In the given figure, the charge stored in 6 μF capacitor, when points A and B are joined by a connecting wire is μC. [Figure: two vertical branches between a top node held at 9 V and a grounded bottom node. Left branch: 6 Ω from the top to point A, then 3 μF from A to ground. Right branch: 6 μF fro…

Question

In the given figure, the charge stored in 6 μF capacitor, when points A and B are joined by a connecting wire is μC. [Figure: two vertical branches between a top node held at 9 V and a grounded bottom node. Left branch: 6 Ω from the top to point A, then 3 μF from A to ground. Right branch: 6 μF from the top to point B, then 3 Ω from B to ground.]

Accepted Answer

36. With A and B joined, steady current flows 9 V→6 Ω→ A→ B→3 Ω→ ground; the capacitors carry none. I=9/(6+3)=1 A, so V_A=V_B=9-6×1=3 V. Voltage across the 6 μF capacitor: 9-3=6 V. Q=6×6=36 μC.

Step-by-step solution