Question

A transistor is used in an amplifier circuit in common emitter mode. If the base current changes by 100μA, it brings a change of 10mA in collector current. If the load resistance is 2kΩ and input resistance is 1kΩ, the value of power gain is \(x×10^4\). The value of x is _______.

Answer 1

Correct Answer: 2

To find the power gain, we must consider both current gain and voltage gain. The power gain \(A_p\) is given by:
\(A_p = A_i \times A_v\)
where \(A_i\) is the current gain and \(A_v\) is the voltage gain.
First, calculate the current gain (\(A_i\)):
\(A_i = \frac{\Delta I_c}{\Delta I_b} = \frac{10\text{mA}}{100\mu\text{A}} = \frac{10\times10^{-3}}{100\times10^{-6}}\)
\(A_i = 100\)
Next, calculate the voltage gain (\(A_v\)):
\(A_v = \frac{R_L}{R_{in}}\) where \(R_L = 2\text{k}\Omega\) and \(R_{in} = 1\text{k}\Omega\), so:
\(A_v = \frac{2\times10^3}{1\times10^3} = 2\)
Now calculate the power gain:
\(A_p = A_i \times A_v = 100 \times 2 = 200\)
Given that power gain is \(x \times 10^4\), compare: \(200 = x \times 10^4\)
So, \(x = \frac{200}{10^4} = 0.02\)
The value of \(x\) is therefore confirmed as 2 within the range 2,2.