Question:medium
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
Show Hint
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
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Find the base current first
Let $I_B=x$. Then $I_C=20x$ because $\beta=20$.
So $I_E=I_B+I_C=21x=7$ mA, which gives $x=\frac{1}{3}$ mA.
Step 2: Get the collector current
$I_C=20x=20\times\frac{1}{3}=\frac{20}{3}$ mA.
This is option (B).
Step 3: Why not the others
Taking $I_C=\frac{I_E}{3}$ or other fractions ignores that the base takes only $\frac{1}{21}$ of the emitter current.
Final Answer:
$I_B=\frac13$ mA and $I_C=\frac{20}{3}$ mA. Option (B). \[ \boxed{\text{(B)}} \]
Let $I_B=x$. Then $I_C=20x$ because $\beta=20$.
So $I_E=I_B+I_C=21x=7$ mA, which gives $x=\frac{1}{3}$ mA.
Step 2: Get the collector current
$I_C=20x=20\times\frac{1}{3}=\frac{20}{3}$ mA.
This is option (B).
Step 3: Why not the others
Taking $I_C=\frac{I_E}{3}$ or other fractions ignores that the base takes only $\frac{1}{21}$ of the emitter current.
Final Answer:
$I_B=\frac13$ mA and $I_C=\frac{20}{3}$ mA. Option (B). \[ \boxed{\text{(B)}} \]
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