Question

Two batteries with e.m.f. $12\, V$ and $13\, V$ are connected in parallel across a load resistor of $10 \, \Omega$. The internal resistances of the two batteries are $1 \, \Omega$ and $2 \, \Omega$ respectively. The voltage across the load lies between :

• A. 11.6 V and 11.7 V
• B. 11.5 V and 11.6 V
• C. 11.4 V and 11.5 V
• D. 11.7 V and 11.8 V

Answer 1

Answer: (B)

11.5 V and 11.6 V

Answer 2

Applying $KVL$ in loops,

$12−x−10(x+y)=0$

$⇒12=11x+10y$ …….(i)

$13=10x+12y$ …….. (ii)

Solving $x=\frac{7}{15}A, y=\frac{23}{32}A$

$V=10(x+y)=11.56V$

Aliter: $r_{eq}=32Ω$, $R=10Ω$

$\frac{E_{eq}}{r_{eq}}=\frac{E_1}{r_1} + \frac{E_2}{r_2}$

$⇒E_{eq}=\frac{37}{3}V$

$V=\frac{E_{eq}}{R+r_{eq}} R=11.56V$

$\text{The Correct Option is (B):} 11.5 \;V \;and \;11.6 V$