Question

The charge on the 5 $\mu$F capacitor is

• A. 60 $\mu$C
• B. 24 $\mu$C
• C. 12 $\mu$C
• D. 20 $\mu$C

Answer 1

Answer: (D)

20 $\mu$C

Answer 2

Here's how to determine the charge on the 5 $\mu$F capacitor:

1. Calculate the equivalent capacitance of the series combination:

   • The 10 $\mu$F and 5 $\mu$F capacitors are in parallel, so their equivalent capacitance ($C_{AB}$) is: $C_{AB} = 10 \, \mu F + 5 \, \mu F = 15 \, \mu F$
   • The 6 $\mu$F and 4 $\mu$F capacitors are in parallel, so their equivalent capacitance ($C_{BC}$) is: $C_{BC} = 6 \, \mu F + 4 \, \mu F = 10 \, \mu F$
   • $C_{AB}$ and $C_{BC}$ are in series. The equivalent capacitance ($C_{eq}$) of the entire circuit is: $\frac{1}{C_{eq}} = \frac{1}{15 \, \mu F} + \frac{1}{10 \, \mu F} = \frac{2 + 3}{30 \, \mu F} = \frac{5}{30 \, \mu F}$ $C_{eq} = 6 \, \mu F$

2. Determine the total charge supplied by the battery:

   • $Q_{total} = C_{eq} \times V_{AC} = 6 \, \mu F \times 10 \, V = 60 \, \mu C$

3. Find the potential difference across the 5 $\mu$F capacitor:

   • Since $C_{AB}$ and $C_{BC}$ are in series, they have the same charge: $Q_{AB} = Q_{total} = 60 \, \mu C$
   • The potential difference across $C_{AB}$ is: $V_{AB} = \frac{Q_{AB}}{C_{AB}} = \frac{60 \, \mu C}{15 \, \mu F} = 4 \, V$
   • The 5 $\mu$F capacitor is in parallel with the 10 $\mu$F capacitor, so it has the same potential difference: $V_{5\mu F} = V_{AB} = 4 \, V$

4. Calculate the charge on the 5 $\mu$F capacitor:

   • $Q_{5\mu F} = C_{5\mu F} \times V_{5\mu F} = 5 \, \mu F \times 4 \, V = 20 \, \mu C$

Therefore, the charge on the 5 $\mu$F capacitor is 20 $\mu$C.