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JEE Main 2021 (22 Jul, Shift 2) · Cells: EMF, Internal Resistance & Combinations

Cells: EMF, Internal Resistance & Combinations question from JEE Main 2021 (22 Jul, Shift 2)

Q26JEE Main 2021 · 22 JUL · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations

In an electric circuit, a cell of certain emf provides a potential difference of 1.25 V across a load resistance of 5Ω\Omega. However, it provides a potential difference of 1 V across a load resistance of 2Ω\Omega. The emf of the cell is given by x10 V\frac{x}{10}\text{ V}. Then the value of x is ______________.

Numerical answer type. Work it out, then open the answer.

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Answer: 15

  1. V=ERR+rV = \frac{ER}{R + r}, so E=V(1+rR)E = V\left(1 + \frac{r}{R}\right)
  2. 1.25(1+r5)=1(1+r2)⇒1.25+0.25r=1+0.5r⇒r=1 Ω1.25\left(1 + \frac{r}{5}\right) = 1\left(1 + \frac{r}{2}\right) \Rightarrow 1.25 + 0.25r = 1 + 0.5r \Rightarrow r = 1\ \Omega
  3. E=1×(1+0.5)=1.5 V=1510 VE = 1 \times (1 + 0.5) = 1.5\text{ V} = \frac{15}{10}\text{ V}
  4. x=15x = 15
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