Question

Two cells having the same emf are connected in series through an external resistance $R$. Cells have internal resistances ${r_1}$​ and ${r_2}$ such that ${r_1} > {r_2}$ respectively. When the circuit is closed, the potential difference across the first cell is zero, then the value of $R$ is
A. $({r_1} + {r_2})$
B. $({r_1} - {r_2})$
C. $\dfrac{{({r_1} + {r_2})}}{2}$
D. $\dfrac{{({r_1} - {r_2})}}{2}$

Answer 1

Hint:To solve this question, we just need complete understanding of Ohm’s law, firstly, we will find out the complete net resistance and e.m.f and then using ohm's law and then comparing the values after the circuit is closed, we will get the answer.

Formula used:
\[V = I \times R\]
Where, $V$ is the voltage across the circuit, $I$ is the current and $R$ is the net resistance.

Complete step by step answer:
According to the question,
Net resistance of the circuit is ${r_1} + {r_2} + R$
Net e.m.f in series is $E + E = 2E$
Therefore, from Ohm’s law, current in the circuit is
$I = \dfrac{{{\text{Net e}}{\text{.m}}{\text{.f}}}}{{{\text{Net resistance}}}}$
$ \Rightarrow I = \dfrac{{2E}}{{{r_1} + {r_2} + R}}$ -----(1)
Now, it is given that, as the circuit is closed, the potential difference across the cell is zero.
i.e., $V = E - I \times {r_1} = 0$
$ \Rightarrow I = \dfrac{E}{{{r_1}}}$ -----(2)
Now comparing both the equation, we get
$\dfrac{E}{{{r_1}}} = \dfrac{{2E}}{{{r_1} + {r_2} + R}} \\
\Rightarrow 2{r_1} = {r_1} + {r_2} + R \\
\therefore R = {r_1} - {r_2} \\ $
Hence the correct option is B.

Note:Electromotive force is defined as the electric potential produced by either electrochemical cell or by changing the magnetic field. EMF is the commonly used acronym for electromotive force. A generator or a battery is used for the conversion of energy from one form to another. In these devices, one terminal becomes positively charged while the other becomes negatively charged. Therefore, an electromotive force is a work done on a unit electric charge.