Question

Two sources of current of equal e.m.f. are connected in series and have different internal resistances $r_1$ and $r_2$. An external resistance $R$ is connected in series with these sources of e.m.f

• A. R = (r_1 - r_2), for terminal potential difference to be zero across source 1
• B. R = \sqrt{r_1r_2}, for terminal potential difference to be zero across both sources
• C. R = (r_2 - r_1), for terminal potential difference to be zero across source 2
• D. data insufficient to arrive at a conclusion

Answer 1

Answer: (A), (C)

R = (r_1 - r_2), for terminal potential difference to be zero across source 1 R = (r_2 - r_1), for terminal potential difference to be zero across source 2

Answer 2

The circuit consists of two sources of equal e.m.f. ($E$) connected in series, along with an external resistance $R$. The internal resistances are $r_1$ and $r_2$.

1. Calculate the total e.m.f. and total resistance: When two identical e.m.f. sources are connected in series (aiding each other), the total e.m.f. is $E_{total} = E + E = 2E$. The total internal resistance is $r_{total} = r_1 + r_2$. The total resistance in the circuit is $R_{circuit} = R + r_1 + r_2$.

2. Calculate the current flowing in the circuit: Using Ohm's law for the entire circuit: $I = \frac{E_{total}}{R_{circuit}} = \frac{2E}{R + r_1 + r_2}$

3. Analyze the terminal potential difference across a source: For a source with e.m.f. $E$ and internal resistance $r$, when it is discharging (current flows out of its positive terminal), the terminal potential difference $V$ is given by $V = E - Ir$.

   Let's evaluate each option:

   • Option 1: $R = (r_1 - r_2)$, for terminal potential difference to be zero across source 1 If the terminal potential difference across source 1 ($V_1$) is zero: $V_1 = E - I r_1 = 0$ This implies $E = I r_1$. Substitute the expression for $I$: $E = \left(\frac{2E}{R + r_1 + r_2}\right) r_1$ Since $E \neq 0$, we can cancel $E$ from both sides: $1 = \frac{2 r_1}{R + r_1 + r_2}$ $R + r_1 + r_2 = 2 r_1$ $R = 2 r_1 - r_1 - r_2$ $R = r_1 - r_2$ This statement is a correct derivation. For $R$ to be a physically positive external resistance, this condition requires $r_1 > r_2$.

   • Option 2: $R = \sqrt{r_1r_2}$, for terminal potential difference to be zero across both sources If the terminal potential difference is zero across both sources, then: $V_1 = E - I r_1 = 0 \implies E = I r_1$ $V_2 = E - I r_2 = 0 \implies E = I r_2$ For both conditions to hold simultaneously, we must have $I r_1 = I r_2$. Since $I \neq 0$ (as $E \neq 0$), this implies $r_1 = r_2$. However, the problem states that $r_1$ and $r_2$ are different. Therefore, it is impossible for the terminal potential difference to be zero across both sources simultaneously. This option is incorrect.

   • Option 3: $R = (r_2 - r_1)$, for terminal potential difference to be zero across source 2 If the terminal potential difference across source 2 ($V_2$) is zero: $V_2 = E - I r_2 = 0$ This implies $E = I r_2$. Substitute the expression for $I$: $E = \left(\frac{2E}{R + r_1 + r_2}\right) r_2$ Since $E \neq 0$, we can cancel $E$ from both sides: $1 = \frac{2 r_2}{R + r_1 + r_2}$ $R + r_1 + r_2 = 2 r_2$ $R = 2 r_2 - r_1 - r_2$ $R = r_2 - r_1$ This statement is a correct derivation. For $R$ to be a physically positive external resistance, this condition requires $r_2 > r_1$.

   • Option 4: data insufficient to arrive at a conclusion We have arrived at conclusions for options 1, 2, and 3. So, this option is incorrect.

Conclusion: Given that $r_1$ and $r_2$ are different, either $r_1 > r_2$ or $r_2 > r_1$. If $r_1 > r_2$, then $R = r_1 - r_2$ is a positive resistance that makes $V_1=0$. If $r_2 > r_1$, then $R = r_2 - r_1$ is a positive resistance that makes $V_2=0$. Both options 1 and 3 represent valid physical scenarios that can be achieved depending on the relative values of $r_1$ and $r_2$. Since the question asks to identify correct statements and uses checkboxes, it implies that multiple statements can be correct. Both statements 1 and 3 are correct derivations for the respective conditions.