Question:
In an NPN transistor, the collector current is 28 mA. If 80% electrons reach the collector, its base current in mA is
In an NPN transistor, the collector current is 28 mA. If 80% electrons reach the collector, its base current in mA is
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Collector current is 80 percent of the emitter current. Then I_B = I_E - I_C.
Updated On: Oct 1, 2026
- \(7\) mA
- \(14\) mA
- \(28\) mA
- \(35\) mA
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
In a transistor, the emitter current splits into the base current and the collector current: \(I_E = I_B + I_C\). The fraction of carriers that reach the collector is \(\alpha = \dfrac{I_C}{I_E}\).
Step 2: Key Formula or Approach:
\[ \alpha = 0.8 = \frac{I_C}{I_E},\qquad I_B = I_E - I_C \]
Step 3: Detailed Explanation:
\[ I_E = \frac{I_C}{\alpha} = \frac{28}{0.8} = 35 \text{ mA} \]
\[ I_B = 35 - 28 = 7 \text{ mA} \]
Option (B) 14 mA and (C) 28 mA are not consistent with an alpha of 0.8, and option (D) 35 mA is the emitter current, not the base current.
Final Answer:
The base current is 7 mA, option (A). \[ \boxed{7\text{ mA (A)}} \]
In a transistor, the emitter current splits into the base current and the collector current: \(I_E = I_B + I_C\). The fraction of carriers that reach the collector is \(\alpha = \dfrac{I_C}{I_E}\).
Step 2: Key Formula or Approach:
\[ \alpha = 0.8 = \frac{I_C}{I_E},\qquad I_B = I_E - I_C \]
Step 3: Detailed Explanation:
\[ I_E = \frac{I_C}{\alpha} = \frac{28}{0.8} = 35 \text{ mA} \]
\[ I_B = 35 - 28 = 7 \text{ mA} \]
Option (B) 14 mA and (C) 28 mA are not consistent with an alpha of 0.8, and option (D) 35 mA is the emitter current, not the base current.
Final Answer:
The base current is 7 mA, option (A). \[ \boxed{7\text{ mA (A)}} \]
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