Three independent parallel branches A, B and C draw 0.02, 0.06, 0.18 A respectively from a regulated 9 V battery when connected. After a fault, total current is 0.26 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 2 h, and estimate runtime from a usable battery capacity of 12 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: none; remaining power 2.34 W; charge in 2 h = 0.52 Ah; ideal runtime 46.1538 h. At a parallel junction, supply current is the sum of branch currents. The healthy value is 0.02 + 0.06 + 0.18 = 0.26 A, so the missing current is 0 A. Match that missing amount to the supplied branch currents. No branch current is missing; the other combinations do not match under the stated model. The remaining circuit still receives 9 V, so its total power is VI = 9 × 0.26 = 2.34 W. Current conservation does not imply that every parallel branch draws equal current. Over 2 h the remaining branches use 0.26 × 2 = 0.52 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: 12/0.26 = 46.1538 h. Real self-discharge, voltage changes or standby loads would require a more detailed model.

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