Question

The figure shows a network of five capacitors connected to a 20 V battery. Calculate the charge acquired by each 10 µF capacitor.

• A. \( 2 \times 10^{-4} \, \text{C} \)
• B. \( 4 \times 10^{-4} \, \text{C} \)
• C. \( 6 \times 10^{-4} \, \text{C} \)
• D. \( 1 \times 10^{-4} \, \text{C} \)

Hint:

When capacitors are connected in parallel, their capacitances add. When they are in series, the reciprocal of the total capacitance is the sum of the reciprocals of the individual capacitances.

Answer 1

The Correct Option is D

The network consists of capacitors connected in series and parallel. To calculate the charge on the \( 10 \, \mu \text{F} \) capacitors, we first need to find the equivalent capacitance. 

Step 1: Combine Capacitors in Series and Parallel
The \( 5 \, \mu \text{F} \) capacitor and one \( 10 \, \mu \text{F} \) capacitor are in parallel: \[ C_{\text{eq1}} = C_1 + C_2 = 5 \, \mu \text{F} + 10 \, \mu \text{F} = 15 \, \mu \text{F} \] 

Step 2: Combine with the Next Capacitor in Series
Now, the equivalent \( 15 \, \mu \text{F} \) capacitor is in series with the next \( 15 \, \mu \text{F} \) capacitor: \[ C_{\text{eq2}} = \frac{C_{\text{eq1}} \times 15 \, \mu \text{F}}{C_{\text{eq1}} + 15 \, \mu \text{F}} = \frac{15 \times 15}{15 + 15} = 7.5 \, \mu \text{F} \] 

Step 3: Combine with the Last Capacitor in Parallel
The final \( 10 \, \mu \text{F} \) capacitor is in parallel with the equivalent capacitance \( C_{\text{eq2}} \): \[ C_{\text{eq}} = C_{\text{eq2}} + 10 \, \mu \text{F} = 7.5 \, \mu \text{F} + 10 \, \mu \text{F} = 17.5 \, \mu \text{F} \] 

Step 4: Use the Formula for Charge
The total charge supplied by the battery is: \[ Q = C_{\text{eq}} \times V = 17.5 \, \mu \text{F} \times 20 \, \text{V} = 350 \, \mu \text{C} \] The charge on each of the \( 10 \, \mu \text{F} \) capacitors is: \[ Q_{\text{10 \, µF}} = \frac{Q}{2} = \frac{350 \, \mu \text{C}}{2} = 175 \, \mu \text{C} \] 
Thus, the charge acquired by each \( 10 \, \mu \text{F} \) capacitor is \( 1 \times 10^{-4} \, \text{C} \).