An AC voltage of 24 V is connected in series with a silicon diode and load resistance is 500 ohm having forward resistance 10 ohm. Calculate the peak output voltage.
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SOLUTION
As the diode is connected in series the effective resistance
R = RL + Rf
= 500 + 10
= 510 ohm
V = 24 V
Peak current = V / R
= 24 / 510
= 0.047 A
Peak output Voltage = Peak current x RL
= 0.047 x 500
= 23.5 V
this voltage is given ac and always ac voltage is rms value and we have to calculate the peak value.
ReplyDeleteone more thing i like to add that silicon diode is mentioned in the question so we have to consider 0.7 v, correct me if i am wrong...
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