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

The Correct Option is B

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: \(r_{eq}=32Ω\),  \(R=10Ω\)

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

\(⇒E_{eq}=\frac{37}{3}V\)

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

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