Current Electricity Part 2

Given, value of each resistance, R = 10 Ω Emf of battery, e = 20V Internal resistance of battery, r = 10 Ω Current in parallel connection is 20 times current in series combination, $i_{p}=20 i_{s}$. Net resistance in parallel combination will be given as $R_{p}=r+\left[\frac{1}{\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\ldots+n}\right]$ $=r+\frac{R}{n}$ $R_{p}=10+\frac{10}{n}$ (∵ r = 10 Ω and R = 10 Ω )...(i) In series combination, The net resistance of circuit will be equivalent to sum of all resistances as all are connected in series. $R_{s}=\left[R+R+\ldots+n\right]+r=nR+r$ $R_{s}=10n+10$ (∵ r = 10 Ω and R = 10 Ω)...(ii) By Ohm’s law, current flowing in circuit is given as $i=\frac{V}{R}$ As, $i_{p}=20i_{s}$ $\frac{V_{p}}{R_{p}}=20\frac{V_{s}}{R_{s}}$ $\Rightarrow \frac{20}{10+\frac{10}{n}}=\frac{20\times20}{10n+10}$ $[\because V_{p}=V_{s}=e=20\,\mathrm{V}]$ $\Rightarrow \frac{20n}{10n+10}=\frac{400}{10n+10}$ ⇒ 20n = 400 n = 20 Thus, the value of number of resistances n is 20.