Question

Three conductors draw currents of $1A,2A$ and $3A$ 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 $1A,2A$ and $3A$ respectively when connected in turn across a battery.
$\therefore V=R_1,V=2R_2\text{ and }V=3R_3$
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 $I=\frac{6}{11}A$