For the transistor circuit shown below,if bet | vedclass

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(C) Applying Kirchhoff's Voltage Law $(KVL)$ to the input loop:$V_{BB} - I_B R_B - V_{BE} = 0$$5\ V - I_B (8.6\ k\Omega) - 0.7\ V = 0$$I_B = \frac{4.3

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$13$

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(C) Applying Kirchhoff's Voltage Law $(KVL)$ to the input loop:$V_{BB} - I_B R_B - V_{BE} = 0$$5\ V - I_B (8.6\ k\Omega) - 0.7\ V = 0$$I_B = \frac{4.3\ V}{8.6 	imes 10^3\ \Omega} = 0.5 	imes 10^{-3}\ A = 0.5\ mA$Now,calculate the collector current $I_C$:$I_C = \beta I_B = 100 	imes 0.5\ mA = 50\ mA = 0.05\ A$Applying $KVL$ to the output loop:$V_{CC} - I_C R_L - V_{CE} = 0$$18\ V - (0.05\ A 	imes 100\ \Omega) - V_{CE} = 0$$18\ V - 5\ V - V_{CE} = 0$$V_{CE} = 13\ V$

For the transistor circuit shown below,if $\beta = 100$ and the voltage drop between the emitter and base is $0.7\ V$,then the value of $V_{CE}$ will be.....$V$.

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