The network shown in the figure is a part of a circuit (the battery has negligible internal resistance). At a certain instant the current is I = 5 A and it is decreasing at the rate of 10³ A s⁻¹. What is the potential difference V_B - V_A between points B and A?

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(3) 15 V. Given: from A to B the branch is a 1 Ω resistor, a 15 V battery with its negative terminal towards A, and a 5 mH inductor; I = 5 A from A to B; (dI)/(dt) = -10³ A s⁻¹. Idea: walk from A to B adding up the changes in potential: V_A - IR + ε - L (dI)/(dt) = V_B. Resistor, along the current: a drop of IR = 5 × 1 = 5 V. Battery, from its negative to its positive terminal: a rise of 15 V. Inductor: L (dI)/(dt) = 5×10⁻³ × (-10³) = -5 V, so the drop across it is negative, a rise of 5 V: the inductor opposes the fall in current by pushing the current forward. V_B = V_A - 5 + 15 + 5, so V_B - V_A = 15 V. So the potential difference V_B - V_A is 15 V.

The network shown in the figure is a part of a circuit (the battery has negligible internal resistance). At a certain instant the current is I = 5 A and it is decreasing at the rate of 10³ A s⁻¹. What is the potential difference V_B - V_A between points B and A?

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