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Current Electricity question

2020 · 4 Sep · Shift 1 · Q48
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Current Electricity question

2020 · 4 Sep · Shift 1 · Q48

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated within the internal resistance is :
  1. A
    0.50 W
  2. B
    0.072 W
  3. C
    0.10 W
  4. D
    0.125 W
View written solutionFree

Correct answer: C

  1. Given data

    • Emf of battery: E=3.0 VE = 3.0\,\text{V}E=3.0V
    • Terminal voltage: V=2.5 VV = 2.5\,\text{V}V=2.5V
    • Power dissipated in external resistor: P=0.5 WP = 0.5\,\text{W}P=0.5W

  2. Find the current in the circuit

    The external resistor has terminal voltage 2.5 V2.5\,\text{V}2.5V across it, and it dissipates 0.5 W0.5\,\text{W}0.5W.

    Using P=VIP = VIP=VI 0.5=2.5 I0.5 = 2.5\, I0.5=2.5I I=0.52.5=0.2 AI = \frac{0.5}{2.5} = 0.2\,\text{A}I=2.50.5​=0.2A

  3. Find the voltage across the internal resistance

    The drop inside the battery is E−V=3.0−2.5=0.5 VE - V = 3.0 - 2.5 = 0.5\,\text{V}E−V=3.0−2.5=0.5V

  4. Find power dissipated in internal resistance

    Power lost inside the battery is Pinternal=(E−V)IP_{\text{internal}} = (E-V)IPinternal​=(E−V)I Pinternal=0.5×0.2=0.10 WP_{\text{internal}} = 0.5 \times 0.2 = 0.10\,\text{W}Pinternal​=0.5×0.2=0.10W

  5. Match with options

    • A: 0.50 W0.50\,\text{W}0.50W
    • B: 0.072 W0.072\,\text{W}0.072W
    • C: 0.10 W0.10\,\text{W}0.10W
    • D: 0.125 W0.125\,\text{W}0.125W

    So the correct option is C.

  6. Comparison with stored answer

    Stored correct answer: C

    My derived answer: C

    They agree.

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