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Two identical capacitors A and B, charged to the same potential 5 V are connected in two different circuits as shown below at time t=0. If the charge on capacitors A and B at time t=CR is Q_A and Q_B respectively, then (Here e is the base of natural logarithm) [Figure: two separate loops. In each, a charged capacitor sits in the top wire with its positive plate on the left, a resistor R forms the left side and a diode forms the right side. In circuit A the diode points down; in circuit B the diode points up.]

Asked in JEE Main 9th Jan 2nd Shift 2020 · Circuits with diodes

Figure: Circuits with diodes
Answer: (3) Q_A=VC, Q_B=(VC)/e

Step-by-step solution

Each capacitor tries to discharge from its positive (left) plate, down through R, along the bottom and up the right side through the diode.

Circuit A: the diode points down, against this current. It blocks, so A keeps its charge: Q_A=CV.

Circuit B: the diode points up, along this current, so B discharges through R: Q=CVe^-t/RC.

At t=RC: Q_B=(CV)/e.

Note: the printed key marks this question as having no correct option; option (c) matches this analysis. See DEFECTS.md.

Why the other options are wrong

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