Question

In the combination of capacitors shown in figure the potential difference across the plates of the capacitor $A$ will be

• A. 4.8 V
• B. 6 V
• C. 1.2 V
• D. 2.4 V

Answer 1

Answer: (C)

In figure,
$C_P=1+1+2=4\mu F$
This combination is in series with $1\mu F$ capacitor.
$\frac{1}{C_S}=\frac{1}{4}+\frac{1}{1}=\frac{5}{4},C_S=\frac{4}{5}\mu F$
$Q=CV=\frac{4}{5}\times 6=4.8\mu C$
Potential difference across $1\mu F$ capacitor,
$V_1=\frac{Q}{C}=\frac{4.8\mu C}{1\mu F}=4.8V$
Potential difference across the combination of three capacitors,
$V_2=V-V_1=6-4.8=1.2V$