In an NPN transistor circuit,the collector cu | vedclass

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(D) Given that the collector current $I_C = 10\, mA$.According to the problem,$90\%$ of the electrons emitted reach the collector,which means the coll

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$i_E = 11\, mA, i_B = 1\, mA$

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(D) Given that the collector current $I_C = 10\, mA$.According to the problem,$90\%$ of the electrons emitted reach the collector,which means the collector current is $90\%$ of the emitter current:$I_C = 0.90 	imes I_E$Substituting the value of $I_C$:$10\, mA = 0.90 	imes I_E$$I_E = \frac{10}{0.9} \approx 11.11\, mA \approx 11\, mA$.Using the relation $I_E = I_B + I_C$,we can find the base current:$I_B = I_E - I_C$$I_B = 11\, mA - 10\, mA = 1\, mA$.Thus,$I_E = 11\, mA$ and $I_B = 1\, mA$.

In an $NPN$ transistor circuit,the collector current is $10\, mA$. If $90\%$ of the electrons emitted reach the collector,the emitter current $(i_E)$ and base current $(i_B)$ are given by

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