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An a.c. voltage of 100 V, 50 Hz is connected across a 20 Omegaresistance and a 2 mH inductor in series. Calculate (1) impedance of the circuit, (ii) rms current in the circuit. |
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Answer» Solution :Here, `V_(RMS) = 100 V`, v= 50 Hz, `R = 20 Omega` and L = 2mH `=2 xx 10^(-3)` H (i) IMPEDANCE of the circuit `Z = sqrt(R^(2) + X_(L)^(2)) = sqrt(R^(2) + (2pi vL)^(2))` `= sqrt((20)^(2) + (2pi xx 50 xx 2 xx 10^(-3))^(2)) = 20 Omega` (ii) The rms current in the circuit `I_(rms) = V_(rms)/Z = (100 V)/(20 Omega) = 5A` |
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