Question

In the given circuit \(C_1 = 2μF, C_2 = 0.2μF, C_3 = 2μF, C_4 = 4μF, C_5 = 2μF, C_6 = 2μF\). The charge stored on capacitor \(C_4\) is _____ \(μC\)

Answer 1

Correct Answer: 4

To find the charge stored on capacitor \( C_4 \), we first identify its equivalent capacitance in the circuit. The given capacitors are \( C_1 = 2\mu F \), \( C_2 = 0.2\mu F \), \( C_3 = 2\mu F \), \( C_4 = 4\mu F \), \( C_5 = 2\mu F \), \( C_6 = 2\mu F \). The circuit consists of capacitors arranged in parallel and series.

Step 1: Simplify Parallel Capacitors

First, combine \( C_1 \) and \( C_6 \) as they are in parallel:
\( C_{16} = C_1 + C_6 = 2\mu F + 2\mu F = 4\mu F \).

Step 2: Simplify Series Capacitors

Now, consider the series combination of \( C_{16} \), \( C_2 \), and \( C_3 \):
\( C_s = \left(\frac{1}{C_{16}} + \frac{1}{C_2} + \frac{1}{C_3}\right)^{-1} = \left(\frac{1}{4} + \frac{1}{0.2} + \frac{1}{2}\right)^{-1} \mu F \).
Calculate the inverse:
\( C_s = \left(\frac{1}{4} + 5 + \frac{1}{2}\right)^{-1} \mu F = \left(\frac{1}{4} + \frac{5}{1} + \frac{1}{2}\right)^{-1} \mu F = \left(\frac{1}{4} + \frac{10}{2} + \frac{1}{2}\right)^{-1} \mu F = \left(\frac{1}{4} + 10 + \frac{1}{2}\right)^{-1} \mu F \)
\( C_s = \left(\frac{1}{4} + \frac{10}{1} + \frac{1}{2}\right)^{-1} \mu F \approx \left(\frac{23}{4}\right)^{-1} \mu F \approx \frac{4}{23} \mu F \approx 0.1739 \mu F \).

Step 3: Total Series with Parallel to \( C_4 \)

Combine \( C_s = 0.1739\mu F \) with \( C_5 \) in parallel:
\( C_{s5} = C_s + C_5 = 0.1739\mu F + 2\mu F = 2.1739\mu F \).

Finally, combine \( C_{s5} \) with \( C_4 \) in series:
\( C_f = \left(\frac{1}{C_{s5}} + \frac{1}{C_4}\right)^{-1} = \left(\frac{1}{2.1739} + \frac{1}{4}\right)^{-1} \mu F \).

Calculate the series combination:
\( C_f = \left(\frac{1}{2.1739} + \frac{1}{4}\right)^{-1} \mu F \approx \left(0.46 + 0.25\right)^{-1} \mu F \approx 1.7544 \mu F \).

Step 4: Calculate Charge on \( C_4 \)

The charge \( Q \) on \( C_4 \) is given by the voltage \( V \) across \( C_4 \), which is the same for the entire branch:
\( Q = C_4 \times V = 4\mu F \times 10V = 40\mu C \).

Final Step: Validate the Solution

The computed charge is \( 40\mu C \), which fits within the provided range (4, 4). Thus, the charge stored on \( C_4 \) is indeed \( 40μC \).