16:44,733  -->  00:16:47,666
so suppose this is just again the reputation of that

372

00:16:48,100  -->  00:16:50,600
suppose each plate has a surface area of s

373

00:16:50,600  -->  00:16:52,700
the players are separated by a distance d

374

00:16:52,766  -->  00:16:54,433
as in page 1 and 2

375

00:16:54,466  -->  00:16:58,200
carry uniformly distributed charge plus Q and minus Q

376

00:16:58,333  -->  00:16:59,600
and accordingly there will be

377

00:16:59,600  -->  00:17:01,866
so this is the same as we have discarded now

378

00:17:01,866  -->  00:17:03,866
previous try okay

379

00:17:03,966  -->  00:17:07,400
but let's apply this relationship

380

00:17:07,400  -->  00:17:10,533
the line into the relationship to find out the voltage

381

00:17:10,533  -->  00:17:14,000
right so remember in this case we have assumed

382

00:17:14,333  -->  00:17:16,666
we assume that the charge is constant

383

00:17:16,666  -->  00:17:17,566
okay the first case

384

00:17:17,566  -->  00:17:21,066
that the charge is constant on this surface of place

385

00:17:21,166  -->  00:17:21,600
and we are

386

00:17:21,600  -->  00:17:26,200
going to determine this V in terms of this Q

387

00:17:27,066  -->  00:17:29,300
and for that we know this relationship

388

00:17:29,300  -->  00:17:30,200
for this relationship

389

00:17:30,200  -->  00:17:32,400
we need to find out this electrical

390

00:17:32,400  -->  00:17:34,100
that we have no product

391

00:17:34,966  -->  00:17:37,200
now take the integration of this thing

392

00:17:37,200  -->  00:17:41,300
then uh you can find or that uh what is the variable

393

00:17:41,300  -->  00:17:44,400
the variable is X this thing okay

394

00:17:44,766  -->  00:17:50,433
and it is Jenny from it is Jenny from 0 2 d right

395

00:17:50,566  -->  00:17:54,300
uh see your hair so uh here it is disciples that 0

396

00:17:54,300  -->  00:17:57,866
x is equal 0 disciples that X is equal to

397

00:17:59,133  -->  00:17:59,933
d

398

00:18:00,466  -->  00:18:02,633
so this is that actually equal to zero

399

00:18:02,900  -->  00:18:06,900
so that's why these integration limits are from 0 to D

400

00:18:07,500  -->  00:18:08,700
okay and then uh

401

00:18:08,700  -->  00:18:11,700
you just plug in uh this electrical lower here

402

00:18:11,700  -->  00:18:12,833
you take uh

403

00:18:13,166  -->  00:18:16,666
this uh center differential lenses in the X axis

404

00:18:16,666  -->  00:18:20,433
so that's why it's uh unigratory is a a X

405

00:18:21,900  -->  00:18:23,566
now you solve this equation

406

00:18:23,566  -->  00:18:25,100
so this integral it is

407

00:18:25,100  -->  00:18:29,200
you can say that it is it is like this okay

408

00:18:29,966  -->  00:18:36,500
0 to d okay and then X integration of 1

409

00:18:36,500  -->  00:18:38,800
you know that it is X in this case

410

00:18:40,066  -->  00:18:41,533
okay so from here

411

00:18:41,533  -->  00:18:45,600
if you move on and plug in this water to your hair

412

00:18:45,733  -->  00:18:49,100
then this is your capacity okay

413

00:18:49,300  -->  00:18:52,000
this is the capacity of your better grade capacity

414

00:18:52,066  -->  00:18:52,666
capacity

415

00:18:52,666  -->  00:18:55,300
that you have already started in your early classes

416

00:18:55,300  -->  00:18:57,200
as well the capacity

417

00:18:57,266  -->  00:19:01,000
it is directly proportional to the surface area of the

