In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes q1μC. Then S is switched to position Q. After a long time, the charge on the capacitor is q2μC.
The magnitude of q2 is ____.
Answer: 0.67
Solution
When switch is at position Q, in steady state:
Applying KVL: VA−1⋅i2+2−2i2=VA, giving 3i2=2, so i2=32 A.
Potential difference: VA−VB=i2×1=32 V.
Charge on capacitor: q2=CΔV=1×32=0.67 μC.
JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Numerical answer