Two batteries of different e.m.f.’s and internal resistance connected in series with each other and with an external load resistor. The current is 2.0 amp. When the polarity of one battery is reversed, the current becomes 1.0 amp. The ratio of the e.m.f.’s of the two batteries is

Question

Two batteries of different e.m.f.’s and internal resistance connected in series with each other and with an external load resistor. The current is 2.0 amp. When the polarity of one battery is reversed, the current becomes 1.0 amp. The ratio of the e.m.f.’s of the two batteries is (A) 2.5 (B) 2.0 (C) 3.0 (D) 1.0

Tardigrade Admin · Accepted Answer

(ε1+ε2/ r 1+ r 2+ R )=2 and (ε1-ε2/ r 1+ r 2+ R )=1 ⇒ (ε1+ε2/ε1-ε2)=(2/1) or (ε1/ε2)=(3/1)

Q. Two batteries of different e.m.f.’s and internal resistance connected in series with each other and with an external load resistor. The current is 2.0 amp. When the polarity of one battery is reversed, the current becomes 1.0 amp. The ratio of the e.m.f.’s of the two batteries is

2.5

2.0

3.0

1.0

$\frac{ε _{1} + ε _{2}}{r _{1} + r _{2} + R} = 2$ and
$\frac{ε _{1} - ε _{2}}{r _{1} + r _{2} + R} = 1$
$\Rightarrow \frac{ε _{1} + ε _{2}}{ε _{1} - ε _{2}} = \frac{2}{1}$ or $\frac{ε _{1}}{ε _{2}} = \frac{3}{1}$

Q. Two batteries of different e.m.f.’s and internal resistance connected in series with each other and with an external load resistor. The current is 2.0 amp. When the polarity of one battery is reversed, the current becomes 1.0 amp. The ratio of the e.m.f.’s of the two batteries is

2.5

2.0

3.0

1.0

r1​+r2​+Rε1​+ε2​​=2 and r1​+r2​+Rε1​−ε2​​=1 ⇒ε1​−ε2​ε1​+ε2​​=12​ or ε2​ε1​​=13​

$\frac{ε _{1} + ε _{2}}{r _{1} + r _{2} + R} = 2$ and
$\frac{ε _{1} - ε _{2}}{r _{1} + r _{2} + R} = 1$
$\Rightarrow \frac{ε _{1} + ε _{2}}{ε _{1} - ε _{2}} = \frac{2}{1}$ or $\frac{ε _{1}}{ε _{2}} = \frac{3}{1}$