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Current Electricity question

2022 · 25 Jun · Shift 1 · Q67
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  5. /2022 · 25 Jun · Shift 1 · Q67

Current Electricity question

2022 · 25 Jun · Shift 1 · Q67

JEE MainPhysicsCurrent ElectricityNumerical+4 / −1
A resistor develops 300 J of thermal energy in 15 s, when a current of 2 A is passed through it. If the current increases to 3 A, the energy developed in 10 s is ‾\underline{\hspace{2cm}}​ J.
Numerical answer
View written solutionFree

Correct answer: 450

  1. Use Joule’s law of heating:

H=I2RtH = I^2 R tH=I2Rt

where:

  • HHH = thermal energy developed,
  • III = current,
  • RRR = resistance,
  • ttt = time.
  1. From the first condition:

300=(2)2R(15)300 = (2)^2 R (15)300=(2)2R(15)

300=4⋅15⋅R=60R300 = 4 \cdot 15 \cdot R = 60R300=4⋅15⋅R=60R

R=30060=5 ΩR = \frac{300}{60} = 5\,\OmegaR=60300​=5Ω

  1. Now for the second condition, current is 3 3\,3A and time is 10 10\,10s:

H=I2RtH = I^2RtH=I2Rt

H=(3)2(5)(10)H = (3)^2 (5)(10)H=(3)2(5)(10)

H=9⋅5⋅10=450 JH = 9 \cdot 5 \cdot 10 = 450\text{ J}H=9⋅5⋅10=450 J

  1. Therefore, the energy developed is:

450 J\boxed{450\text{ J}}450 J​

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