Discussion
Assuming ideal diode and no ripple, what is the dc load current in the circuit shown below ?
Answer: D
The dc load current is about 6.8 mA.
With an ideal diode and no ripple, the capacitor charges to the peak of the secondary voltage and holds it, so the load sees a steady dc voltage equal to that peak.
\[ \begin{aligned} V_{2} &= \frac{120}{5} = 24\ \text{V rms} \\ V_{m} &= 24\sqrt{2} \approx 33.9\ \text{V} \\ I_{dc} &= \frac{33.9}{5000} \approx 6.8\ \text{mA} \end{aligned} \]
4.8 mA is the rms secondary voltage divided by the load, \(24/5000\), which forgets that the capacitor charges to the peak. 0.9 A and 1.28 A are hundreds of times too large for a 5 kΩ load.
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