In a full-wave P-N junction diode rectifier, | vedclass

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$(D)$ The peak current $I_m$ is given by $I_m = \frac{V_m}{R_f + R_L} = \frac{30}{10 + 1500} = \frac{30}{1510} \approx 0.01987 \ A = 19.87 \ mA$.The $

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$240.4 \ mW, 295.9 \ mW$

Vedclass · Answer

$(D)$ The peak current $I_m$ is given by $I_m = \frac{V_m}{R_f + R_L} = \frac{30}{10 + 1500} = \frac{30}{1510} \approx 0.01987 \ A = 19.87 \ mA$.The $D.C.$ load current is $I_{dc} = \frac{2I_m}{\pi} = \frac{2 	imes 19.87 \ mA}{3.1416} \approx 12.65 \ mA$.The $D.C.$ power output is $P_{dc} = I_{dc}^2 	imes R_L = (12.65 	imes 10^{-3})^2 	imes 1500 \approx 240.1 \ mW$.The $R.M.S.$ current for a full-wave rectifier is $I_{rms} = \frac{I_m}{\sqrt{2}} = \frac{19.87}{\sqrt{2}} \approx 14.05 \ mA$.The $A.C.$ power input is $P_{in} = I_{rms}^2 	imes (R_f + R_L) = (14.05 	imes 10^{-3})^2 	imes (10 + 1500) = 1.974 	imes 10^{-4} 	imes 1510 \approx 298.1 \ mW$.Comparing with the given options, the closest values are $240.4 \ mW$ and $295.9 \ mW$.

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