A 64kΩ resistor has a specified maximum power dissipation of 1000 watts. The maximum current that may be passed through the resistor is ________.

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SSC JE Electrical 11 Oct 2023 Shift 3 Official Paper-I
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  1. 64 A
  2. 32 A
  3. \(\frac{1}{8} A\)
  4. 8 A

Answer (Detailed Solution Below)

Option 3 : \(\frac{1}{8} A\)
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Detailed Solution

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Concept

The power across the resistance is given by:

\(P={I^2\times R}\)

where, P = Power

I = Current

R = Resistance

Calculation

Given, P = 1000 W

R = 64 kΩ 

\(1000={I^2\times 64\times 10^3}\)

\(R={1\over 8}\space \Omega\)

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