In the circuit shown in the figure,the input | vedclass

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(C) Given: $V_{BE} = 0$,$V_{CE} = 0$,$V_i = 20\, V$,$R_B = 500\, k\Omega$,$R_C = 4\, k\Omega$.For the output circuit (collector-emitter loop):$V_{CC}

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$I_B=40\,\mu A$,$I_C=5\,mA$,$\beta=125$

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(C) Given: $V_{BE} = 0$,$V_{CE} = 0$,$V_i = 20\, V$,$R_B = 500\, k\Omega$,$R_C = 4\, k\Omega$.For the output circuit (collector-emitter loop):$V_{CC} = I_C R_C + V_{CE}$Since $V_{CE} = 0$,we have:$20\, V = I_C 	imes (4 	imes 10^3\,\Omega) + 0$$I_C = \frac{20}{4 	imes 10^3} = 5 	imes 10^{-3}\, A = 5\, mA$.For the input circuit (base-emitter loop):$V_i = I_B R_B + V_{BE}$Since $V_{BE} = 0$,we have:$20\, V = I_B 	imes (500 	imes 10^3\,\Omega) + 0$$I_B = \frac{20}{500 	imes 10^3} = 40 	imes 10^{-6}\, A = 40\,\mu A$.The current gain $\beta$ is given by:$\beta = \frac{I_C}{I_B} = \frac{5 	imes 10^{-3}\, A}{40 	imes 10^{-6}\, A} = \frac{5000}{40} = 125$.

In the circuit shown in the figure,the input voltage $V_i$ is $20\, V$,$V_{BE} = 0$,and $V_{CE} = 0$. The values of $I_B$,$I_C$,and $\beta$ are given by:

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