As shown in the figure, in the circuit, the power supply voltage is constant, R1 = 6 Ω, R2 = 4 Ω. When S1 and S2 are closed, the indication of ammeter A1 is 1a, and the indication of ammeter A is 2.5A. Calculate: (1) the resistance value of resistance R3 and the indication of voltmeter. (2) when S1 and S2 are both open, the indication of ammeter A1, a and voltmeter v

As shown in the figure, in the circuit, the power supply voltage is constant, R1 = 6 Ω, R2 = 4 Ω. When S1 and S2 are closed, the indication of ammeter A1 is 1a, and the indication of ammeter A is 2.5A. Calculate: (1) the resistance value of resistance R3 and the indication of voltmeter. (2) when S1 and S2 are both open, the indication of ammeter A1, a and voltmeter v

Given: R1 = 6 Ω, R2 = 4 Ω, I = 2.5A, I1 = 1a, find: R3, u, I ′, I ″, u ′ (1) u = i1r1, R1 = 1a × 6 Ω = 6V; R3 = UI − I1 = 6v2.5a − 1A = 4 Ω; because the voltmeter is short circuited, the voltage representation is zero. (2) when S1 and S2 are disconnected, RL and R2 are in series, and the current in the circuit is zero I ′ = ur1 + R2 = 6v6 Ω + 4 Ω = 0.6A; therefore, the circuit is a series circuit, and both ammeters measure the current in the series circuit, so the indication of ammeter is 0.6A; the indication of voltage is u ′ = I ′ R2 = 0.6A × 4 Ω = 2.4V

As shown in the figure, the power supply voltage remains unchanged and the filament resistance re remains unchanged

When S1 and S2 are closed, R1 is connected in parallel with RE, the current represents the total current, u = 0.9x [rexr1 / (re + R1)] when S1 and S2 are disconnected, R2 is connected in series with RE, the voltage represents the voltage on R2, uxr2 / (re + R2) = 4 (where u is the power supply voltage), R1 = R2 = 20 Ω is substituted, the solution is: RE1 = 40 Ω (different from the condition Re < R1)

R1 = 60 ohm, R2 = 20 ohm, R3 = 30 ohm, the power supply voltage remains unchanged, when S1 and S2 are closed, the ammeter is 0.6 a
(2) What is the current in R2? (3) when S1 and S2 are disconnected, what is the work done by the current passing through R1 within 1 minute?

There are too many unknown factors. Whether R1, R2 and R3 are connected in parallel or in series, and which switches S1 and S2 control are unknown

In the circuit shown in the figure, R2 = 40 Ω, the power supply voltage remains unchanged, only close the switch s, the ammeter indication is 0.2A, and then close the switch S1, the ammeter increases to 1a
The resistance of R1 is Ω and the power supply voltage is v

Answer: 10 Ω, 8V analysis: the circuit is parallel: u = IR2 = 0.2A × 40 Ω = 8V. After S1 is closed, the current increases to 1a, indicating that the current of R1 is: I1 = 1a-0.2a = 0.8A, resistance value: R1 = u / I1 = 8V / 0.8A = 10 Ω