The charge stored by the capacitor C in the given circuit in the steady state is . [figure]
- 7.5
- 5
- 10
- 12.5
Answer
(3) 10
Solution
The longer plate of the cell (+) is on the left, so the left junction is at the higher potential and current leaves the cell into the left junction.
Diode with 3 : points from right to left reverse biased, no current.
Diode with 4 : points from left to right forward biased (ideal), conducts.
Capacitor branch: no current in steady state.
I = 2.51 + 4 = 0.5 A
Potential difference between the junctions: V = I 4 = 2 V
(no current through 2 , so the whole 2 V is across C)
Q = CV = 5 2 V = 10
Answer: (3)