Question: 11. AB is a part of an electrical circuit (see figure). The potential difference “V_{A}-V_{B}“, at the instant when current i=2\,\mathrm{A} and is increasing at a rate of 1 amp/second is:
(1) 5 volt
(2) 6 volt
(3) 9 volt
(4) 10 volt
Answer: Option (4)
Explanation:
Along the path from A to B, the elements are an inductor of 1 H, a 5 V battery, and a 2 Ω resistor.
1. Potential across the inductor: The induced emf across an inductor is
V_L = L\,\dfrac{di}{dt}.Given L = 1\,\mathrm{H} and \dfrac{di}{dt} = 1\,\mathrm{A/s}:
V_L = 1 \times 1 = 1\,\mathrm{V}.2. **Potential across the resistor:**
V_R = iR = 2 \times 2 = 4\,\mathrm{V}.3. **Battery:** Moving from its negative to positive terminal, it raises potential by
5\,\mathrm{V}.4. **Total potential difference:** Going from A → inductor → battery → resistor → B,
V_A - V_B = V_L + 5 + V_R.Substitute values:
V_A - V_B = 1 + 5 + 4 = 10\,\mathrm{V}.Thus, the potential difference is 10\,\mathrm{V}.