Three independent parallel branches A, B and C draw 0.04, 0.08, 0.16 A respectively from a regulated 5 V battery when connected. After a fault, total current is 0.2 A. Assume faults can only open whole branches; surviving branch currents and source voltage do not change. Identify the disconnected branch combination, find the remaining total power and charge used in 5 h, and estimate runtime from a usable battery capacity of 10 Ah. Use P = VI and charge in Ah = current in A × time in h. Ignore standby drain and battery ageing.

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Disconnected: B; remaining power 1 W; charge in 5 h = 1 Ah; ideal runtime 50 h. At a parallel junction, supply current is the sum of branch currents. The healthy value is 0.04 + 0.08 + 0.16 = 0.28 A, so the missing current is 0.08 A. Match that missing amount to the supplied branch currents. The combination B contributes exactly 0.08 A; the other combinations do not match under the stated model. The remaining circuit still receives 5 V, so its total power is VI = 5 × 0.2 = 1 W. Current conservation does not imply that every parallel branch draws equal current. Over 5 h the remaining branches use 0.2 × 5 = 1 Ah of charge. This is not a value in watts or watt-hours; it is charge expressed in the supplied battery unit. Ideal runtime is capacity divided by current: 10/0.2 = 50 h. Real self-discharge, voltage changes or standby loads would require a more detailed model.

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