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Draw the dc load line for above circuit.
1 Answer
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$-4\times 10^3I_C+20-V_{CB}=0\\ I_C=\dfrac{-V_{CB}}{4\times10^3}+\dfrac{20}{4\times10^3}\cdots\cdots(1)$

When IC=0

VCB=20

When VCB=0

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

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