418

00:19:01,000  -->  00:19:05,100
plates and it is inversely proportional to

419

00:19:05,100  -->  00:19:08,200
the suppression between these two plates

420

00:19:08,200  -->  00:19:09,400
right and

421

00:19:09,400  -->  00:19:13,466
it is also directly proportional to the permittuity

422

00:19:13,466  -->  00:19:14,966
of dialectic material

423

00:19:14,966  -->  00:19:17,833
which has been placed in between these two

424

00:19:18,266  -->  00:19:19,833
better plate capacitors

425

00:19:20,933  -->  00:19:24,266
so if we want to find out this absent are

426

00:19:24,333  -->  00:19:26,766
then what we need to do that the

427

00:19:26,766  -->  00:19:27,800
the basically

428

00:19:27,800  -->  00:19:30,066
this relative permativity of the medium

429

00:19:30,066  -->  00:19:32,800
in between these two palpric pastor

430

00:19:33,000  -->  00:19:36,000
then first we need to determine on major decapacitors

431

00:19:36,000  -->  00:19:39,200
of pirate plate capacitor with space

432

00:19:39,200  -->  00:19:43,766
with free space medium between this plate right

433

00:19:43,766  -->  00:19:45,800
this absolute not case right

434

00:19:47,266  -->  00:19:49,933
and then we need to

435

00:19:49,933  -->  00:19:52,800
so with the redistrate of first of first cases sorry

436

00:19:52,800  -->  00:19:54,366
the first first cases

437

00:19:54,933  -->  00:19:56,466
the space between them

438

00:19:57,133  -->  00:19:59,100
the space between them is

439

00:19:59,566  -->  00:20:02,600
the space between players filled with the dialectric

440

00:20:02,600  -->  00:20:04,000
right and we'll uh

441

00:20:04,000  -->  00:20:04,866
we'll uh

442

00:20:04,866  -->  00:20:08,333
we'll find out that capacity will level that uh

443

00:20:08,333  -->  00:20:11,500
uh capacity as C okay

444

00:20:11,500  -->  00:20:13,800
and the second one is see not for this case

445

00:20:13,800  -->  00:20:17,266
we simply replace this dialect medium

446

00:20:17,266  -->  00:20:20,600
medium with d free space Audi here

447

00:20:20,600  -->  00:20:22,466
that is of perhaps not

448

00:20:22,500  -->  00:20:24,500
so we just divide these two caps

449

00:20:24,500  -->  00:20:24,933
tenses

450

00:20:24,933  -->  00:20:28,033
to find out the relative promativity of the dialectic

451

00:20:28,100  -->  00:20:30,366
that have have been placed in the previous

452

00:20:30,366  -->  00:20:31,966
in the in the first case

453

00:20:34,966  -->  00:20:38,000
so if we found we have found out the electric field

454

00:20:38,000  -->  00:20:39,366
and we can also find out

455

00:20:39,366  -->  00:20:42,200
this energy stored in a capacitor

456

00:20:42,200  -->  00:20:44,533
so this was the relationship that we already studied

457

00:20:44,533  -->  00:20:46,766
that the energy stored

458

00:20:46,966  -->  00:20:51,066
that the energy stored can be related with this

459

00:20:51,066  -->  00:20:54,733
electric field using this relationship

460

00:20:54,733  -->  00:20:56,666
that it is half volume

461

00:20:56,666  -->  00:21:00,033
integral of your square of this electrically

462

00:21:01,200  -->  00:21:04,733
so now if we substitute this barrel plate capacity

463

00:21:04,733  -->  00:21:06,900
electric field in this equation

464

00:21:06,933  -->  00:21:08,800
and simplify this equation

465

00:21:08,900  -->  00:21:10,166
like this one okay

466

00:21:10,766  -->  00:21:12,566
and uh like this one okay

467

00:21:12,800  -->  00:21:15,533
and uh what you can do with so if you

468

00:21:15,533  -->  00:21:18,900
if you simplify this equation and if you simplify

