Question

two cells having emf $4V,$ $2V$ and internal resistances $1\Omega ,$ $1\Omega$ are connected as shown in figure below. Current through $6\Omega$ resistance is

• A. $\frac{1}{3}A$
• B. $\frac{2}{3}A$
• C. $1A$
• D. $\frac{2}{9}A$

Answer 1

Answer: (A)

The emf of the circuit is $E=E_1+E_2$
$=4V+2V=6V$
In the given circuit, $3\Omega$ . and $6\Omega$ are connected in parallel,
hence equivalent resistance is
$\frac{1}{R^{\prime }}=\frac{1}{3}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$
$\Rightarrow$ $i=\frac{V}{R}=\frac{6}{6}=1A$
Total resistance of circuit is
$R=1\Omega +1\Omega +2\Omega +2\Omega =6\Omega$
From Ohm's law $V=iR$
$\Rightarrow$ $i=\frac{V}{R}=\frac{6}{6}=1A$
The $3\Omega$ and $6\Omega$ resistors are in parallel,
hence $i_1{R}_1=i_2{R}_2=V$
$\therefore$ $i_1\times 3=i_2\times 6$
$\Rightarrow$ $i_1=2i_2$
and $i_1+i_2=1$
$2i_2+i_2=1$
$3i_2=1$
$\Rightarrow$ $i_2=\frac{1}{3}A$