Question

Two cells of emf ${E_1}$​ and ${E_2}$​ are joined in opposition (such that ${E_1} > {E_2}$ ​). If ${r_1}$​ and ${r_2}$​ be the internal resistance and $R$ be the external resistance, then the terminal potential difference is:
A) $\dfrac{{{E_1} + {E_2}}}{{{r_1} + {r_2}}} \times R$
B) $\dfrac{{{E_1} + {E_2}}}{{{r_1} + {r_2} + R}} \times R$
C) $\dfrac{{{E_1} - {E_2}}}{{{r_1} + {r_2}}} \times R$
D) $\dfrac{{{E_1} - {E_2}}}{{{r_1} + {r_2} + R}} \times R$

Answer 1

We solve this problem by finding the equivalent emf and equivalent resistance of the circuit. Then use ohm's law to find current flowing in the circuit. Using ohm's law again with the external resistance we find the terminal potential difference.

Formula used:
Ohms law $V = iR$
Emf or voltage is represented by $V$
Current is represented by $i$
Resistance is represented by $R$

Complete step by step solution:
The terminals of the two cells are in opposition hence the equivalent emf is the difference of both the emfs.
${E_{eq}} = {E_1} - {E_2}$
${E_1} > {E_2}$Hence ${E_2}$is subtracted from${E_1}$. Emf cannot be negative.
Here the emf of cell 1 and cell 2 are represented by${E_1}$, ${E_2}$ respectively.
Equivalent emf is represented by ${E_{eq}}$
All the internal and external resistances in the circuit are connected in series so the equivalent resistance in series is the sum of all the resistances
${R_{eq}} = {r_1} + {r_2} + R$
Internal resistances are represented by ${r_1}$​ and ${r_2}$
External resistance is represented by $R$
Equivalent resistance is represented by ${R_{eq}}$
Using ohm's law, we find the current flowing in the circuit
${E_{eq}} = i{R_{eq}}$
$i = \dfrac{{{E_{eq}}}}{{{R_{eq}}}} = \dfrac{{{E_1} - {E_2}}}{{{r_1} + {r_2} + R}}$
The terminal potential difference is the difference in potential between the two ends of the terminals of the two cells.

We find this by using ohm's law again but taking only external resistance. Because only external resistance is present between the two terminals when current flows from one terminal end to the other.
$V = iR$
$V = \dfrac{{{E_1} - {E_2}}}{{{r_1} + {r_2} + R}} \times R$
The terminal potential difference is $V = \dfrac{{{E_1} - {E_2}}}{{{r_1} + {r_2} + R}} \times R.$

Hence option (D), $V = \dfrac{{{E_1} - {E_2}}}{{{r_1} + {r_2} + R}} \times R$ is the correct answer.

Note: While finding the current in the circuit we also consider the internal resistances for finding equivalent resistance this is because the current flows through the cells and the circuit. Whereas when we find the potential difference between terminals we take only external resistance as the path between the two terminals of both the cells together has only external resistance.