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JEE Main 2019 (10 Jan, Shift 1) · Ohm's Law, Resistivity & Temperature

Ohm's Law, Resistivity & Temperature question from JEE Main 2019 (10 Jan, Shift 1)

Q49JEE Main 2019 · 10 JAN · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature

A 2 W carbon resistor is color coded with green, black, red and brown respectively. The maximum current which can be passed through this resistor is -

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Answer: (b)

  1. Green = 5, black = 0, red = ×102\times10^2: R=50×102=5000 ΩR = 50\times10^2 = 5000\ \Omega (brown = 1% tolerance).
  2. P=I2R⇒I=P/R=2/5000=0.02P = I^2R \Rightarrow I = \sqrt{P/R} = \sqrt{2/5000} = 0.02 A
  3. Imax=20I_{max} = 20 mA. The answer is (b).
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