Question

Two batteries of different e.m.f. and internal
resistance are connected in series with each
other and with an external load resistor. The
current is 3.0 amp. When the polarity of one
battery is reversed, the current becomes 1.0
amp. The ratio of the e.m.f. of the two batteries
is -
(1) 2.5 (2) 2.0 (3) 1.5 (4) 1.0​

Answer 1

Given that,

The  current is 3 A. When the polarity of one  battery is reversed, the current becomes 1.

To find,

The ratio of the e.m.f. of the two batteries when they are connected in series with each  other and with an external load resistor and when the  polarity of one  battery is reversed.

Solution,

Current flowing on first condition, $I_1=\dfrac{E_1+E_2}{R}$

Current flowing on second condition, $I_2=\dfrac{E_1-E_2}{R}$

Dividing current 1 by 2. So,

$\dfrac{I_1}{I_2}=\dfrac{E_1+E_2}{E_1-E_2}\\\\\dfrac{I_1}{I_2}=\dfrac{3+1}{3-1}\\\\\dfrac{I_1}{I_2}=\dfrac{4}{2}\\\\\dfrac{I_1}{I_2}=2$

So, the ratio of the e.m.f. of the two batteries  is 2. So, the correct option is (2).