Question

Three capacitors 6 $\mu$F, $3$ $\mu$F and 4 $\mu$F are connected to a battery of emf $12V$ as shown in the Figure. If the point b is grounded, then :

• A. the charge on $3$ $\mu$F capacitor is 24 $\mu$C.
• B. the potential at a is $4V$.
• C. the potential at c is $-8V$.
• D. the energy dissipated in the battery is $432$ $\mu$J

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A the potential at a is $4V$.
B the energy dissipated in the battery is $432$ $\mu$J
C the charge on $3$ $\mu$F capacitor is 24 $\mu$C.
D the potential at c is $-8V$.

Potential diff across $ac$ is $V_{ac}=12V$
We know when capacitors are in series charge present on them must be equal.
$\therefore$ charge on $6\mu F=chargeon3\mu F$
$\Rightarrow \frac{q}{6\mu F}+\frac{q}{3\mu F}=12$
$\Rightarrow q\left(\frac{3}{6\mu F}\right)=12$
$\Rightarrow q=24\mu FC$
$V_{diff}$ across $6\mu F$ capacitor $V_a-V_b=\frac{24\times {10}^{-6}}{6\times {10}^{-6}}$
$V_a=4.$
$V_{diff}across3\mu Fcapacitor{V}_0-V_c=\frac{24\times {10}^{-6}}{3\times {10}^{-6}}=8.\Rightarrow V_c=-8$
total charge flown from battery $=72\mu c.$
work done by battery
$=72\times {10}^{-6}\times 12$
$=864\mu J$
energy stored $=\frac{1}{2}C{V}^2$
$=\frac{1}{2}(12{)}^2\times (4+2)$
$=3\times 44=432.$
energy dissipated $=864-432=432\mu J.$