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A capacitor and a resistor are connected in series with an a.c. source. If the potential difference across C, R are 120 V, 90 V, respectively and if the rms current of the circuit is 3 A, calculate the (i) impedance, (ii) power factor of the circuit. |
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Answer» Solution :Here, `V_(C) = 120 V` and `V_(R) = 90 V` and `I_(rms) = 3A` (i) VOLTAGE across R-C circuit `V_(rms) = SQRT(V_(C)^(2)+V_(R)^(2)) = sqrt((120)^(2) + (90)^(2)) = 150 V` `therefore` IMPEDANCE `Z = V_(rms)/I_(rms) = (150 V)/(3 A) = 50 Omega` (ii) Resistance of the circuit `R = V_(R)/I_(rms) = (90V)/(3 A) = 30 Omega` `therefore` Power factor `cos phi = R/Z = 30/50 = 0.60` |
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