Question

2 A half wave rectifier is fed from a supply of 230V, 50Hz with step down transformer of ratio 3:1. Resistive load connected is 10KΩ. The diode forward resistance is 75Ω and transformer secondary is 10Ω. Solve for DC load current, DC load voltage, Efficiency and Ripple Factor.

L3 CO1

3 Solve

i) DC current through the circuit

ii) AC (rms) value of current through the circuit

iii) DC output voltage

iv) Rectifier efficiency

v) DC power output

For Full wave rectifier with input V(t) = 200 sin50t. If RL is 1KΩ and forward resistance of diode is 50Ω.

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Answer 1

Answer:

You have the system reversed. The transformer would come first and then the rectifier, since the transformer can’t work on DC. With a turns ratio of 8:1 the voltage on the secondary would be 240/8 = 30 volts.

The peak voltage of a sine wave is the RMS (root mean squared voltage) x square root of 2. That would be 42.5 volts. A typical silicon diode drops about 0.7 volts so the peak voltage would be about 42 volts.

Answer 2

Answer:

Lauderdale

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