Draw the dc load line for above circuit.

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written 3.0 years ago by

teamques10 ★ 64k

modified 2.1 years ago by

DevarenjiniP • 3.1k

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written 3.0 years ago by

teamques10 ★ 64k

$-4\times 10^3I_C+20-V_{CB}=0\\ I_C=\dfrac{-V_{CB}}{4\times10^3}+\dfrac{20}{4\times10^3}\cdots\cdots(1)$

When IC=0

VCB=20

When VCB=0

$I_C=\dfrac{20}{4}\times 10^{-3}=5mA$

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