Question

In the following arrangement of capacitors, the energy stored in the $6\mu$ capacitor is E. Find the value of the following:
(1) Energy stored in $12\mu F$ capacitor.
(2) Energy stored in $3\mu F$ capacitor
(3) Total energy drawn from the battery

Answer 1

(i) Given that energy of the $6\mu F$ capacitor is $E.$

Let $V$ be the potential difference along the capacitor of capacitance $6\mu F$.

Since $\frac{1}{2}C{V}^2=E$

$\therefore \frac{1}{2}\times 6\times {10}^{-6}\times V^2=E$

$\Rightarrow V^2=\frac{E}{3}\times {10}^6$ ___(1)

Since potential is same for parallel connection, the potential through $12\mu F$ capacitor is also V. Hence, energy of $12\mu F$ capacitor is

$E_{12}=\frac{1}{2}\times 12\times {10}^{-6}\times V^2=\frac{1}{2}\times 12\times {10}^{-6}\times \frac{E}{3}\times {10}^6=2E$

(ii) Since charge remains constant in series, the charge on $6\mu F$ and $12\mu F$ capacitors combined will be equal to the charge on $3\mu F$ capacitor. Using the formula, $Q=CV$, we can write

$(6+12)\times {10}^{-6}\times V=3\times {10}^{-6}\times V^{\prime }$

Using (1) and squaring both sides, we get

$(V^{\prime }{)}^2=12E\times {10}^6$

$\therefore E_3=\frac{1}{2}\times 3\times {10}^{-6}\times 12E\times {10}^6=18E$

(iii) Total energy drawn from battery is

$E_{total}=E+E_{12}+E_3=E+2E+18E=21E$