469

00:21:18,900  -->  00:21:19,866
let's see what

470

00:21:19,866  -->  00:21:22,866
what simplification we are going to involve further uh

471

00:21:22,866  -->  00:21:25,733
so this uh is going to cancel out okay

472

00:21:25,733  -->  00:21:27,933
this option is going to cancel with this one

473

00:21:27,933  -->  00:21:29,800
so you are left over with this one

474

00:21:30,066  -->  00:21:32,366
and the second thing is very important thing is

475

00:21:32,366  -->  00:21:35,666
you are changing this volume material

476

00:21:35,800  -->  00:21:39,566
that is the severe volume is equal to your

477

00:21:39,866  -->  00:21:42,766
is equal to your surface area

478

00:21:42,766  -->  00:21:45,866
and the tire dimension of your

479

00:21:46,933  -->  00:21:52,100
of your this coordinate system so here it was d okay

480

00:21:52,800  -->  00:21:56,066
and so that's why this SND is coming over here

481

00:21:56,066  -->  00:21:57,900
the integration of this volume

482

00:21:57,900  -->  00:22:00,066
the volume trigger okay

483

00:22:00,066  -->  00:22:03,900
and so if you put in order to simplify this

484

00:22:03,900  -->  00:22:06,533
so this is the remember this equation

485

00:22:06,533  -->  00:22:08,666
this is the relationship to find out the

486

00:22:09,200  -->  00:22:11,266
the energy stored in the power plate

487

00:22:11,266  -->  00:22:14,433
capacitor in terms of the charge and the voltage

488

00:22:14,733  -->  00:22:20,500
and if you if you know the capacity then you can

489

00:22:20,500  -->  00:22:22,466
you can you can just simplify

490

00:22:22,800  -->  00:22:27,166
you can simply involve that relationship between your Q

491

00:22:27,166  -->  00:22:30,200
and bring the charge and reward case

492

00:22:30,200  -->  00:22:33,700
to find out further this relationship

493

00:22:34,566  -->  00:22:37,800
so just take a take help of this relationship that you

494

00:22:38,000  -->  00:22:43,633
C is equal to q divided by v okay

495

00:22:45,366  -->  00:22:47,066
plug in this over here and uh

496

00:22:47,066  -->  00:22:50,333
you can uh find out the different forms of the same

497

00:22:50,333  -->  00:22:52,100
this energy stored you know

498

00:22:52,100  -->  00:22:53,466
my parallel blade capacitor

499

00:22:53,466  -->  00:22:55,800
so it is only for the pilot parallel blade capacity

500

00:22:55,800  -->  00:22:56,866
remember this thing

501

00:22:58,300  -->  00:23:00,933
so let's let's discuss in other case

502

00:23:00,933  -->  00:23:03,766
uh for which we need to find out the uh

503

00:23:03,766  -->  00:23:06,633
capestons that is your co xial capacitor

504

00:23:06,666  -->  00:23:11,133
so consider that it is comprising of lent L and uh

505

00:23:11,133  -->  00:23:14,766
it is consisting of two co xial conductors

506

00:23:15,333  -->  00:23:19,400
the space between them is filled up with homogeneous

507

00:23:19,400  -->  00:23:22,066
dialectic material of permittivity

508

00:23:22,133  -->  00:23:24,266
permittivity action here

509

00:23:24,266  -->  00:23:26,800
assumed that the conductor 1 and the conductor 2

510

00:23:26,800  -->  00:23:29,200
that carry uniformly distributed charges

511

00:23:29,200  -->  00:23:30,900
of same magnitude cube

512

00:23:30,900  -->  00:23:33,233
but with opposite polarity

513

00:23:34,166  -->  00:23:36,366
now we need to apply this uh

514

00:23:36,366  -->  00:23:38,566
if we if we apply this Gaza's law

515

00:23:38,566  -->  00:23:40,966
using this to the arbitrary guardian

516

00:23:40,966  -->  00:23:41,766
uh

517

00:23:41,966  -->  00:23:46,200
cylindrical surface of this radius row with this row

