A transistor is connected in common emitter circuit configuration, the collector supply voltage is 10 V and the voltage drop across a resistor of 1000 Ω in the collector circuit is 0.6 V. If the current gain factor (β) is 24, then the base current is _____________ μA. (Round off to the Nearest Integer)

Solution

Physics sir · 3 steps

Ohm's law on the collector resistor gives I_C; divide by β and the base current is done.

1. Collector current. The 0.6 V drop is across the 1000 Ω collector resistor, so Ohm's law gives the current through it. The 10 V supply also drops across the transistor itself, so it does not give this current.

   I_C = V_(R_C)/R_C = (0.6 V)/(1000 Ω) = 6 × 10⁻⁴ A = 600 μA

2. Use the current gain. In common emitter mode, β = I_C/I_B, so I_B = I_C/β = (600 μA)/24.
3. So. I_B = 25 μA, so the answer is 25.

The common slip: 417. Dividing the full 10 V supply by 1000 Ω gives I_C = 10 mA and I_B ≈ 417 μA, not 25. Only 0.6 V drops across the resistor.

The idea to keep. Find the collector current from the drop across the collector resistor itself, not from the supply voltage. Then I_B = I_C/β in common emitter mode.

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