In the circuit shown here, assuming threshold voltage of diode is negligibly small, then voltage V_(AB) is correctly represented by:

Solution

Physics sir · 4 steps

Follow the arrow of the diode: current can only go round one way, so only one half-cycle reaches R.

1. Read the diode. The diode's arrow points from the source's lower terminal towards B. Current can flow only in that direction: out of the lower terminal, through the diode to B, up through R to A, and back into the upper terminal.
2. First half-cycle. For 0 < ω t < π, V = V₀ sin ω t is positive, so the upper terminal is at the higher potential. That pushes current the wrong way through the diode, which is reverse biased. No current flows in R, so V_(AB) = 0.
3. Second half-cycle. For π < ω t < 2π, the lower terminal is at the higher potential. The diode is forward biased and acts like a plain wire. Current flows up through R from B to A, so B is at the higher potential and V_(AB) = V₀ sin ω t, which is negative.
4. So. V_(AB) stays at zero for the first half of every cycle and follows the negative sine loops in the second half. That is a half-wave output made of only negative pulses, which is graph (b).

The idea to keep. In a half-wave rectifier, find which half-cycle forward biases the diode. Then check which way current runs through the load. Current enters the load at its higher-potential end, and that fixes the sign of the output.

Answer from the NTA final answer key. Solution by PrepWiser.