Question
Two cells, having the same emf are connected in series through an external resistance $$R.$$ Cells have internal resistances $${r_1}$$ and $${r_2}\left( {{r_1} > {r_2}} \right)$$ respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of $$R$$ is
A.
$${r_1} - {r_2}$$
B.
$$\frac{{{r_1} + {r_2}}}{2}$$
C.
$$\frac{{{r_1} - {r_2}}}{2}$$
D.
$${r_1} + {r_2}$$
Answer :
$${r_1} - {r_2}$$
Solution :
Net resistance of the circuit $$ = {r_1} + {r_2} + R$$
Net emf in series $$ = E + E = 2E$$
Therefore, from Ohm’s law, current in the circuit
$$\eqalign{ & i = \frac{{{\text{Net emf}}}}{{{\text{Net resistance}}}} \cr & \Rightarrow i = \frac{{2E}}{{{r_1} + {r_2} + R}}\,.......\left( {\text{i}} \right) \cr} $$
It is given that, as circuit is closed, potential difference across the first cell is zero.
i.e., $$V = E - i{r_1} = 0$$
$$ \Rightarrow i = \frac{E}{{{r_1}}}\,......\left( {{\text{ii}}} \right)$$
Equating Eqs. (i) and (ii), we get
$$\eqalign{ & \frac{E}{{{r_1}}} = \frac{{2E}}{{{r_1} + {r_2} + R}} \Rightarrow 2{r_1} = {r_1} + {r_2} + R \cr & \therefore R = {\text{external resistance}} \cr & \Rightarrow R = {r_1} - {r_2} \cr} $$
NOTE
The question is wrong as the statement is when the circuit is closed, the potential difference across the first cell is zero which implies that in a series circuit, one part cannot conduct current which is wrong, Kirchhoff 's law is violated. The question must have been modified.
Net resistance of the circuit $$ = {r_1} + {r_2} + R$$
Net emf in series $$ = E + E = 2E$$
Therefore, from Ohm’s law, current in the circuit
$$\eqalign{ & i = \frac{{{\text{Net emf}}}}{{{\text{Net resistance}}}} \cr & \Rightarrow i = \frac{{2E}}{{{r_1} + {r_2} + R}}\,.......\left( {\text{i}} \right) \cr} $$
It is given that, as circuit is closed, potential difference across the first cell is zero.
i.e., $$V = E - i{r_1} = 0$$
$$ \Rightarrow i = \frac{E}{{{r_1}}}\,......\left( {{\text{ii}}} \right)$$
Equating Eqs. (i) and (ii), we get
$$\eqalign{ & \frac{E}{{{r_1}}} = \frac{{2E}}{{{r_1} + {r_2} + R}} \Rightarrow 2{r_1} = {r_1} + {r_2} + R \cr & \therefore R = {\text{external resistance}} \cr & \Rightarrow R = {r_1} - {r_2} \cr} $$
NOTE
The question is wrong as the statement is when the circuit is closed, the potential difference across the first cell is zero which implies that in a series circuit, one part cannot conduct current which is wrong, Kirchhoff 's law is violated. The question must have been modified.
