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Question

The diagram shows the biasing of an n-p-n transistor in common emitter configuration used in an amplifier. The design of the transistor is such that 98% of the charge carriers from emitter reach the collector. If base current changes from 50 μA to 100 μA, then the corresponding change in the voltage across the load resistance RL will be

A
0.25 V
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B
0.25 V
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C
24.5 V
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D
49.0 V
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Solution

The correct option is C 24.5 V

From the given data
IC=0.98 IE
ΔIB=50×106 A

voltage across the Rl=RL×ΔIC
where, β=ΔICΔIBΔIC=β×IB
here, β=ICIB

For CE-configuration,
IE=0.98 IE
IE=IB+IC
so, IB=0.02 IE

so β=0.987 IE0.02 IC=49
form eq.(1)
ΔIC=49×50×106

So the voltage across RL=10×103×49×50×106RL=24.5 V

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