Question:
In an n-p-n transistor the collector current is \(9\) mA. If \(90\%\) of the electrons from the emitter reach the collector then
In an n-p-n transistor the collector current is \(9\) mA. If \(90\%\) of the electrons from the emitter reach the collector then
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If 90 percent of emitter electrons reach the collector, alpha is 0.9, and beta equals alpha divided by (1 - alpha).
Updated On: Nov 12, 2026
- \(\beta =9.0\)
- the base current is \(10\) mA
- the emitter current is \(1\) mA
- \(\beta =0.9\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
The fraction of emitter current reaching the collector is the current ratio \(\alpha=\dfrac{I_C}{I_E}\).
The current gain is \(\beta=\dfrac{I_C}{I_B}=\dfrac{\alpha}{1-\alpha}\).
Step 2: Calculation
\(\alpha=0.9\), \(I_C=9\) mA.
\(I_E=\dfrac{I_C}{\alpha}=\dfrac{9}{0.9}=10\) mA
\(I_B=I_E-I_C=1\) mA
Step 3: Current gain
\(\beta=\dfrac{I_C}{I_B}=\dfrac{9}{1}=9.0\)
Step 4: Compare the options
Only \(\beta=9.0\) is correct. The base current is 1 mA (not 10 mA) and the emitter current is 10 mA (not 1 mA). This is option (A).
Final Answer:
The current gain beta equals 9.0. \[ \boxed{\text{(A) }\beta=9.0} \]
The fraction of emitter current reaching the collector is the current ratio \(\alpha=\dfrac{I_C}{I_E}\).
The current gain is \(\beta=\dfrac{I_C}{I_B}=\dfrac{\alpha}{1-\alpha}\).
Step 2: Calculation
\(\alpha=0.9\), \(I_C=9\) mA.
\(I_E=\dfrac{I_C}{\alpha}=\dfrac{9}{0.9}=10\) mA
\(I_B=I_E-I_C=1\) mA
Step 3: Current gain
\(\beta=\dfrac{I_C}{I_B}=\dfrac{9}{1}=9.0\)
Step 4: Compare the options
Only \(\beta=9.0\) is correct. The base current is 1 mA (not 10 mA) and the emitter current is 10 mA (not 1 mA). This is option (A).
Final Answer:
The current gain beta equals 9.0. \[ \boxed{\text{(A) }\beta=9.0} \]
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