Chapter 17, Problem 40P

To determine

The value of $C_2$ will be determined.

Answer to Problem 40P

The answer is $5.6\times {10}^{-5}F$ .

Explanation of Solution

Given data:

Capacitor $C_1=7.7\times {10}^{-6}F$

Initial voltage $V_{initial}=125V$

Final voltage $V_{final}=15V$

Formula used:

It is known that: $Q=CV$

Where,

  $Q$ is the charge.

  $\text{C}$ is the capacitance of capacitor.

  $V$ is the potential.

Calculation:

Using the total charge and the final potential difference to calculate the value of the second capacitor,

  $Q_{total}=Q_{1_{final}}+Q_2{}_{{}_{final}}$

  $C_1{V}_1=\left(C_1+C_2\right)V_{final}$

  $C_2=C_1\left(\frac{V_{1_{initial}}}{V_{2_{final}}}-1\right)$

  $C_2=\left(7.7\times {10}^{-6}F\right)\left(\frac{125V}{15V}-1\right)$

  $C_2=5.6\times {10}^{-5}F$

Conclusion: The value of $C_2$ is $5.6\times {10}^{-5}F$ .