Practice portal › Current Electricity › Kirchhoff's Laws and Circuit Analysis
Asked in GSEB Board March 2019 · Kirchhoff's laws in networks
Given (figure): from A to B the current I = 2 A passes through 0.5 Ω, then through a 1 V cell entered at its + plate, then through 0.5 Ω.
Idea: walk from A to B adding up potential changes. Across a resistor, in the direction of the current, the potential falls by IR; across a cell from + to - it falls by the emf.
V_A - 2(0.5) - 1 - 2(0.5) = V_B.
V_B - V_A = -1 - 1 - 1 = -3 V.
So V_B - V_A = -3 V: B is 3 V below A.
Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.
Practise this with a timer