The circuit has ideal diodes, a 5 V Zener breakdown voltage, a 10 sin(100πt) V input through 50 Ω, and a 100 Ω branch to a +5 V source as shown. The power dissipated in the 100 Ω resistor, in watts, is
The ordinary diode admits positive input excursions. The Zener clamps the junction to at most +5 V, and the 5 V source biases the other end of the 100 Ω resistor to +5 V. With the diode off, that same source holds the junction at 5 V through the resistor. Thus both ends remain at 5 V, no current flows through the 100 Ω resistor, and its power is zero, A.
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GATE 2020 IN Q61– Assume all diodes in the circuit are ideal. Find current Iₖ through the 3 kΩ resistor in…
