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Figure shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80 muF, R= 40Omega a. Determine the source frequency which drives the circuit in resonanceb. Obtain the impedance of the circuit and the amplitude of current at the resonating frequency. c. Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the rasonating frequency. |
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Answer» Solution :a. `omega = (1)/(sqrtLC) = (1)/(sqrt(5 xx 80 xx 10^(-6) )`= 50 rad/sec b. At RESONANCE `Z = R = 40 Omega` `I_(RMS) = (V_(rms) )/(R ) = (230 sqrt2)/(40) = 8A` c.`V_L = IX_L = I omega L = 8 xx 50 xx 5= 2000 V` `V_C = IX_C = (I)/(omega C) = (8)/(50 xx 80 xx 10^(-6) ) = 200 V, V_R = 230 V , V_L- V_C = 0` i.e., potential drop across LC combination is zero . |
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