Question

A set of $n$ equal Resistances of value R each are connected in series to a battery of emf $E$, and internal Resistance $R$, then the current drawn is $I$. Now when the $n$ resistors are connected in parallel to same battery, then current drawn is $I^{\prime }$, Find the relation between $I$ and $I^{\prime }$ ?

• A. $I=I^{\prime }$
• B. $I=n{I}^{\prime }$
• C. $I=n^2{I}^{\prime }$
• D. $I^{\prime }=n^{2}I$

Answer 1

The correct option is B $I=n{I}^{\prime }$


$R_{series}=nR+R=(n+1)R$

$I=\frac{v}{R(n+1)}$

$R_{parllel}=R+\frac{R}{n}=\left(\frac{(n+1)R}{n}\right)$

$I^{\prime }=\frac{V.n}{(n+1)R}$

$\frac{I}{I^{\prime }}=\frac{1}{n}\Rightarrow I^{\prime }=nI$