(B) Given that the collector current $I_C = 10 \ mA$. Since $90\%$ of the electrons emitted from the emitter reach the collector,we have the relation:

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(B) Given that the collector current $I_C = 10 \ mA$. Since $90\%$ of the electrons emitted from the emitter reach the collector,we have the relation: $I_C = 0.90 \times I_E$. Rearranging for $I_E$: $I_E = \frac{I_C}{0.90} = \frac{10 \ mA}{0.9} = 11.11 \ mA \approx 11 \ mA$. Using the Kirchhoff's current law for a transistor: $I_E = I_B + I_C$. Therefore,the base current is $I_B = I_E - I_C = 11 \ mA - 10 \ mA = 1 \ mA$. Thus,$I_E = 11 \ mA$ and $I_B = 1 \ mA$.

Class 12 Physics Semiconductor Electronics Junction Transistor

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In an $N-P-N$ transistor circuit,the collector current is $10 \ mA$. If $90\%$ of the electrons emitted reach the collector,then the emitter current $(I_E)$ and base current $(I_B)$ are respectively:

A
$I_E = 1 \ mA; I_B = 11 \ mA$
B
$I_E = 11 \ mA; I_B = 1 \ mA$
C
$I_E = -1 \ mA; I_B = 9 \ mA$
D
$I_E = 9 \ mA; I_B = -1 \ mA$

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Semiconductor Electronics

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Junction Transistor

If $\alpha = 0.98$ and the current through the emitter $i_e = 20 \, mA$,the value of $\beta$ is

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MediumAP EAMCET 2023

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In an $N-P-N$ transistor circuit,the collector current is $10 \ mA$. If $90\%$ of the electrons emitted reach the collector,then the emitter current $(I_E)$ and base current $(I_B)$ are respectively:

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