Electromagnetic Induction and Alternating Current 6 Question 10

10. At time t=0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I(t)=I0cos(ωt), with I0=1A and ω=500rads1 starts flowing in it with the initial direction shown in the figure. At t=7π/6ω, the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C=20μF,R=10Ω and the battery is ideal with emf of 50V, identify the correct statement(s).

(2014 Adv.)

(a) Magnitude of the maximum charge on the capacitor before t=7π6ω is 1×103C

(b) The current in the left part of the circuit just before t=7π6ω is clockwise

(c) Immediately after A is connected to D, the current in R is 10A

(d) Q=2×103C

Show Answer

Answer:

Correct Answer: 10. (c, d)

Solution:

  1. dQdt=I

Q=Idt=(I0cosωt)dtQmax=I0ω=1500=2×103C

Just after switching

In steady state

At t=7π6ω or ωt=7π6

Current comes out to be negative from the given expression. So, current is anti-clockwise. Charge supplied by source from t=0 to t=7π6ω

Apply Kirchhoff’s loop law, just after changing the switch to position D

50+Q1CIR=0

Substituting the values of Q1,C and R, we get

I=10A

In steady state Q2=CV=1mC

Net charge flown from battery =2mC



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