Question:

An experiment is performed to determine the \(I\) - V characteristics of a Zener diode, which has a protective resistance of \(R = 100\ \Omega\), and a maximum power of dissipation rating of 1 W. The minimum voltage range of the \(DC\) source in the circuit is :

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The 1 W rating limits the current to \(\sqrt{P/R} = 0.1\) A, giving about 10 V across R; the source must cover this plus the Zener drop, so the smallest sufficient range among the options is 0 - 12 V.
Updated On: Nov 11, 2026
  • 0 - 5 V
  • 0 - 8 V
  • 0 - 12 V
  • 0 - 24 V
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The Correct Option is C

Solution and Explanation

Maximum permitted power 1 W with \(R = 100\ \Omega\): \(P = I^2R\Rightarrow I_{max} = \sqrt{\frac{1}{100}} = 0.1\) A.
Voltage across the protective resistor at this current: \(V_R = I_{max}R = 0.1\times 100 = 10\) V.
The source must supply this 10 V and in addition the Zener drop, so a source of range 0 - 5 V or 0 - 8 V is inadequate to drive the circuit up to the rated dissipation; of the options the minimum suitable range is 0 - 12 V.
Answer: (3) 0 - 12 V
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