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# The base current in a transistor circuit changes from $45\mu A$ to $140\mu A$. Accordingly, the collector current changes from $0.2 mA$ to $4.00 mA$. The gain in current is:A) 9.5B) 1C) 40D) 20

Last updated date: 21st Jun 2024
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Hint: For a transistor in common base configuration the current gain is defined as the ratio of change in collector current and the change in emitter current when the voltage between base and collector is kept constant.

Formulae used:
Current gain of a transistor circuit is given by:
${g_I} = \dfrac{{\Delta {I_C}}}{{\Delta {I_B}}}$..............(1)
Where,
${g_I}$denotes the current gain of the transistor,
$\Delta {I_C}$denotes change in collector current of the transistor,
$\Delta {I_B}$ denotes change in base current of the transistor.

Complete step by step solution:
Here, base current changes from $45\mu A$ to $140\mu A$.
Here, collector current changes from 0.2 mA to 4 mA.
To find: Gain in current for the transistor.
From the given information get the change in base current as:
$\Delta {I_B} = 140\mu A - 45\mu A = 95\mu A \\ \therefore \Delta {I_B} = 9.5 \times {10^{ - 5}}A \\$....................(2)
Similarly, get the change in collector current as:
$\Delta {I_C} = 4mA - 0.2mA = 3.8mA \\ \therefore \Delta {I_C} = 3.8 \times {10^{ - 3}}A \\$...................(3)
Substitute the obtained value of $\Delta {I_B}$and $\Delta {I_C}$from eq.(2) and eq.(3) respectively into the eq.(1) to obtain current gain ${g_I}$as:
${g_I} = \dfrac{{3.8 \times {{10}^{ - 3}}A}}{{9.5 \times {{10}^{ - 5}}A}} = \dfrac{{3800}}{{95}} \\ \therefore {g_I} = 40 \\$
Gain in current for the transistor is got to be (C), 40.