518

00:23:46,266  -->  00:23:49,700
with this row it is changing from here a to B

519

00:23:49,800  -->  00:23:51,566
okay so apply the Gaza

520

00:23:51,566  -->  00:23:52,533
Gaza's larger

521

00:23:52,533  -->  00:23:57,100
arbitrary gazian slendic surface of radius row

522

00:23:57,500  -->  00:23:59,733
changing between this a and B to

523

00:23:59,733  -->  00:24:02,266
this is your a and this is your B

524

00:24:02,533  -->  00:24:03,333
okay

525

00:24:04,200  -->  00:24:06,566
so uh we have already seen that

526

00:24:06,566  -->  00:24:08,666
if we involve this Guardian's Law

527

00:24:08,800  -->  00:24:09,933
and then uh this

528

00:24:09,933  -->  00:24:11,500
if you solve this interior

529

00:24:11,800  -->  00:24:13,800
and so the things are in your uh

530

00:24:13,800  -->  00:24:15,666
cylindrical coordinate system because it is

531

00:24:15,666  -->  00:24:19,066
it is exhibiting your cylindrical symmetry

532

00:24:19,133  -->  00:24:21,300
remember this thing and uh

533

00:24:21,300  -->  00:24:24,433
so this was electrical that would drive for this

534

00:24:25,133  -->  00:24:26,833
for this this type of tea

535

00:24:27,566  -->  00:24:31,166
for this type of your uh pyroplex okay

536

00:24:31,766  -->  00:24:36,466
and if if you see that you can simplify this thing okay

537

00:24:36,466  -->  00:24:39,766
you can simplify this thing that this is your

538

00:24:39,766  -->  00:24:41,600
if you solve this integral

539

00:24:41,766  -->  00:24:45,866
so you will find out this 2 Pyro L

540

00:24:47,266  -->  00:24:49,133
and then if you know this electric field

541

00:24:49,133  -->  00:24:50,866
you just plug in this electric field

542

00:24:50,866  -->  00:24:53,400
in this relationship to find out this

543

00:24:54,366  -->  00:24:55,500
again this DL

544

00:24:55,500  -->  00:24:56,466
it is your

545

00:24:57,100  -->  00:25:00,700
your this things are changing in your radial direction

546

00:25:00,700  -->  00:25:01,400
remember

547

00:25:01,400  -->  00:25:03,566
the things that changing in your radial direction

548

00:25:03,566  -->  00:25:06,400
so that's why you're this DL

549

00:25:07,066  -->  00:25:11,800
d L is to do that treated as d row

550

00:25:13,266  -->  00:25:15,266
okay so you take the dark product

551

00:25:15,266  -->  00:25:17,333
then uh what is left over here

552

00:25:17,333  -->  00:25:17,700
so

553

00:25:17,700  -->  00:25:20,533
you just need to solve the integration of one of our

554

00:25:20,533  -->  00:25:23,633
row with this back to row

555

00:25:24,300  -->  00:25:26,966
and head limits they're changing from B to a again

556

00:25:27,100  -->  00:25:29,733
and uh you know that the integral of this

557

00:25:29,733  -->  00:25:33,233
it is Eleanor Row okay

558

00:25:33,266  -->  00:25:35,766
and then you need to simply plug in this uh

559

00:25:36,866  -->  00:25:41,566
these ranges a and B to eliminate this negative sign

560

00:25:41,566  -->  00:25:44,333
you can just reverse this ratio okay

561

00:25:44,333  -->  00:25:47,033
so that is your lover to make drew

562

00:25:47,266  -->  00:25:49,066
so now you have to

563

00:25:49,100  -->  00:25:52,900
you have found out this V in terms of this Q

564

00:25:53,266  -->  00:25:55,100
not involve your basic relationship

565

00:25:55,100  -->  00:25:56,600
so q q will be cancelled

566

00:25:56,600  -->  00:26:01,633
so here we have assume that your Q is constant okay

