Question

Three conductors draw currents of $1A,2A$ and $3A$ respectively, when connected in a 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. $\frac{6}{11}A$
• B. $\frac{3}{7}A$
• C. $\frac{4}{7}A$
• D. $\frac{5}{7}A$

Answer 1

The correct option is A $\frac{6}{11}A$

Let the conductors are connected individually across a battery of voltage $V$.

Thus resistance of 1st conductor $R_1=\frac{V}{1}$

Similarly, resistances of other conductors $R_2=\frac{V}{2}$ and $R_3=\frac{V}{3}$

Now these are connected in series.

Thus equivalent resistance of circuit $R_{eq}=R_1+R_2+R_3$

$\therefore$ $R_{eq}=\frac{V}{1}+\frac{V}{2}+\frac{V}{3}=\frac{11V}{6}$

Thus current drawn will be $I=\frac{V}{R_{eq}}$

$\therefore$ $I=\frac{V}{11V/6}=\frac{6}{11}$ A