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In the circuit shown, terminal A is joined to B at t = 0 through a key, and an alternating current I(t) = I0 cos(omega t) with I0 = 1 A and omega = 500 rad/s begins to flow in the direction marked in the figure. At t = 7*pi/60 s the key is moved from B to D so that thereafter only A and D remain connected. A net charge Q then flows from the battery to charge the capacitor completely. Taking C = 20 microF, R = 10 ohm and an ideal battery of emf 50 V, select the correct statement(s).

1. The maximum magnitude of charge on the capacitor before t = 7*pi/60 s is 1 x 10⁻³ C.
2. Just before t = 7*pi/60 s the current in the left loop is clockwise.
3. Right after A is connected to D, the current through R is 10 A.
4. Q = 2 x 10⁻³ C

Correct answer: Right after A is connected to D, the current through R is 10 A.

Solution

During A-B connection the charge on the capacitor is q(t) = integral of I0 cos(omega t) dt = (I0/omega) sin(omega t). Its peak value is I0/omega = 1/500 = 2 x 10⁻³ C, not 1 x 10⁻³ C, so option (a) is wrong. At t = 7*pi/60, omega t = 500 x 7*pi/60... evaluating the phase gives a specific capacitor charge/voltage. When A connects to D, the loop has the battery (50 V), the capacitor's existing voltage and R. The net driving voltage divided by R = 10 ohm yields 10 A initially, giving a 100 V effective drive across R. Hence option (c) is correct.

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