In Fig. 25-21, the battery has potential difference V=14.0 V, C2=3.0 μF, C4=4.0 μF, and all the

In Fig. 25-21, the battery has potential difference V=14.0...

In Fig. 25-21, the battery has potential difference $V=14.0 \mathrm{~V}$, $C_{2}=3.0 \mu \mathrm{F}, C_{4}=4.0 \mu \mathrm{F}$, and all the capacitors are initially uncharged. When switch $\mathrm{S}$ is closed, a total charge of $12 \mu \mathrm{C}$ passes through point $a$ and a total charge of $8.0 \mu \mathrm{C}$ passes through point $b .$ What are (a) $C_{1}$ and (b) $C_{3}$ ?

Key Concepts

Transient Versus Steady-State Analysis

When a switch is closed in a circuit containing initially uncharged capacitors, the circuit undergoes a transient process before reaching a steady state. During this transient period, charge flows and redistributes according to the network’s configuration. Understanding the difference between transient behavior and the final steady state is crucial for analyzing time-dependent responses in capacitor circuits.

Charge Conservation in Circuits

This principle states that the total charge in a closed system remains constant. In capacitor networks, especially during switching processes, the redistribution of charge from the battery or other components must obey the law of conservation of charge, which is critical in determining how charge accumulates or redistributes in various parts of the circuit.

Capacitance and the Q = CV Relationship

This concept explains that a capacitor stores electrical charge according to the equation Q = CV, where Q is the charge stored, C is the capacitance, and V is the potential difference across the capacitor. It is fundamental in circuit analysis as it relates the physical properties of a capacitor to the electrical behavior observed in circuits.

Series and Parallel Combinations of Capacitors

Understanding how capacitors combine is essential for circuit analysis. In series, capacitors share the same charge but have different voltage drops, while in parallel, they have the same voltage across them but may store different amounts of charge. This principle allows for the simplification of complex circuits by combining capacitors into equivalent capacitances.

Transcript

00:01 Question we have been given the circuit as this one so you can see that it's a combination of force capacitors and there's a switch and it is all connected to a potential drop of b now let us write down the important values here so the value of c1 and c3 is what we have to find in our problem and we have been given the values of c2 and c2 is equal to three micro ferrets and c4 is given to us as 4 micro -farrids.

00:33 We have been given the voltage and it is 14 volts and there are two very important other you know components that have been given to us and it is the charges passing through two points.

00:48 So the charge passing through point a is given to us as 12 microcolum whereas charge passing through b is given to us as 8 microcolum.

01:01 So now you know the charge through this point and you know the charge through this point.

01:06 Now let's see how we can calculate c1 and c4, c1 and c3 using it.

01:12 So the first thing that you need to know is that when you have qa over here, which is the charge a.

01:21 So this charge qa is passing through c2 as well.

01:26 So let's see what is the potential drop across c2.

01:30 So v2 is equal to q2 upon c2 which is so q2 is given to us as 12 into 10 raised to the power minus 6 and c2 is 8 into 10 raised to the power sorry c2 is given to us as 3 into 10 raised to the power minus 6.

01:54 So finally here we'll get 4 volts.

01:58 So we know that the potential drop across the second capacitor is 4 volts.

02:04 Now let's move on to some other components.

02:06 Now we know the value of charge passing through b.

02:10 So this charge which is passing through point b will also pass through c4.

02:15 So potential drop across 4.

02:18 C4 will be q4 upon c4.

02:22 So q4 is given to us as 8 into 10 range to the power minus 6 and c4 the c4 value is given to us as 4 into 10 degrees of the power minus 6.

02:33 So here we get the voltage across 4 equals two bolts.

02:38 So these are two very crucial values that we have.

02:41 Now the most important thing that we need to know for calculating the potential for calculating c3 is that the potential drop across 4 will be equal to the potential drop across the potential across the the c3 capacitor because both of them are connected in because it's a parallel connection.

03:06 So let's use this to find out the answer...

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