For the transistor circuit shown below,if bet | vedclass

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(C) $1$. Apply Kirchhoff's Voltage Law $(KVL)$ to the base-emitter loop:$V_{BB} - I_B R_B - V_{BE} = 0$$5 \, V - I_B (8.6 \, k\Omega) - 0.7 \, V = 0$$

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$13$

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(C) $1$. Apply Kirchhoff's Voltage Law $(KVL)$ to the base-emitter 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{5 - 0.7}{8.6 	imes 10^3} = \frac{4.3}{8.6 	imes 10^3} = 0.5 	imes 10^{-3} \, A = 0.5 \, mA$.$2$. Calculate the collector current $(I_C)$:$I_C = \beta I_B = 100 	imes 0.5 \, mA = 50 \, mA = 0.05 \, A$.$3$. Apply $KVL$ to the collector-emitter loop:$V_{CC} - I_C R_L - V_{CE} = 0$$18 \, V - (0.05 \, A)(100 \, \Omega) - V_{CE} = 0$$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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