In the circuit shown, given resistance R1 = 4 Ω, R2 = 8 Ω, R3 = 6 Ω, R4 = 12 Ω, voltage US1 = 1.2V, US2 = 3V, use superposition theorem to calculate current I

In the circuit shown, given resistance R1 = 4 Ω, R2 = 8 Ω, R3 = 6 Ω, R4 = 12 Ω, voltage US1 = 1.2V, US2 = 3V, use superposition theorem to calculate current I

1. When US1 acts alone, the current I1 = - US1 ÷ (R1 / / r2 + R3 / / R4) = - 1.2 △ 4 × 8 ^ (4 + 8) + 6 × 12 ^ (6 + 12)} = - 1.2 ^ (2.67 + 4) = - 0.18a2, when US2 acts alone, the current is2 = US2 ^ (R2 / / R4 + R1 / / R3) = 3 ^ ^ {8 × 12 ^ (8 + 12) + 4 × 6 ^ (4

Given the resistance R1 = 6 Ω, R2 = 3 Ω, R3 = 1 Ω, R4 = 2 Ω, the power supply e = 3V, r = 0, find the voltage between a and B

I guess it's bridge connection?
If so, R1: R2 = R4: R3 = 2
The two points AB are equipotential, and the voltage is zero

If the indication of power supply voltage and voltmeter is 3V and 1V respectively, the voltage applied to both ends of resistance R1 and R2 is () V and () V respectively

Why don't you have a circuit diagram here. 1: if the voltmeter measures the voltage of R1, it doesn't consider the internal resistance of the power supply. If the indication of the power supply voltage and the voltmeter is 3V and 1V respectively, then the voltage applied to both ends of R1 and R2 is (1) V and (2) v. 2: if the voltmeter measures the voltage of R2, it doesn't consider the internal resistance of the power supply