Discussion
In a power supply circuit, the a.c. voltage of 100-0-100 Volts r.m.s. is applied to diodes of the full wave rectifier circuit. The output of the diodes is connected to a load of 5 kΩ. The D.C. current through the load resistance of 5 kΩ is nearly :
Answer: B
The load current is nearly 18 mA.
In the centre-tapped full-wave rectifier, each half of the secondary gives 100 V r.m.s., a peak of \(V_{m} = 100\sqrt{2} = 141.4\) V. The average output of a full-wave rectifier is:
\[ \begin{aligned} V_{dc} &= \frac{2V_{m}}{\pi} = \frac{2 \times 141.4}{\pi} \\ &\approx 90\ \text{V} \end{aligned} \]
so \(I_{dc} = 90/5000 = 18\) mA. Using the r.m.s. value of 100 V would give 20 mA, and using the full 200 V across both halves would double the answer.
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