In Fig. 25-18, a potential difference of V=65.0 V is applied across a capacitor arrangement wit

In Fig. 25-18, a potential difference of V=65.0 V is...

Key Concepts

Capacitor Combinations

This concept involves understanding how capacitors behave when arranged in series and parallel. In a network of capacitors, the overall voltage distribution, effective capacitance, and charge stored depend on how the individual capacitors are connected. When one capacitor shifts from its designed behavior (e.g., undergoing electrical breakdown), the configuration effectively alters, changing the way voltages and charges are distributed across the remaining components.

Capacitance and Charge Relationship

The fundamental relationship Q = CV describes how the charge stored on a capacitor is directly proportional to both its capacitance and the voltage across it. This relationship is crucial when analyzing how changes in the circuit – such as a capacitor being shorted out – affect the stored charge on other capacitors. A change in the effective voltage across a capacitor will directly modify the charge stored on it.

Electrical Breakdown and Short Circuit Effects

Electrical breakdown in a capacitor causes it to act like a conducting wire, effectively removing its impedance from the circuit. This sudden change can alter the voltage drop across other capacitive elements in the circuit, thereby increasing their potential differences, stored charge, and stored energy. Understanding this effect is key in analyzing transient and steady state responses in capacitor networks when one element fails.

Energy Stored in a Capacitor

The energy stored in a capacitor is given by the equation U = ½ CV². This relationship shows that energy depends on both the capacitance and the square of the voltage across the capacitor. When the breakdown of one capacitor causes an increase in voltage across another capacitor, the stored energy in that capacitor increases more significantly, making it essential to consider the quadratic dependency of energy on voltage during such analyses.

Transcript

00:01 Been given a series we are given a combination of capacitors so this is how the combination looks the first two capacitors are in parallel with each other whereas the third one is in series and they are connected to a potential voltage potential source which has v now we let's label this one is c1 this is c2 and this is c3 and this is the voltage so the values that have been given to is the potential difference is given as 65 volts.

00:35 The first capacitors capacitance is given as 10 micro -farrows.

00:42 C2 is given to us as 5 micro -fairds and c3 is given to us as 4 mu -f.

00:50 Now we have to calculate three things in this question.

00:55 The first thing is the increase in charge, then potential difference and then stored energy.

00:59 So let us basically firstly go through about and understand the question now they're saying that what happens in this circuit is that this capacitor which is c3 undergoes an electrical breakdown and gets converted into a wire so now it will be considered as a wire so what we have to do is that there are two cases here the first case is this original case and the second cases where c3 gets converted into a conducting wire now we have to find out the change in charge, change in potential difference and change in stored energy in both the two cases.

01:36 Now let's see how we can do it.

01:38 So firstly let's calculate the equivalent capacitance.

01:43 So eq initially when c3 is in the picture then it will be so c1 and c2 are in parallel so they'll get added so if we write down the equivalent resistance so this is equivalent capacity.

01:59 So this is so it's going to be 3 .15 mu f.

02:07 So this is the equivalent capacitance of this entire system.

02:11 Now let us find out the original voltage.

02:16 So now let's see how we'll find out the original voltage.

02:21 Then we'll see the charge.

02:23 So the original voltage here will be.

02:27 So v original will be.

02:34 Now, if we see, so we have to keep in mind, we have to first find out the original potential across v1 because they are talking on capacitor 1 over here.

02:48 So we have to keep in mind that we have to take out all the values for the first capacitor.

02:53 So we original here we are referring to v1.

02:57 So this is the potential drop across the first capacitor, which is c1 capacitor.

03:02 And this can be written as c equivalent into total potential drop upon c1 plus c2 because this is in a parallel combination.

03:13 So c equivalent we have as 3 .15 and into the voltage which is given to us as 65 volts upon c1 plus c2 will give us c1 plus c2 is 15.

03:26 And since the micro -fired and micro -fired on both the denominator and the numerator gets cancelled, so v1 we can finally write it as 13 .65 volts on calculation.

03:45 So this is the potential difference across the first capacitor, here c1, when this c3 is in existence...

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