Question

two cells having emf $ 4\,V, $ $ 2\,V $ 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. $ 1\,A $
• 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'}=\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 $