JEE PYQ: Current Electricity Question 13
Question 13 - 2021 (24 Feb 2021 Shift 1)
A cell $E_1$ of emf 6 V and internal resistance $2,\Omega$ is connected with another cell $E_2$ of emf 4 V and internal resistance $8,\Omega$ (as shown in the figure). The potential difference across points X and Y is -
(1) 3.6 V
(2) 10.0 V
(3) 5.6 V
(4) 2.0 V
Show Answer
Answer: (3)
Solution
$E_{\text{eff}} = 6 - 4 = 2$ V, $R_{\text{eq}} = 2 + 8 = 10,\Omega$. $I = \frac{2}{10} = 0.2$ A. $|V_x - V_y| = E + iR = 4 + 0.2 \times 8 = 5.6$ V.
BETA
