The network shown in the figure is a part of a circuit. (The battery has negligible resistance.) At a certain instant the current is I=2 A and it is decreasing at the rate of 10² A s⁻¹. What is the potential difference V_B-V_A between the points B and A?

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(1) 8.5 V. Given: from A to B the network has R=2 Ω, a 12 V cell with its negative terminal towards A, and L=5 mH; I=2 A flows from A to B and (dI)/(dt)=-10² A s⁻¹. Idea: walk from A to B and add up every rise and fall of potential. Resistor: a fall of IR=2×2=4 V in the direction of the current. Cell: entering at - and leaving at + is a rise of 12 V. Inductor: the falling current induces an emf that tries to keep it flowing, so the potential rises along the current by L|(dI)/(dt)|=5×10⁻³×10²=0.5 V. V_B-V_A=-4+12+0.5. So V_B-V_A=8.5 V.

The network shown in the figure is a part of a circuit. (The battery has negligible resistance.) At a certain instant the current is I=2 A and it is decreasing at the rate of 10² A s⁻¹. What is the potential difference V_B-V_A between the points B and A?

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