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Current Electricity question

2021 · 22 Jul · Shift 2 · Q71
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Current Electricity question

2021 · 22 Jul · Shift 2 · Q71

JEE MainPhysicsCurrent ElectricityNumerical+4 / −1
In an electric circuit, a cell of certain emf provides a potential difference of 1.25 V across a load resistance of 5 Ω\OmegaΩ. However, it provides a potential difference of 1 V across a load resistance of 2 Ω\OmegaΩ. The emf of the cell is given by x10V{x \over {10}}V10x​V. Then the value of x is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 15

  1. Let the emf of the cell be EEE and its internal resistance be rrr.

  2. When the load resistance is R=5 ΩR = 5\,\OmegaR=5Ω, the terminal voltage is V=1.25 VV = 1.25\,\text{V}V=1.25V.

    Using current through the load: I=VR=1.255=0.25 AI = \frac{V}{R} = \frac{1.25}{5} = 0.25\,\text{A}I=RV​=51.25​=0.25A

    For a cell, E=V+IrE = V + IrE=V+Ir So, E=1.25+0.25r...(1)E = 1.25 + 0.25r \qquad ...(1)E=1.25+0.25r...(1)

  3. When the load resistance is R=2 ΩR = 2\,\OmegaR=2Ω, the terminal voltage is V=1 VV = 1\,\text{V}V=1V.

    Then current is I=VR=12=0.5 AI = \frac{V}{R} = \frac{1}{2} = 0.5\,\text{A}I=RV​=21​=0.5A

    Again, E=V+IrE = V + IrE=V+Ir So, E=1+0.5r...(2)E = 1 + 0.5r \qquad ...(2)E=1+0.5r...(2)

  4. Since emf is same in both cases, equate (1) and (2): 1.25+0.25r=1+0.5r1.25 + 0.25r = 1 + 0.5r1.25+0.25r=1+0.5r

    0.25=0.25r0.25 = 0.25r0.25=0.25r

    r=1 Ωr = 1\,\Omegar=1Ω

  5. Substitute r=1r=1r=1 into (1): E=1.25+0.25(1)=1.5 VE = 1.25 + 0.25(1) = 1.5\,\text{V}E=1.25+0.25(1)=1.5V

  6. Given E=x10 VE = \frac{x}{10}\,\text{V}E=10x​V Therefore, x10=1.5\frac{x}{10} = 1.510x​=1.5 x=15x = 15x=15

Therefore, the required integer is 15\boxed{15}15​

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