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# The input resistance of a common emitter transistor amplifier, if the output resistance is $500\,k\Omega$, the current gain $\alpha = 0.98$ and power gain is $6.0625 \times {10^6}$, is :A. $198\Omega$B. $300\Omega$C. $100\Omega$D. $400\Omega$

Last updated date: 22nd Jul 2024
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Hint:The ratio of the output divided by the input is known as amplifier gain.Gain is a ratio of no units. $\beta$ denotes the current gain in a common-emitter circuit. The relationship between collector current (output current) and base current is known as beta (input current).

Power gain = voltage gain $\times$ current gain –(i)
Voltage gain $({A_v}) = \beta \times \dfrac{{{R_2}}}{{{R_1}}}$
Here, ${R_2}$ is the output resistance, ${R_1}$ is the input resistance and $\beta$ is the current gain.
We know that $\beta = \dfrac{\alpha }{{1 - \alpha }}$
Its given in the question that $\alpha = 0.98$, substituting these values to the equation we get,
$\beta = \dfrac{{0.98}}{{1 - 0.98}} \\ \Rightarrow \beta = \dfrac{{0.98}}{{0.02}} \\ \Rightarrow \beta = 49 \\$
We also know that ${R_2} = 500 \times {10^3}\Omega$, and power gain $= 6.0625 \times {10^6}$. Substituting these values to the (i) equation we get,
Power gain = voltage gain $\times$ current gain
$6.0625 \times {10^6} = 49 \times \dfrac{{500 \times {{10}^3}}}{{{R_1}}} \times 49 \\ \Rightarrow {R_1} = 49 \times \dfrac{{500 \times {{10}^3}}}{{6.0625 \times {{10}^6}}} \times 49 \\ \therefore {R_1} = 198\Omega \\$
Hence, the correct option is A.

Note:The transistor parameters that characterise the current gain in a transistor are $\alpha$ and $\beta$ . The ratio of the collector current to the emitter current is known as $\alpha$. $\beta$ is the current gain, which is calculated by dividing the collector current by the base current.