Three capacitors of capacity $2 μF, 2 μF, 1 μF$ and two ideal cells of potential difference 10V and 5V are connected in a circuit through a switch ‘S’. Initially switch is opened for a long time and then it is closed. Choose the correct option(s) after switch is closed.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Charge passing through 10V cell is $\frac{4}{3} μC$

Work done by 5V cell is $10 μJ$

Charge on $1 μF$ capacitor is $6 μC$

Total heat dissipated in the circuit after switching is $\frac{5}{3} μJ$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Before switch is closed

$10 - q_{0} - \frac{q_{0}}{2} = 0 q_{0} = \frac{20}{3} μC$
After switch is closed
From KVL, $5 - \frac{q - q_{1}}{2} - \frac{q_{0} + q}{1} = 0$
$10 - \frac{q_{0} + q}{1} - \frac{q_{0} + q}{2} = 0 On solving q = - \frac{2}{3} μC, q_{1} = \frac{4}{3} μC$
$U_{i} = \frac{1}{2} \frac{{(20 / 3)}^{2}}{1} + \frac{1}{2} \frac{{(20 / 3)}^{2}}{2} = \frac{100}{3}$
$U_{f} = \frac{1}{2} \frac{{(6)}^{2}}{1} + \frac{1}{2} \frac{{(8)}^{2}}{2} + \frac{1}{2} \frac{{(2)}^{2}}{2} = 35 μJ$
$W_{B} = 5 (- 2) + 10 \times \frac{4}{3} = \frac{10}{3} μJ$
So, heat required $= \frac{5}{3} μJ$