Han Yong designed the circuit as shown in the figure to measure the resistance value of resistance R1. When he contacts the switch with a and B respectively, the ratio of ammeter indication is 5:2. Given that R2 = 6 Ω, what is the resistance value of R1?

Han Yong designed the circuit as shown in the figure to measure the resistance value of resistance R1. When he contacts the switch with a and B respectively, the ratio of ammeter indication is 5:2. Given that R2 = 6 Ω, what is the resistance value of R1?

If the power supply voltage is u, then when s is connected to a, IA = ur1 ① When s is connected to B, IB = ur1 + R2 ② Compared with 2, we can get: iaib = R1 + r2r1, substituting the data, we can get: 52 = R1 + 6 Ω, R1 solution: R1 = 4 Ω answer: the resistance of R1 is 4 Ω

In the circuit as shown in the figure, the power supply voltage is 3 V and remains unchanged, and the resistance value of setting resistor R1 is 30 Ω. When the key s is closed, the indication of ammeter A1 is 0.4 a (1) (2) the resistance value of the sliding rheostat R2 connected into the circuit. (3) when the position of the slide P is changed properly so that the change of the number of the ammeter A1 is just equal to the number of the ammeter A2, the resistance value of the R2 connected into the circuit is calculated

It can be seen from the circuit diagram that the setting resistance R1 is connected in parallel with the sliding rheostat R2, the ammeter A1 measures the main circuit current, and the ammeter A2 measures the R1 branch current. (1) ∵ the voltage at both ends of each branch in the parallel circuit is equal, ∵ according to Ohm's law, the current through the R1 branch is the indication of the ammeter A2: I1 = ur1 = 3v30, Ω = 0.1A; (2) ∵ the main circuit current in the parallel circuit is equal to the sum of the branch currents If the current of R2 branch is I2 = i-i1 = 0.4a-0.1a = 0.3A, then the resistance value of sliding rheostat R2 connected to the circuit is R2 = ui2 = 3v0.3a = 10 Ω; (3) ∵ all branches in the parallel circuit work independently and do not affect each other, ∵ when the slide moves, the current passing through R1 branch remains unchanged, ∵ the change of ammeter A1 is just equal to that of A2, ∵ then the main current I ′ = 0.4a-0.1a = 0.3A or I ″ = 0.4A + 0.3A 1A = 0.5A, when I ′ = 0.3A, the current through R2 branch: I2 ′ = I ′ - I1 = 0.3a-0.1a = 0.2A, at this time, the resistance value of R2 connected to the circuit: R2 ′ = ui2 ′ = 3v0.2a = 15 Ω; when I ″ = 0.5A, the current through R2 branch: I2 ″ = I ″ - I1 = 0.5a-0.1a = 0.4A, at this time, the resistance value of R2 connected to the circuit: R2 ″ = ui2 ″ = 3v0.4a = 7.5 Ω The resistance value of the input circuit is 10 Ω; (3) the resistance value of R2 connected to the circuit is 7.5 Ω or 15 Ω when the change of the indication of ammeter A1 is just equal to the indication of A2

As shown in the figure, the voltage at both ends of AB is u, when the switch is open and closed, the ratio of ammeter indication is 1:4, then the ratio of resistance value of two resistors R1: R2=______ ．

Because the switch is a parallel circuit after closing, the current of the original branch is unchanged, and the ratio of the two current representations is 1:4, so I1: I2 = 1:3; that is, R1: R2 = I2: I1 = 3:1

In the circuit as shown in the figure, the resistance R1 = 8 Ω. When the key s is off, the indication of the voltmeter V1 is 5.7v, the indication of the ammeter is 0.75a, and the total power of the power supply is 9W; when the key s is closed, the indication of the voltmeter V2 is 4V. If the ratio of the electric power loss inside the power supply when the key s is off and closed is 9:16, calculate the electromotive force and resistance R2, R3 of the power supply

(1) When the key s is off, e = 90.75 = 12V is obtained from P = EI. (2) when the key s is off, R3 + R4 = UI = 5.70.75 = 7.6 Ω the ratio of power loss inside the power supply is 9:16, i.e. I