Two sources of equal emf are connected in series. This combination is connected to an external resistance R. The internal resistance of two sources are $r_1$ and $r_2$ ($r_2>r_1$). If the potential difference across the source of internal resistance $r_2$ is zero, then R equals to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $R = r_2-r_1$
Let the EMFs of the sources be $E$ each, internal resistances $r_1$ and $r_2$, and external resistance $R$.
Current in series: $I = \frac{E + E}{R + r_1 + r_2} = \frac{2E}{R + r_1 + r_2}$
Potential difference across source with $r_2$:
$V_2 = E - I r_2$
Given $V_2 = 0$:
$0 = E - I r_2 \Rightarrow I r_2 = E \Rightarrow I = \frac{E}{r_2}$
Also, $I = \frac{2E}{R + r_1 + r_2}$
Equate currents:
$\frac{E}{r_2} = \frac{2E}{R + r_1 + r_2} \Rightarrow R + r_1 + r_2 = 2 r_2$
$R = 2 r_2 - r_1 - r_2 = r_2 - r_1$