In an NPN transistor,the collector current is | vedclass

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(B) Given that the collector current $I_C = 10\, mA$. Since $90\%$ of the electrons emitted reach the collector,the collector current is $90\%$ of the

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Emitter current will be $11.1\, mA$

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(B) Given that the collector current $I_C = 10\, mA$. Since $90\%$ of the electrons emitted reach the collector,the collector current is $90\%$ of the emitter current $(I_E)$. Therefore,$I_C = 0.90 	imes I_E$. Substituting the value of $I_C$: $10\, mA = 0.90 	imes I_E$. Solving for $I_E$: $I_E = \frac{10}{0.90} = 11.11\, mA$. Using the relation $I_E = I_B + I_C$,we can find the base current $I_B$: $I_B = I_E - I_C = 11.11\, mA - 10\, mA = 1.11\, mA$. Comparing this with the given options,the correct statement is that the emitter current is $11.1\, mA$.

In an $NPN$ transistor,the collector current is $10\, mA$. If $90\%$ of the electrons emitted reach the collector,then:

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