Two fixed resistors, 12 Ω and 24 Ω, are connected in parallel across an ideal 6 V supply. (a) Calculate each branch current, the total current and the combined resistance. (b) The 12 Ω branch becomes an open circuit. State how the current in the other branch and the supply current change. Assume resistor temperatures do not change.

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Branch currents 0.5 A and 0.25 A; total 0.75 A; combined resistance 8 Ω; surviving branch remains 0.25 A and total becomes 0.25 A. Parallel branches share the full supply potential difference. Apply I = V/R separately: 6/12 = 0.5 A and 6/24 = 0.25 A. The voltage is not divided between branches. Charge conservation at the junction requires the supply current to equal the sum of branch currents. It is therefore 0.5 + 0.25 = 0.75 A before the fault occurs. The combined resistance is R = V/I = 6/0.75 = 8 Ω. This is smaller than either branch resistance, as expected when parallel paths are available. An open branch carries no current. The ideal supply maintains the same voltage across the remaining resistor, so its current stays 0.25 A, while the total supply current falls to 0.25 A. This independence relies on the stated ideal-source assumption.

Parallel branches and a broken connection 1

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