question_answer

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)  \[\frac{5}{7}A\]

E)  None of the Above

Correct Answer: E

Solution :

: Let \[{{R}_{1}},{{R}_{2}}\] and \[{{R}_{3}}\] be resistances of three conductors respectively. As per question         \[{{R}_{1}}=\frac{V}{1}=V\] \[{{R}_{2}}=\frac{V}{2}=\frac{V}{2}\] \[{{R}_{3}}=\frac{V}{3}=\frac{V}{3}\] When they are connected in series, their equivalent resistance is \[{{R}_{eq}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}=V+\frac{V}{2}+\frac{V}{3}\] \[=V\left( 1+\frac{1}{2}+\frac{1}{3} \right)=\frac{11}{6}V\] When they are connected in series, across the same, battery, then the current in the circuit is \[I=\frac{V}{{{R}_{eq}}}=\frac{V}{\frac{11}{6}V}=\frac{6}{11}A\] * None of the given options is correct.