Question:
In common emitter mode of transistor, the d.c. current gain is 20, the emitter current is 7 mA. The collector current is
In common emitter mode of transistor, the d.c. current gain is 20, the emitter current is 7 mA. The collector current is
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The emitter current is the sum of the base and collector currents, and the dc current gain beta is Ic over Ib.
Updated On: Oct 1, 2026
- \(\frac{20}{3}\) mA
- \(\frac{14}{5}\) mA
- \(\frac{1}{3}\) mA
- \(\frac{8}{13}\) mA
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
In a transistor, \(I_E = I_B + I_C\), and the dc current gain in common emitter mode is \(\beta = \dfrac{I_C}{I_B}\).
Step 2: Detailed Explanation:
\(\beta = 20\), so \(I_B = \dfrac{I_C}{20}\).
\[ I_E = \frac{I_C}{20} + I_C = \frac{21}{20}I_C \Rightarrow I_C = \frac{20}{21}\times 7 = \frac{20}{3}\text{ mA} \]
Step 3: Check the options.
\(\dfrac{20}{3} \approx 6.67\) mA is slightly less than the emitter current, as it should be. The other options are far too small.
Final Answer:
The collector current is \(\dfrac{20}{3}\) mA, option (A). \[ \boxed{\frac{20}{3}\text{ mA}} \]
In a transistor, \(I_E = I_B + I_C\), and the dc current gain in common emitter mode is \(\beta = \dfrac{I_C}{I_B}\).
Step 2: Detailed Explanation:
\(\beta = 20\), so \(I_B = \dfrac{I_C}{20}\).
\[ I_E = \frac{I_C}{20} + I_C = \frac{21}{20}I_C \Rightarrow I_C = \frac{20}{21}\times 7 = \frac{20}{3}\text{ mA} \]
Step 3: Check the options.
\(\dfrac{20}{3} \approx 6.67\) mA is slightly less than the emitter current, as it should be. The other options are far too small.
Final Answer:
The collector current is \(\dfrac{20}{3}\) mA, option (A). \[ \boxed{\frac{20}{3}\text{ mA}} \]
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