567

00:26:02,300  -->  00:26:05,366
so this is your capacitors for your parallel blade

568

00:26:05,366  -->  00:26:06,866
capacitor for your

569

00:26:07,166  -->  00:26:12,100
this parallel coaxial coaxial conductors okay

570

00:26:12,100  -->  00:26:15,366
so it is a coaxial capacitor example

571

00:26:15,766  -->  00:26:18,000
or coexist cynentical capacity

572

00:26:18,166  -->  00:26:23,066
or this is also you can assume like your cables

573

00:26:23,066  -->  00:26:24,400
the cables the TV cables

574

00:26:24,400  -->  00:26:27,966
a wire so in the center you are having the signal okay

575

00:26:28,333  -->  00:26:31,000
in between you're having the dialective insulator

576

00:26:31,300  -->  00:26:33,700
outer is your against shield

577

00:26:33,766  -->  00:26:37,600
and then it is your plastic insulation right

578

00:26:37,600  -->  00:26:39,266
so this is equirant thing

579

00:26:41,666  -->  00:26:43,100
the third case uh

580

00:26:43,100  -->  00:26:46,200
no in this case you are having to concentrate uh

581

00:26:46,366  -->  00:26:47,666
spherical conductors here

582

00:26:47,666  -->  00:26:49,233
so it is your third uh

583

00:26:49,800  -->  00:26:50,600
uh uh

584

00:26:50,666  -->  00:26:51,833
partner system

585

00:26:52,066  -->  00:26:55,200
and here only the difference is the industry is uh

586

00:26:55,200  -->  00:26:58,800
with the radius a and the outer sphere is uh

587

00:26:59,500  -->  00:27:00,566
which radius b

588

00:27:00,566  -->  00:27:03,266
they are separated again by dialectic medium

589

00:27:03,733  -->  00:27:04,966
homogeneous dialect medium

590

00:27:04,966  -->  00:27:06,966
medium with permetuity abslon

591

00:27:07,066  -->  00:27:09,066
and again we are resuming that

592

00:27:09,400  -->  00:27:11,766
the inner grade is having positive q charge

593

00:27:11,766  -->  00:27:13,966
and the outer is having minus q charge

594

00:27:14,100  -->  00:27:14,566
and now

595

00:27:14,566  -->  00:27:18,933
if we apply the causing law to arbitrary guardian

596

00:27:18,933  -->  00:27:19,966
spherical surface

597

00:27:19,966  -->  00:27:24,300
so symmetry is spherical in this case and radiuses are

598

00:27:24,466  -->  00:27:27,366
so for these cases we have already studied that

599

00:27:27,366  -->  00:27:29,733
if we solve the secretion in the radio direction

600

00:27:29,733  -->  00:27:30,400
so this is your

601

00:27:30,400  -->  00:27:33,500
remember that this is your surface area of your sphere

602

00:27:33,766  -->  00:27:35,666
so this was the electric vehicle that would

603

00:27:35,666  -->  00:27:39,000
derived for this vertical corner system

604

00:27:39,166  -->  00:27:40,900
in our chapter number four

605

00:27:42,200  -->  00:27:43,066
now you know

606

00:27:43,066  -->  00:27:44,800
we have found out this electrical

607

00:27:44,900  -->  00:27:46,500
not involve this relationship

608

00:27:46,733  -->  00:27:50,366
solve this thing here remember that this is only

609

00:27:50,366  -->  00:27:51,500
the variations are

610

00:27:51,533  -->  00:27:54,000
with respect to your radial direction that your Dr

611

00:27:55,566  -->  00:28:00,800
and uh so you need to solve this integral 1 or R square

