Question

Two sources of equal emf are connected in series. This combination is in turn 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 $r_1$ is zero, then R equals to CBSE 2022 (Term-I)

• A. $\frac{r_1 + r_2}{r_2 - r_1}$
• B. $r_2 - r_1$
• C. $\frac{r_1r_2}{r_2 - r_1}$
• D. $\frac{r_1 + r_2}{r_1r_2}$

Answer 1

Answer: (B)

$r_2 - r_1$

Answer 2

For the battery with internal resistance $r_1$, zero terminal potential means

$$E - I\,r_1 = 0 \quad\Longrightarrow\quad I = \frac{E}{r_1}.$$

Using KVL for the entire loop, the drop across the second cell and an external resistor $R$ satisfies:

$$\Bigl(E - I\,r_2\Bigr) + I\,R = 0 \quad\Longrightarrow\quad I(R - r_2) = -E.$$

Substitute $I=\tfrac{E}{r_1}$:

$$\frac{E}{r_1}(R - r_2) = -E \quad\Longrightarrow\quad R - r_2 = -r_1 \quad\Longrightarrow\quad R = r_2 - r_1.$$