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A series combination of a coil of inductance L and a resistor of resistance 12 Omega is connected across a 12 V, 50 Hz supply. Calculate L if the circuit current is 0.5 A. |
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Answer» Solution :Impedance, `Z = (E)/(1) = (12)/(0.5) = 24 OMEGA` and `Z = sqrt(R^(2) + omega^(2)L^(2))` or `Z^(2) = R^(2) + omega^(2)L^(2)` or `L^(2) = (Z^(2) - R^(2))/(omega^(2))` Here `omega = 2pi V = 2 xx(22)/(7) xx 50` `= 31.4 "rad s"^(-1)` and `R = 12 Omega` `L^(2) = ((24^(2))-(12)^(2))/((314)^(2))` or `L = (12sqrt(3))/(314) = 0.066H` |
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