Question

Three conductors draw currents of $1\, A, 2\,A$ and $3\,A$ respectively, when connected in turn across a battery. If they are connected in series and the combination is connected across the same battery, the current drawn will be

• A. $\frac{2}{7} \,A$
• B. $\frac{3}{7}A$
• C. $\frac{4}{7}A$
• D. $\frac{6}{11} \,A$

Answer 1

Answer: (D)

Let the three conductors are $R_{1}, R_{2}$ and $R_{3}$ and the current drawn by them $1 A , 2 A$ and $3 A$ respectively when connected in turn across a battery.
$\therefore V=R_{1}, \quad V=2 R_{2} \quad \text { and } \quad V=3 R_{3}$
So, $R_{1}=V_{1} \cdot R_{2}=\frac{V}{2} \text { and } R_{3}=\frac{V}{3}$
When the conductors are connected in series, across the same battery, then
$V=I\left[R_{1}+R_{2}+R_{3}\right]$
$\Rightarrow V=1\left[V+\frac{V}{2}+\frac{V}{3}\right] $
$\Rightarrow 1=1\left[\frac{6+3+2}{6}\right]$
or $ I=\frac{6}{11} A $