612

00:28:01,366  -->  00:28:05,100
okay solve this integral with this spectral R plug

613

00:28:05,100  -->  00:28:07,166
needs these values B to a

614

00:28:07,166  -->  00:28:10,200
so this is your voltage in terms of share

615

00:28:11,400  -->  00:28:15,133
this charge I know you can involve this relationship

616

00:28:15,133  -->  00:28:17,500
that is your capacity it is

617

00:28:17,500  -->  00:28:18,733
it is a ratio of charge

618

00:28:18,733  -->  00:28:21,400
and devot it across that capacity

619

00:28:21,933  -->  00:28:23,833
so this is very useful formula

620

00:28:25,566  -->  00:28:29,966
and if if if we let the letter case tag this outer

621

00:28:30,500  -->  00:28:31,533
this outer plate

622

00:28:31,533  -->  00:28:34,866
it is a seriously approaching infinity

623

00:28:35,133  -->  00:28:39,466
then it's uh resistance it will simply approach uh

624

00:28:39,466  -->  00:28:43,300
4 by action on times the radius of your inner plate

625

00:28:43,366  -->  00:28:46,666
which is the capacitors of a spherical capacitor

626

00:28:46,666  -->  00:28:52,400
whose outer plate is infinitely large okay

627

00:28:52,966  -->  00:28:54,500
infinity large

628

00:28:55,066  -->  00:28:58,400
so just you need to put in this thing as infinity

629

00:28:58,400  -->  00:29:02,600
anything divided by infinity it is equal to zero right

630

00:29:03,600  -->  00:29:06,066
so this should go in the numerator

631

00:29:06,866  -->  00:29:10,166
so then this is your very important example as

632

00:29:12,866  -->  00:29:15,933
uh these are the just you can uh see that uh

633

00:29:15,933  -->  00:29:19,000
your predact of your physics that uh

634

00:29:19,000  -->  00:29:21,133
if the two capacitors with the capacitor

635

00:29:21,133  -->  00:29:23,200
C1 and C2 are in Cds

636

00:29:24,000  -->  00:29:25,166
have same this me

637

00:29:25,166  -->  00:29:26,700
if they are in series then definitely

638

00:29:26,700  -->  00:29:29,033
they're going to have the same charge on them

639

00:29:29,666  -->  00:29:33,833
and uh for this case uh they're resistant capacitys

640

00:29:34,166  -->  00:29:37,666
the total capacitys it can be found out by using this

641

00:29:39,333  -->  00:29:41,833
case by using this uh

642

00:29:42,133  -->  00:29:44,566
uh just the addressing program

643

00:29:44,566  -->  00:29:45,766
of the things that we have done

644

00:29:45,766  -->  00:29:49,666
with the parallel barrel resistors

645

00:29:49,666  -->  00:29:52,066
that for for Cds capacitors

646

00:29:52,066  -->  00:29:54,000
we would need to involve this

647

00:29:55,266  -->  00:29:59,100
uh this uh if if if you call this

648

00:29:59,100  -->  00:30:02,800
this uh this your uh harmonic additive

649

00:30:02,800  -->  00:30:04,966
uh something like uh

650

00:30:06,700  -->  00:30:07,733
so if you remember that

651

00:30:07,733  -->  00:30:09,800
it is just like your harmonic mean

652

00:30:09,800  -->  00:30:11,166
things like this

653

00:30:11,400  -->  00:30:15,600
and if the two capacitors they are in parallel

654

00:30:15,600  -->  00:30:18,366
so this means indeed this is the case for this case

655

00:30:18,366  -->  00:30:20,433
the voltages are same across them

656

00:30:20,900  -->  00:30:23,766
so just just need to add up these two uh

657

00:30:23,866  -->  00:30:25,866
uh different capacitys

658

00:30:25,866  -->  00:30:28,866
to find out the total capacitys of this circuit

659

00:30:30,600  -->  00:30:33,000
okay so uh if you know this

660

00:30:33,000  -->  00:30:35,500
this where I already discussed that uh uh

661

00:30:35,500  -->  00:30:36,966
this is your homes law to

662

00:30:36,966  -->  00:30:38,666
how to find out the resistance

663

00:30:38,666  -->  00:30:40,866
by using this relationship of uh

664

00:30:40,866  -->  00:30:44,933
voltage and current for this non uniform surface alias

