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Question

Consider the circuit shown in figure. When switch S1 closed and the other two switches open, the circuit has a time constant 0.05 sec. When switch S2 closed and the other two switches open, the circuit has a time consatnt 2 sec. When switch S3 closed an the other two switches open,the circuit oscillates with a period T. Find T(in sec). Take π2=10

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Solution

When switch S1 is closed and the others are open, the inductor is escentially out of the circuit and what remains is an RC circuit The time consatnt is τC=RC. Given τC=0.05 sec When switch S2 is closed and the others are open, the capacitor is essentially out of the circuit, in this case, what we have an LR circuit with the time constant τL=LR Given τL=2 sec Finally, when switch S3 is closed and the others are open,the resistor is essentially out of then circuit and what remains is an LCcircuit that oscillates with time period. ⇒T=2π√LC Here LC=τCτL=0.05×2=0.1 ⇒T=2π√110=2 sec

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