Choose the correct waveform that can represent the voltage across R of the following circuit, assuming the diode is ideal one :

Solution

Physics sir · 3 steps

The 3 V cell pushes back on the diode, so current flows only once the input climbs past 3 V.

1. Walk the loop. Current through the diode goes clockwise: through D, down through R, and into the + terminal of the 3 V cell. So the cell opposes the source. With an ideal diode, Kirchhoff's loop rule gives:

   Vᵢ - iR - 3 = 0 ⇒ V_R = iR = Vᵢ - 3

2. When it conducts. The current must flow forward through the diode, so i > 0, which needs Vᵢ > 3 V. Below that the diode is reverse biased, no current flows, and V_R = iR = 0.

   Input range	Diode	Voltage across R
   Vᵢ > 3 V	ON	Vᵢ - 3, up to 10 - 3 = 7 V
   Vᵢ ≤ 3 V (including the whole negative half)	OFF	0, never negative

3. So. R carries a voltage only during the part of each cycle where the input is above the +3 V line. Its height is measured above that line, so the peak is 7 V, and R shows nothing for the rest of the cycle. That is the waveform in (a), where only the part above +3 is drawn solid. The answer is (a).

The idea to keep. With a diode and a cell in series, first find the input at which the diode just turns on. Below it no current flows, so R reads zero. Above it, R gets the input minus the cell voltage.

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