StreakPeaked· Practice

ExamsNEETPhysics › Electronics

NEET Physics: Electronics questions with solutions

4 questions with worked solutions.

Questions

Q1. The typical operating current for a lightemitting diode is approximately:

  1. 1 to \( 5 m A \)
  2. 5 to \( 20 m A \)
  3. 20 to \( 500 m A \)
  4. 500 to \( 2000 m A \)

Answer: 5 to \( 20 m A \)

A standard light-emitting diode is normally operated at a modest current so it emits light without being damaged. The common range is about 5 to 20 mA, which matches typical indicator LED specifications.

Q2. In a semiconductor, which of the following statement is correct?

  1. At \( 0 K \), Si is a super conductor.
  2. In a p-type semiconductor the acceptor level lies near the conduction band.
  3. Each donor atom contributes one hole.
  4. \( p-n \) junction is electrically neutral

Answer: \( p-n \) junction is electrically neutral

In a p-n junction, the uncovered ionized donors and acceptors in the depletion region create equal and opposite charges, so the junction as a whole is electrically neutral. The other options are incorrect because Si is not superconducting at 0 K, acceptor levels are near the valence band, and donor atoms contribute electrons, not holes.

Q3. For a common emitter amplifier, current gain is \( 60 . \) If the emitter current is \( 6.6 ~ m A \) the collector current is :

  1. \( 0.108 \mathrm{mA} \)
  2. 6.492 m \( A \)
  3. \( 1.1 \mathrm{m} \) А
  4. 3.3 \( m A \)

Answer: 6.492 m \( A \)

In a common-emitter transistor, current gain is typically \(\beta = I_C/I_B\). With \(\beta = 60\), the base current is \(I_B = I_C/60\), and since \(I_E = I_C + I_B\), the collector current is slightly less than the emitter current. That gives \(I_C = 6.492\,\text{mA}\).

Q4. If the input frequency of a full wave rectifier is \( 50 H z \) ac. Its output frequency is

  1. \( 50 H z \) pulsating dc
  2. \( 100 H z \) pulsating do
  3. \( 200 H z \) pulsating dc
  4. \( 500 H z \) pulsating dc

Answer: \( 100 H z \) pulsating do

In a full-wave rectifier, both the positive and negative half-cycles are converted into positive output pulses. That doubles the ripple frequency, so a 50 Hz input becomes 100 Hz at the output.

⚔️ Practice NEET Physics free + battle 1v1 →