Question

In the given circuit after closing switch s

• A. Work done by 3 V cell is $-27\mu J$
• B. Energy stored in $2\mu F$ capacitor increases
• C. Energy stored in $4\mu F$ capacitor increases
• D. Energy stored in $2\mu F$ capacitor decreases.

Answer 1

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

The correct options are
A Work done by 3 V cell is $-27\mu J$
B Energy stored in $2\mu F$ capacitor increases
C Energy stored in $4\mu F$ capacitor increases


$q=(\frac{2\times 4}{2+4})\times 3\mu C$
$q=4\mu C$


By K . V . L
$10=\frac{q_1}{6}+\frac{q_1-q_2}{4}$
$\Rightarrow 120=5q_1-3q_2$ ---(1)
and $3=\frac{q_1-q_2}{4}-\frac{q_2}{2}\Rightarrow 12=q_1-3q_2$ --(2)
By (1) and (2) $\Rightarrow q_1=27\mu Cand{q}_2=5\mu C$
work done, $W=qV=3\times -(5+4)=-27\mu J$