In a part of a circuit, a current of 2 A flows from point A to point B. Between A and B there is a 3 Ω resistor followed by a cell of emf 10 V and internal resistance 0.5 Ω. The negative terminal of the cell faces A. The potential difference \( V_A-V_B \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(-3\ \mathrm{V}\)
Across the resistor, the potential falls by \( IR \) along the current. Inside the cell the current flows from the negative terminal N to the positive terminal P. So the potential difference across the cell is \( V_P-V_N=\varepsilon-Ir \). Point B is joined to P.
\( \begin{aligned} V_A-V_N &= IR \\ &= 2\times3 \\ &= 6\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} V_P-V_N &= \varepsilon-Ir \\ &= 10-2\times0.5 \\ &= 9\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} V_A-V_B &= (V_A-V_N)-(V_P-V_N) \\ &= 6-9 \\ &= -3\ \mathrm{V} \end{aligned} \)
From the above, \( V_A-V_B=-3\ \mathrm{V} \). Hence, point B is 3 V higher in potential than point A.
NCERT (2026-27):
Kirchhoff’s Rules – “… This will tell us the potential difference, V = V (P) – V (N) = ε – I r … between the positive terminal P and the negative terminal N; I here is the current flowing from N to P through the cell.”