Current Electricity Part 2

Given, number of cells, n = 5 Internal resistance of each cell, r = 1Ω Emf of each cell, e = 5 V When all cells are connected in parallel as shown below. Potential will remain same as, $V_{P}=5 \mathrm{V}$ Net resistance in parallel combination will be given as   $R_{p}=R+\frac{1}{5}$ (∵ r = 1 Ω) ...(i) When all cells are connected in series as shown below The net potential will increase as cells are connected in series, $V_{S}=5+5+5+5+5=25 \mathrm{V}$ The net resistance of circuit will be equivalent of sum of all resistances as all are connected in series. $R_{s}=r+r+r+r+r+R=5r+R$ $R_{s}=5+R$ ( ∵ r = 1Ω) ...(ii) By Ohm’s law, current flowing in circuit is given as $I=\frac{V}{R}$ As current in both series and parallel combination is same, $\frac{V_{p}}{R_{p}}=\frac{V_{s}}{R_{s}}$ $\Rightarrow \frac{5}{R+\frac{1}{5}}=\frac{25}{5+R}$ [From Eqs. (i) and (ii)] $\Rightarrow 25+5R=25R+5$ ⇒ R = 1 Ω