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Current Electricity question

2024 · 5 Apr · Shift 2 · Q74
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  5. /2024 · 5 Apr · Shift 2 · Q74

Current Electricity question

2024 · 5 Apr · Shift 2 · Q74

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
The ratio of heat dissipated per second through the resistance 5Ω5 \Omega5Ω and 10Ω10 \Omega10Ω in the circuit given below is: JEE Main 2024 (Online) 5th April Evening Shift Physics - Current Electricity Question 26 English
  1. A
    4 : 1
  2. B
    2 : 1
  3. C
    1 : 2
  4. D
    1 : 1
View written solutionFree

Correct answer: B

  1. Interpret the circuit

    The two resistors of 5 Ω5\,\Omega5Ω and 10 Ω10\,\Omega10Ω are connected in parallel across the same potential difference.

    Hence, if the voltage across each resistor is VVV, then the heat dissipated per second (i.e. power) in each resistor is:

    P=V2RP = \frac{V^2}{R}P=RV2​

  2. Power in each resistor

    • For 5 Ω5\,\Omega5Ω resistor: P5=V25P_5 = \frac{V^2}{5}P5​=5V2​

    • For 10 Ω10\,\Omega10Ω resistor: P10=V210P_{10} = \frac{V^2}{10}P10​=10V2​

  3. Take the ratio

    P5P10=V2/5V2/10=105=2\frac{P_5}{P_{10}} = \frac{V^2/5}{V^2/10} = \frac{10}{5} = 2P10​P5​​=V2/10V2/5​=510​=2

    Therefore,

    P5:P10=2:1P_5 : P_{10} = 2 : 1P5​:P10​=2:1

  4. Match with options

    This corresponds to Option B.

  5. Comparison with stored correct answer

    Stored correct answer: B

    Derived answer: B

    So, the derived answer agrees with the stored correct answer.

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