(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

Difficult

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

In a $CE$ circuit,the collector of a transistor is kept at a constant voltage $V_C = 1.5 \ V$. When the base current changes from $100 \ \mu A$ to $150 \ \mu A$,the collector current changes from $5 \ mA$ to $10 \ mA$. The current gain $\beta$ for this transistor is:

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An $NPN$ transistor circuit is arranged as shown in the figure. It is:

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In a common emitter amplifier, a change of $0.2 \text{ mA}$ in the base current causes a change of $5 \text{ mA}$ in the collector current. If input resistance is $2 \text{ k}\Omega$ and voltage gain is $75$, the load resistance used in the circuit is

DifficultMHT CET 2026

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In an $npn$ transistor, $10^{10}$ electrons enter the emitter in $10^{-6} \; s$. If $4\%$ of the electrons are lost in the base, the current transfer ratio $(\alpha)$ will be:

MediumAIIMS 2017

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$A$ transistor is connected in a common emitter configuration with $R_o = 4 \text{ k}\Omega$,$R_i = 1 \text{ k}\Omega$,$I_C = 1 \text{ mA}$,and $I_b = 20 \text{ }\mu\text{A}$. The voltage gain is:

Medium

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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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In a $CE$ circuit,the collector of a transistor is kept at a constant voltage $V_C = 1.5 \ V$. When the base current changes from $100 \ \mu A$ to $150 \ \mu A$,the collector current changes from $5 \ mA$ to $10 \ mA$. The current gain $\beta$ for this transistor is:

An $NPN$ transistor circuit is arranged as shown in the figure. It is:

In a common emitter amplifier, a change of $0.2 \text{ mA}$ in the base current causes a change of $5 \text{ mA}$ in the collector current. If input resistance is $2 \text{ k}\Omega$ and voltage gain is $75$, the load resistance used in the circuit is

In an $npn$ transistor, $10^{10}$ electrons enter the emitter in $10^{-6} \; s$. If $4\%$ of the electrons are lost in the base, the current transfer ratio $(\alpha)$ will be:

$A$ transistor is connected in a common emitter configuration with $R_o = 4 \text{ k}\Omega$,$R_i = 1 \text{ k}\Omega$,$I_C = 1 \text{ mA}$,and $I_b = 20 \text{ }\mu\text{A}$. The voltage gain is:

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