In the following transistor circuit,R_i is th | vedclass

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(B) The input current $I_i$ is given by the ratio of input voltage $V_i$ to input resistance $R_i$:$I_i = \frac{V_i}{R_i} = \frac{200 	ext{ mV}}{20 \

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$250$

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(B) The input current $I_i$ is given by the ratio of input voltage $V_i$ to input resistance $R_i$:$I_i = \frac{V_i}{R_i} = \frac{200 	ext{ mV}}{20 	ext{ } \Omega} = \frac{0.2 	ext{ V}}{20 	ext{ } \Omega} = 10 	ext{ mA}$.For a common-base transistor configuration,the collector current $I_C$ is approximately equal to the emitter current $I_E$ (i.e.,$I_C \approx I_E = I_i$):$I_L \approx I_i = 10 	ext{ mA}$.The output load voltage $V_L$ across the load resistance $R = 5 	ext{ k}\Omega$ is:$V_L = I_L 	imes R = (10 	ext{ mA}) 	imes (5 	ext{ k}\Omega) = (10 	imes 10^{-3} 	ext{ A}) 	imes (5 	imes 10^3 	ext{ } \Omega) = 50 	ext{ V}$.The voltage gain $A_V$ is defined as the ratio of output voltage $V_L$ to input voltage $V_i$:$A_V = \frac{V_L}{V_i} = \frac{50 	ext{ V}}{200 	ext{ mV}} = \frac{50}{0.2} = 250$.

In the following transistor circuit,$R_i$ is the input resistance and $R_0$ is the output resistance. The approximate voltage gain for the circuit is:

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