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

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Current Electricity

In an electric circuit, a cell of certain emf provides a potential difference of 1.25 V across a load resistance of

5Ω

. However, it provides a potential difference of 1 V across a load resistance of

2Ω

. The emf of the cell is given by

\frac{x}{10} V

. Then the value of x is ______.

Answer: 15

Let the emf be

E

and internal resistance be

r

. For the first case:

V_{1} = 1.25

R_{1} = 5Ω

. Current

I_{1} = V_{1} / R_{1} = 0.25

A. Using

E = V_{1} + I_{1} r

E = 1.25 + 0.25 r

. For the second case:

V_{2} = 1

R_{2} = 2Ω

I_{2} = 0.5

E = 1 + 0.5 r

. Equating:

1.25 + 0.25 r = 1 + 0.5 r

, giving

0.25 = 0.25 r

, so

r = 1Ω

. Therefore

E = 1.25 + 0.25 = 1.5

= \frac{15}{10}

V. Thus

x = 15

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Integer answer