Question

A circuit consists of three batteries of emf $E_1=1V,E_2=2V$ and $E_3=3V$ and internal resistances $1\Omega ,2\Omega$ and $1\Omega$ respectively which are connected in parallel as shown in the figure. The potential difference between points $P$ and $Q$ is:

• A. $1.0V$
• B. $2.0V$
• C. $2.2V$
• D. $3.0V$

Answer 1

The correct option is B $2.0V$

Given : $E_1=1V$ $E_2=2V$ $E_3=3V$

$r_1=1\Omega$ $r_2=2\Omega$ $r_3=1\Omega$

Equivalent resistance $\frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}$

$\frac{1}{r_{eq}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{1}$ $⟹r_{eq}=0.4\Omega$

Equivalent potential $E_{eq}=[\frac{E_1}{r_1}+\frac{E_2}{r_2}+\frac{E_3}{r_3}]r_{eq}$

$E_{eq}=[\frac{1}{1}+\frac{2}{2}+\frac{3}{1}]\times 0.4$ $⟹E_{eq}=2.0V$