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In a $\mathrm{n}-\mathrm{p}-\mathrm{n}$ transistor circuit, the collector current is 10 $\mathrm{mA}$. If 95 per cent of the electrons emitted reach the collector, which of the following statements are true?
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Verified Answer
The correct answers are:
The emitter current will be $10.53 \mathrm{~mA}$
,
The base current will be $0.53 \mathrm{~mA}$
The emitter current will be $10.53 \mathrm{~mA}$
,
The base current will be $0.53 \mathrm{~mA}$
As given that
$$
I_c=10 \mathrm{~mA}
$$
$\mathrm{So}, \mathrm{I}_{\mathrm{c}}=95 \%$ of $\mathrm{I}_e$
$$
\mathrm{I}_{\mathrm{c}}=\frac{95}{100} \mathrm{I}_{\mathrm{e}} \Rightarrow \mathrm{I}_{\mathrm{e}}=\frac{10 \times 100}{95}=10.53 \mathrm{~mA}
$$
We know that $I_E=\left(I_b+I_c\right)$
Also, $\mathrm{I}_{\mathrm{b}}=\mathrm{I}_{\mathrm{e}}-\mathrm{I}_{\mathrm{c}}=10.53-10=0.53 \mathrm{~mA}$
$$
I_c=10 \mathrm{~mA}
$$
$\mathrm{So}, \mathrm{I}_{\mathrm{c}}=95 \%$ of $\mathrm{I}_e$
$$
\mathrm{I}_{\mathrm{c}}=\frac{95}{100} \mathrm{I}_{\mathrm{e}} \Rightarrow \mathrm{I}_{\mathrm{e}}=\frac{10 \times 100}{95}=10.53 \mathrm{~mA}
$$
We know that $I_E=\left(I_b+I_c\right)$
Also, $\mathrm{I}_{\mathrm{b}}=\mathrm{I}_{\mathrm{e}}-\mathrm{I}_{\mathrm{c}}=10.53-10=0.53 \mathrm{~mA}$
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