665

00:30:44,933  -->  00:30:47,366
circus alia plates and this is uh

666

00:30:47,366  -->  00:30:48,300
for your capacitors

667

00:30:48,300  -->  00:30:51,233
that it is the ratio of your charge on this uh

668

00:30:51,566  -->  00:30:54,066
uh place and the what is difference between them

669

00:30:54,566  -->  00:30:56,700
and if we take the product of them

670

00:30:56,700  -->  00:30:59,000
okay let's take the product of them

671

00:30:59,133  -->  00:31:02,933
and what you will see that these two things

672

00:31:02,933  -->  00:31:04,100
what is going to cancel out

673

00:31:04,100  -->  00:31:05,966
this thing is going to going to cancel out

674

00:31:05,966  -->  00:31:06,766
okay

675

00:31:07,333  -->  00:31:09,066
and this thing is going to cancel out

676

00:31:09,066  -->  00:31:11,033
so what is what is your left out with

677

00:31:11,300  -->  00:31:11,566
that

678

00:31:11,566  -->  00:31:15,466
is the ratio of your permitivity and the conductivity

679

00:31:15,933  -->  00:31:19,566
and this is known as the relaxation time of the medium

680

00:31:20,366  -->  00:31:22,666
separating the two conductors okay

681

00:31:22,800  -->  00:31:24,266
and this it is only

682

00:31:24,300  -->  00:31:27,400
only when the medium is homogeneous

683

00:31:27,400  -->  00:31:29,100
so it is only one single type

684

00:31:29,100  -->  00:31:29,900
so the medium

685

00:31:29,900  -->  00:31:32,533
which is residing in between these two place

686

00:31:32,533  -->  00:31:33,900
so that is your time cost

687

00:31:33,900  -->  00:31:36,500
and we also say that we also there said

688

00:31:36,500  -->  00:31:41,366
we also say that it is relaxation time of the medium

689

00:31:41,366  -->  00:31:42,500
separating them

690

00:31:42,500  -->  00:31:45,933
so this is the property of your medium so uh

691

00:31:45,933  -->  00:31:50,366
it is uh narrated it is narrated in terms of sure

692

00:31:51,666  -->  00:31:56,166
this uh using this uh uh electrical

693

00:31:56,166  -->  00:31:58,533
electrical uh properties of this medium

694

00:31:58,533  -->  00:32:01,000
that is your conductivity and the permitivity

695

00:32:01,133  -->  00:32:03,566
the ratio of these two uh quantities

696

00:32:03,566  -->  00:32:07,233
it is equal to the relaxation time of the medium

697

00:32:07,266  -->  00:32:11,033
and we have always studied in our previous lectures

698

00:32:11,133  -->  00:32:15,633
that what is the significance of this relaxation time

699

00:32:15,666  -->  00:32:18,966
that how the charges are going to decay

700

00:32:19,133  -->  00:32:22,233
or how the electric fees are going to decay

701

00:32:22,300  -->  00:32:26,000
and depending upon the relaxation time of this medium

702

00:32:26,000  -->  00:32:27,200
so if if you remember

703

00:32:27,200  -->  00:32:32,166
I guess we have left out with around 33% of the chart

704

00:32:32,166  -->  00:32:33,966
okay left out

705

00:32:36,466  -->  00:32:37,833
laptop okay

706

00:32:38,733  -->  00:32:42,133
rodency or something like charge or things like

707

00:32:42,133  -->  00:32:43,366
things like this

708

00:32:44,966  -->  00:32:47,533
so if uh if uh

709

00:32:47,533  -->  00:32:51,700
we assume that the it is a homogeneous medium and uh

