Three independent parallel branches A, B and C draw 0.03, 0.09, 0.27 A respectively from a regulated 6 V battery when connected. After a fault, total current is 0 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 4 h, and estimate runtime from a usable battery capacity of 6 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: A and B and C; remaining power 0 W; charge in 4 h = 0 Ah; no finite discharge time in the stated zero-drain model. At a parallel junction, supply current is the sum of branch currents. The healthy value is 0.03 + 0.09 + 0.27 = 0.39 A, so the missing current is 0.39 A. Match that missing amount to the supplied branch currents. The combination A and B and C contributes exactly 0.39 A; the other combinations do not match under the stated model. The remaining circuit still receives 6 V, so its total power is VI = 6 × 0 = 0 W. Current conservation does not imply that every parallel branch draws equal current. Over 4 h the remaining branches use 0 × 4 = 0 Ah of charge. This is not a value in watts or watt-hours; it is charge expressed in the supplied battery unit. The ideal current is zero, so capacity/current has no finite discharge-time value. It would be misleading to report zero hours: the simplified battery does not discharge. Real self-discharge, voltage changes or standby loads would require a more detailed model.

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