JEE PYQ: Current Electricity Question 7
Question 7 - 2021 (17 Mar 2021 Shift 2)
Two cells of emf $2E$ and $E$ with internal resistance $r_1$ and $r_2$ respectively are connected in series to an external resistor R (see figure). The value of R, at which the potential difference across the terminals of the first cell becomes zero is
(1) $r_1 + r_2$
(2) $\frac{r_1}{2} - r_2$
(3) $\frac{r_1}{2} + r_2$
(4) $r_1 - r_2$
Show Answer
Answer: (2)
Solution
$i = \frac{3E}{R + r_1 + r_2}$. TPD across first cell $= 2E - ir_1 = 0$. So $2E = \frac{3E \times r_1}{R + r_1 + r_2}$. Solving: $2R + 2r_1 + 2r_2 = 3r_1$, $R = \frac{r_1}{2} - r_2$.
BETA
