Question:
When a diode is forward biased, it has a voltage drop of 0.5 V. The safe limit of current through the diode is 10 mA. If a battery of emf 1.5 V is used in the circuit, the value of minimum resistance to be connected in series with the diode so that the current does not exceed the safe limit is
When a diode is forward biased, it has a voltage drop of 0.5 V. The safe limit of current through the diode is 10 mA. If a battery of emf 1.5 V is used in the circuit, the value of minimum resistance to be connected in series with the diode so that the current does not exceed the safe limit is
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Voltage across the series resistor is the battery emf minus the diode drop; divide by the safe current.
Updated On: Nov 5, 2026
- \(50\ \Omega\)
- \(200\ \Omega\)
- \(300\ \Omega\)
- \(100\ \Omega\)
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The Correct Option is D
Solution and Explanation
Voltage across the series resistor: \(V_R = 1.5 - 0.5 = 1.0\) V.
Maximum current \(I = 10\) mA \(= 10^{-2}\) A.
Minimum resistance \(R = \frac{V_R}{I} = \frac{1.0}{10^{-2}} = 100\ \Omega\).
Answer: (4) \(100\ \Omega\)
Maximum current \(I = 10\) mA \(= 10^{-2}\) A.
Minimum resistance \(R = \frac{V_R}{I} = \frac{1.0}{10^{-2}} = 100\ \Omega\).
Answer: (4) \(100\ \Omega\)
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