Question:
The power gain and voltage gain of a transistor connected in common emitter configuration are 1800 and 60 respectively. If the change in the emitter current is 0.62 mA, then the change in the collector current is
The power gain and voltage gain of a transistor connected in common emitter configuration are 1800 and 60 respectively. If the change in the emitter current is 0.62 mA, then the change in the collector current is
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Power Gain = Voltage Gain \(\times\) Current Gain. Always check if the given current is base, emitter, or collector. \( I_E \) is the largest current.
Updated On: Mar 30, 2026
- 0.60 mA
- 0.58 mA
- 0.52 mA
- 0.48 mA
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
We are given Power Gain (\( A_p \)) and Voltage Gain (\( A_v \)). We can find the Current Gain (\( \beta \)). Using \( \beta \), we relate base current and collector current. However, we are given emitter current change. We need to use the relation \( \Delta I_C = \alpha \Delta I_E \) or find \( \beta \) first and use \( \Delta I_C = \frac{\beta}{1+\beta} \Delta I_E \).
Step 2: Key Formula or Approach:
1. \( A_p = \beta \times A_v \implies \beta = \frac{A_p}{A_v} \). 2. Current gain in CE is \( \beta = \frac{\Delta I_C}{\Delta I_B} \). 3. Relation between currents: \( I_E = I_C + I_B \). 4. \( \Delta I_C = \frac{\beta}{1+\beta} \Delta I_E \) (since \( \alpha = \frac{\beta}{1+\beta} \)).
Step 3: Detailed Explanation:
Calculate \( \beta \): \( \beta = \frac{1800}{60} = 30 \). Calculate \( \alpha \) (current gain for CB, relation between \( I_C \) and \( I_E \)): \( \alpha = \frac{\beta}{1+\beta} = \frac{30}{31} \). We know \( \Delta I_C = \alpha \Delta I_E \). Given \( \Delta I_E = 0.62 \) mA. \( \Delta I_C = \frac{30}{31} \times 0.62 \) \( \Delta I_C = 30 \times \frac{0.62}{31} = 30 \times 0.02 = 0.60 \) mA.
Step 4: Final Answer:
The change in collector current is 0.60 mA.
We are given Power Gain (\( A_p \)) and Voltage Gain (\( A_v \)). We can find the Current Gain (\( \beta \)). Using \( \beta \), we relate base current and collector current. However, we are given emitter current change. We need to use the relation \( \Delta I_C = \alpha \Delta I_E \) or find \( \beta \) first and use \( \Delta I_C = \frac{\beta}{1+\beta} \Delta I_E \).
Step 2: Key Formula or Approach:
1. \( A_p = \beta \times A_v \implies \beta = \frac{A_p}{A_v} \). 2. Current gain in CE is \( \beta = \frac{\Delta I_C}{\Delta I_B} \). 3. Relation between currents: \( I_E = I_C + I_B \). 4. \( \Delta I_C = \frac{\beta}{1+\beta} \Delta I_E \) (since \( \alpha = \frac{\beta}{1+\beta} \)).
Step 3: Detailed Explanation:
Calculate \( \beta \): \( \beta = \frac{1800}{60} = 30 \). Calculate \( \alpha \) (current gain for CB, relation between \( I_C \) and \( I_E \)): \( \alpha = \frac{\beta}{1+\beta} = \frac{30}{31} \). We know \( \Delta I_C = \alpha \Delta I_E \). Given \( \Delta I_E = 0.62 \) mA. \( \Delta I_C = \frac{30}{31} \times 0.62 \) \( \Delta I_C = 30 \times \frac{0.62}{31} = 30 \times 0.02 = 0.60 \) mA.
Step 4: Final Answer:
The change in collector current is 0.60 mA.
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