Question:
\(10^{10}\) electrons enter the emitter of a junction transistor in a time of \(0.4\ \mu\text{s}\). If \(5%\) of the electrons are lost in the base, then the collector current is
\(10^{10}\) electrons enter the emitter of a junction transistor in a time of \(0.4\ \mu\text{s}\). If \(5%\) of the electrons are lost in the base, then the collector current is
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Current transfer ratio \(\alpha = \frac{I_C}{I_E}\). If base loss is \(x%\), then \(\alpha = 1 - \frac{x}{100}\). Here \(\alpha = 0.95\).
Updated On: Mar 26, 2026
- \(3.0\ \text{mA}\)
- \(3.2\ \text{mA}\)
- \(3.6\ \text{mA}\)
- \(3.8\ \text{mA}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
In a transistor, the emitter current (\(I_E\)) is the sum of the base current (\(I_B\)) and the collector current (\(I_C\)): \[ I_E = I_B + I_C \] The electrons lost in the base due to recombination constitute the base current. The remaining electrons reach the collector, forming the collector current.
Step 2: Calculate Emitter Current (\(I_E\)):
Given: Number of electrons, \(n = 10^{10}\). Time, \(t = 0.4\ \mu\text{s} = 0.4 \times 10^{-6}\ \text{s}\). Charge of an electron, \(e = 1.6 \times 10^{-19}\ \text{C}\). \[ I_E = \frac{Q}{t} = \frac{n \times e}{t} \] \[ I_E = \frac{10^{10} \times 1.6 \times 10^{-19}}{0.4 \times 10^{-6}} \] \[ I_E = \frac{1.6 \times 10^{-9}}{4 \times 10^{-7}} = \frac{16 \times 10^{-10}}{4 \times 10^{-7}} = 4 \times 10^{-3}\ \text{A} = 4\ \text{mA} \]
Step 3: Calculate Collector Current (\(I_C\)):
Given that \(5%\) of electrons are lost in the base, this implies \(I_B = 5%\) of \(I_E\). Therefore, the remaining \(95%\) constitutes the collector current. \[ I_C = \frac{95}{100} \times I_E \] \[ I_C = 0.95 \times 4\ \text{mA} \] \[ I_C = 3.8\ \text{mA} \] Final Answer:
The collector current is \(3.8\ \text{mA}\).
In a transistor, the emitter current (\(I_E\)) is the sum of the base current (\(I_B\)) and the collector current (\(I_C\)): \[ I_E = I_B + I_C \] The electrons lost in the base due to recombination constitute the base current. The remaining electrons reach the collector, forming the collector current.
Step 2: Calculate Emitter Current (\(I_E\)):
Given: Number of electrons, \(n = 10^{10}\). Time, \(t = 0.4\ \mu\text{s} = 0.4 \times 10^{-6}\ \text{s}\). Charge of an electron, \(e = 1.6 \times 10^{-19}\ \text{C}\). \[ I_E = \frac{Q}{t} = \frac{n \times e}{t} \] \[ I_E = \frac{10^{10} \times 1.6 \times 10^{-19}}{0.4 \times 10^{-6}} \] \[ I_E = \frac{1.6 \times 10^{-9}}{4 \times 10^{-7}} = \frac{16 \times 10^{-10}}{4 \times 10^{-7}} = 4 \times 10^{-3}\ \text{A} = 4\ \text{mA} \]
Step 3: Calculate Collector Current (\(I_C\)):
Given that \(5%\) of electrons are lost in the base, this implies \(I_B = 5%\) of \(I_E\). Therefore, the remaining \(95%\) constitutes the collector current. \[ I_C = \frac{95}{100} \times I_E \] \[ I_C = 0.95 \times 4\ \text{mA} \] \[ I_C = 3.8\ \text{mA} \] Final Answer:
The collector current is \(3.8\ \text{mA}\).
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