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In a common emitter transistor amplifier, the output resistance is $500{\text{ k}}\Omega$ and the current gain $\beta = 49$. If the power gain of the amplifier is $5 \times {10^6}$, the input resistance isA. $325{\text{ k}}\Omega$B. $165{\text{ k}}\Omega$C. ${\text{198 k}}\Omega$D. $240{\text{ k}}\Omega$

Last updated date: 15th Aug 2024
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Hint:In this question, we need to find the input resistance. For this, we will use the formula of power gain for a common emitter transistor amplifier. After simplification, we will get the final result.

Formula used:
The formula for power gain for a common emitter transistor amplifier is given below.
Power gain $= {\beta ^2} \times \dfrac{{{R_o}}}{{{R_i}}}$
Here, $\beta$ is the current gain, ${R_o}$ is the output resistance and ${R_i}$ is the input resistance.

Complete step by step solution:
We know that the power gain for a common emitter transistor amplifier is Power gain $= {\beta ^2} \times \dfrac{{{R_o}}}{{{R_i}}}$
But the current gain $\left( \beta \right)$is 49.
Also, the output resistance is $\left( {{R_o}} \right) = 500{\text{ k}}\Omega$
But $1{\text{ k}}\Omega = 1000{\text{ }}\Omega = {10^3}{\text{ }}\Omega$
So, $\left( {{R_o}} \right) = 500 \times 1000 = 500 \times {10^3}{\text{ }}\Omega$
Also, power gain is $5 \times {10^6}$
So, we get
$5 \times {10^6} = {\left( {49} \right)^2} \times \dfrac{{500 \times {{10}^3}{\text{ }}}}{{{R_i}}}$
$5 \times {10^6}\left( {{R_i}} \right) = {\left( {49} \right)^2} \times 500 \times {10^3}$

By simplifying, we get
$\left( {{R_i}} \right) = \dfrac{{{{\left( {49} \right)}^2} \times 500 \times {{10}^3}}}{{5 \times {{10}^6}}}$
$\Rightarrow \left( {{R_i}} \right) = {\left( {49} \right)^2} \times 100 \times {10^{3 - 6}}$
$\Rightarrow \left( {{R_i}} \right) = {\left( {49} \right)^2} \times {10^2} \times {10^{ - 3}}$
$\Rightarrow \left( {{R_i}} \right) = {\left( {49} \right)^2} \times {10^{2 - 3}}$

By simplifying further, we get
$\left( {{R_i}} \right) = 2401 \times {10^{2 - 3}}$
$\Rightarrow \left( {{R_i}} \right) = 2401 \times {10^{ - 1}}$
$\Rightarrow \left( {{R_i}} \right) = \dfrac{{2401}}{{10}}$
This gives, $\left( {{R_i}} \right) = 240.1{\text{ }}\Omega$
That is $\left( {{R_i}} \right) \approx 240{\text{ }}\Omega$
Hence, the value of input resistance is approximately $240{\text{ }}\Omega$.

Therefore, the correct option is (D).

Additional information: We know that an amplifier is a type of electronic circuit often used to boost the strength of a poor input signal in terms of voltage, current, or power. So, the common emitter amplifier is a voltage amplifier that consists of three basic single-stage bipolar junction transistors. This amplifier's input is captured from the base terminal, its output is gathered from the collector terminal, and both terminals share the emitter terminal.

Note: Many students generally make mistakes in writing the formula of power gain of an amplifier. They generally write ${\beta ^2} \times \dfrac{{{R_i}}}{{{R_o}}}$ instead of ${\beta ^2} \times \dfrac{{{R_o}}}{{{R_i}}}$. Also, while doing calculations, they may get confused with the power of 10.