Three conductors draw respectively currents of 1A, 2A and 4A when connected separately across a battery. If they are connected in series across the same battery, the current drawn will be

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$\frac{2}{7} A$

$\frac{5}{7} A$

$\frac{3}{7} A$

$\frac{4}{7} A$

answer is C.

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Detailed Solution

Let the potential of the battery be V
$\begin{array}{l} R_{1} = \frac{V}{1} \\ R_{2} = \frac{V}{2} \\ R_{3} = \frac{V}{4} \end{array}$
On connecting in series
$\begin{array}{l} R_{S} = R_{1} + R_{2} + R_{3} \\ \Rightarrow R_{S} = V (1 + \frac{1}{2} + \frac{1}{4}) \\ \Rightarrow R_{S} = \frac{7}{4} V \\ S i n c e, b y d e f i n i t i o n \\ I = \frac{V}{R_{S}} \\ \Rightarrow I = \frac{4}{7} A \end{array}$

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