Practice portal › Current Electricity › Kirchhoff's Laws and Circuit Analysis
Asked in GSEB Board July 2016; GSEB Board March 2018 · Kirchhoff's laws in networks
Given (figure): from A to B the branch has a 0.25 Ω resistor, a 1 V cell with its positive (long) plate towards A, and another 0.25 Ω resistor. A current of 2 A flows from A to B.
Idea: walk from A to B and add up the changes in potential.
Across each resistor, going with the current, the potential falls by IR = 2×0.25 = 0.5 V.
Across the cell, going from its + plate to its - plate, the potential falls by 1 V.
V_A - 0.5 - 1 - 0.5 = V_B, so V_A - V_B = +2 V. Point A is 2 V above B.
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