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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Use $I_C=\frac{\beta}{1+\beta}I_E$ because $I_E=I_B+I_C$.
Updated On: Oct 1, 2026
- \(\frac{14}{3}\) mA
- \(\frac{20}{3}\) mA
- \(\frac{7}{3}\) mA
- \(\frac{8}{3}\) mA
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The Correct Option is B
Solution and Explanation
Step 1: Write the current relations
In a transistor the emitter current is the sum of base and collector currents, \(I_E=I_B+I_C\).
The d.c. current gain in common emitter mode is \(\beta=\frac{I_C}{I_B}\).
Step 2: Express \(I_C\) through \(I_E\)
From \(\beta=I_C/I_B\) we get \(I_B=I_C/\beta\).
Then \(I_E=\frac{I_C}{\beta}+I_C=I_C\frac{1+\beta}{\beta}\).
\[ I_C=\frac{\beta}{1+\beta}I_E \]
Step 3: Put in numbers
With \(\beta=20\) and \(I_E=7\) mA:
\[ I_C=\frac{20}{21}\times 7=\frac{20}{3}\text{ mA} \]
Step 4: Check the other options
\(\frac{14}{3}\) mA would need \(\beta=2\) and \(\frac{7}{3}\) mA would need \(\beta=0.5\). These ignore that \(I_C\) is \(20\) times \(I_B\). Only \(\frac{20}{3}\) mA fits.
Final Answer:
The collector current is \(\frac{20}{3}\) mA, option (B). \[ \boxed{\text{(B) } \frac{20}{3}\text{ mA}} \]
In a transistor the emitter current is the sum of base and collector currents, \(I_E=I_B+I_C\).
The d.c. current gain in common emitter mode is \(\beta=\frac{I_C}{I_B}\).
Step 2: Express \(I_C\) through \(I_E\)
From \(\beta=I_C/I_B\) we get \(I_B=I_C/\beta\).
Then \(I_E=\frac{I_C}{\beta}+I_C=I_C\frac{1+\beta}{\beta}\).
\[ I_C=\frac{\beta}{1+\beta}I_E \]
Step 3: Put in numbers
With \(\beta=20\) and \(I_E=7\) mA:
\[ I_C=\frac{20}{21}\times 7=\frac{20}{3}\text{ mA} \]
Step 4: Check the other options
\(\frac{14}{3}\) mA would need \(\beta=2\) and \(\frac{7}{3}\) mA would need \(\beta=0.5\). These ignore that \(I_C\) is \(20\) times \(I_B\). Only \(\frac{20}{3}\) mA fits.
Final Answer:
The collector current is \(\frac{20}{3}\) mA, option (B). \[ \boxed{\text{(B) } \frac{20}{3}\text{ mA}} \]
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