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Question

Physics Question on cells

When a current of 2A2\,A flows in a battery from negative to positive terminal, the potential difference across it is 12V12 \,V. If a current of 3A3\,A flowing in the opposite direction produces a potential difference of 15V15\, V, the emf of the battery is

A

12.6V12.6\,V

B

13.2V13.2\,V

C

13.5V13.5\,V

D

14.0V14.0\,V

Answer

13.2V13.2\,V

Explanation

Solution

Let ε\varepsilon be emf and rr be internal resistance of the battery. In first case, 12=ε2r...(i)12 = \varepsilon - 2r \quad ...(i) In second case, 15=ε+3r...(ii)15 = \varepsilon + 3r \quad...(ii) Subtract (i)(i) from (ii)(ii), we get r=35Ωr = \frac{3}{5} \Omega Putting this value of rr in eqn. (i)(i), we get ε=12+2×35\varepsilon = 12 + \frac{2\times 3}{5} =60+65 = \frac{60+6}{5} =665=13.2V= \frac{66}{5} = 13.2\,V