Three conductors draw currents of $1 \mathrm{~A}, 2 \mathrm{~A}$ 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

  $\frac{2}{7} \mathrm{~A}$
• Correct answer
  B

  $\frac{3}{7} \mathrm{~A}$
• C

  $\frac{4}{7} \mathrm{~A}$
• D

  $\frac{5}{7} \mathrm{~A}$

Solution

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

Hence, the correct answer is option: 'B'.