In the given figure, the charge stored in 6μF capacitor, when points A and B are joined by a connecting wire is ______ μC.
Answer: 36
Solution
Joining A and B with a wire places them at the same potential. At steady state, no current flows through a capacitor branch, so current flows only along the resistor path: from the 9 V terminal through the 6 Ω resistor to node A(=B), then through the 3 Ω resistor to ground.
Req=6+3=9Ω,i=99=1A
The drop across the 6 Ω resistor is 1×6=6 V, so node A (and hence B) sits at 9−6=3 V above ground.
The 6 μF capacitor connects the 9 V terminal to node B, so the voltage across it is 9−3=6 V:
Q=CV=6μF×6V=36μC
JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Integer answer