Rankers Physics

Combination of Batteries: Practice Problem & Solution

Two cells, having the same emf, are connected in series through an external resistance $R$. Cells have internal resistances $r_1$ and $r_2$ ($r_1 > r_2$) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of $R$ is: (2006)
$r_1 - r_2$
$(r_1 + r_2)/2$
$(r_1 - r_2)/2$
$r_1 + r_2$

Solution Explained:

To solve this problem, we apply the core principles of Combination of Batteries. Understanding the underlying formula is key to arriving at the correct answer below:

Total current $I = \frac{2E}{R + r_1 + r_2}$. The potential difference across the first cell is $V_1 = E - Ir_1 = 0$, meaning $E = Ir_1$. Substituting $I$, we get $E = \frac{2E r_1}{R + r_1 + r_2}$. Simplifying gives $R + r_1 + r_2 = 2r_1$, so $R = r_1 - r_2$.

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