Question

A cell $E_1$ of emf 6 V and internal resistance 2 Ω is connected with another cell $E_2$ of emf 4 V and internal resistance 8 Ω (as shown in the figure). The potential difference across points X and Y is : 

• A. 2.0 V
• B. 3.6 V
• C. 5.6 V
• D. 10.0 V

Hint:

When two cells are in opposition, the terminal voltage of the discharging cell (larger emf) is $E - Ir$, while the charging cell (smaller emf) is $E + Ir$.

Answer 1

The Correct Option is C

Step 1: The cells are in a single loop. Total Emf $E_{net} = 6 - 4 = 2$ V.
Step 2: Total resistance $R_{total} = 2 + 8 = 10$ Ω.
Step 3: Current $I = \frac{2}{10} = 0.2$ A (flowing clockwise/out of 6V cell).
Step 4: Potential difference across $X$ and $Y$ (terminal voltage of $E_1$): $V = E_1 - Ir_1 = 6 - (0.2)(2) = 5.6$ V.
Step 5: Alternatively, across $E_2$: $V = E_2 + Ir_2 = 4 + (0.2)(8) = 5.6$ V.