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Question

# Consider the circuit shown in the figure. Assume base-to -emitter voltage VBE = 0.8 V and common base current gain (α) of the transistor is very close to unity . The value of the collector-to-emitter voltage VCE will be

A
4 V
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B
5 V
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C
6 V
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D
3 V
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Solution

## The correct option is C 6 V By drawing the Thevenin's equivalent between base and ground nodes, the given circuit can be reduced VTh=1616+44×18V=4.8V RTh=(44 k||16 k)Ω Applying KVL in loop 1 we get . IERE=VTh−VBE−IBRTh IB=0 A Given α=1 So IERE=4.8−0.8=4 A IE=42×103 = 2mA IC=IE=2mA VCE=VCC−ICRC−IERE =18 - (2 × 4) - (2× 2 ) = 6 V

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