First slide

Ohm's law

Question

Two cells, having the same emf, are connected in series through an external resistance R. Cells have internal resistances r₁ and r₂ (r₁ > r₂), respectively. Which the circuit is closed, the potential difference across the first cell is zero. The value of R is

Easy

A

$r_{1} - r_{2}$
B

$\frac{r_{1} + r_{2}}{2}$
C

$\frac{r_{1} - r_{2}}{2}$
D

$r_{1} + r_{2}$

Solution

Current in the circuit is given by Ohm's law.
Net resistance of the circuit = r₁ + r₂ + R
Net emf in series - E + E = 2E
Therefore, from Ohm's law, current in the circuit
$i = \frac{Net emf}{Net resistance}$
$\Rightarrow i = \frac{2 E}{r_{1} + r_{2} + R}$ …..(i)
It is given that, as circuit is closed, potential difference across the first cell is zero. That is
$\begin{array}{l} V = E - {ir}_{1} = 0 \\ \Rightarrow i = \frac{E}{r_{1}} \end{array}$ ……(ii)
Equating equations (i) and (ii), we get $\frac{E}{r_{1}} = \frac{2 E}{r_{1} + r_{2} + R}$
$\begin{array}{l} 2 r_{1} = r_{1} + r_{2} + R \\ R = external resistance = r_{1} - r_{2} \end{array}$

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