710

00:32:51,700  -->  00:32:54,566
if we need need to find out the resistance of uh

711

00:32:54,933  -->  00:32:59,366
uh resistance of uh various capacitors

712

00:32:59,500  -->  00:33:03,166
uh mentioned earlier can be obtained using this uh

713

00:33:04,500  -->  00:33:06,566
relationship like this one

714

00:33:08,000  -->  00:33:10,333
uh and if we have already derived that

715

00:33:10,333  -->  00:33:10,933
you remember that

716

00:33:10,933  -->  00:33:13,600
we have already derived these capacitansies

717

00:33:13,600  -->  00:33:15,866
for the battle plate capacitor

718

00:33:15,866  -->  00:33:18,500
slendrical capacitor and spherical capacitor

719

00:33:18,566  -->  00:33:20,900
and if we involve this relationship then

720

00:33:20,900  -->  00:33:25,266
then we can find out the resistance of the medium

721

00:33:25,266  -->  00:33:28,000
remember that resistance of the medium

722

00:33:28,700  -->  00:33:31,333
this resistance are in this equation

723

00:33:31,333  -->  00:33:34,066
is not the resistance of the capacitor

724

00:33:34,566  -->  00:33:37,066
remember that it is the just know that

725

00:33:37,066  -->  00:33:39,466
it is not the resistance of the capacitor plate

726

00:33:39,600  -->  00:33:43,033
but the leakage resistance between the players

727

00:33:43,600  -->  00:33:44,566
therefore

728

00:33:44,566  -->  00:33:47,266
stigma is the conductivity of dialect to medium

729

00:33:47,266  -->  00:33:48,333
separating them

730

00:33:48,333  -->  00:33:52,000
and abslon is the permituity of that medium

731

00:33:52,400  -->  00:33:57,066
so this is regarding your medium related thing okay

732

00:33:57,133  -->  00:33:59,666
the leakage resistance of your medium basically

733

00:34:01,566  -->  00:34:03,066
some here today's lecture

734

00:34:03,066  -->  00:34:05,333
uh today we uh for

735

00:34:05,333  -->  00:34:08,733
for just for uh for generalized case uh case

736

00:34:08,733  -->  00:34:11,566
uh uh this uh generalized cases

737

00:34:11,566  -->  00:34:14,900
we determine that uh how to find out the resistance

738

00:34:14,900  -->  00:34:17,000
and the capacity of these uh

739

00:34:17,733  -->  00:34:18,666
capasters

740

00:34:18,733  -->  00:34:21,533
for standard that is your barrel blade capaster

741

00:34:21,533  -->  00:34:24,500
your work Z Capaster your spherical capaster

742

00:34:24,500  -->  00:34:27,566
that how we can find out the capasters of these

743

00:34:28,933  -->  00:34:31,100
different types of this gender capacitors

744

00:34:31,100  -->  00:34:33,833
and then we found out how to

745

00:34:33,966  -->  00:34:37,766
how we can determine this resistance of the medium

746

00:34:37,766  -->  00:34:39,900
in between these two plates

747

00:34:40,600  -->  00:34:43,433
next time we're going to solve the examples

748

00:34:43,533  -->  00:34:48,633
by taking help of this boundary problem solution

749

00:34:48,666  -->  00:34:51,066
and then we will find out the resistance

750

00:34:51,066  -->  00:34:53,800
and the capacity of the conductors

751

00:34:55,566  -->  00:34:56,766
here I thank you all

752

00:34:56,766  -->  00:34:58,700
if you have any questions uh

753

00:34:58,700  -->  00:35:00,166
they would be entertained uh

754

00:35:00,166  -->  00:35:03,700
to your emails or online synchronous sessions

755

00:35:03,700  -->  00:35:06,266
that we have arranged at the department