As shown in the figure, R1 = 6 Ω, the reading of ammeter A1 is 0.5A, and the degree of ammeter A2 is 0.3A? (2) What is the resistance of R2?

As shown in the figure, R1 = 6 Ω, the reading of ammeter A1 is 0.5A, and the degree of ammeter A2 is 0.3A? (2) What is the resistance of R2?


It can be seen from the circuit diagram that R1 and R2 are connected in parallel. Ammeter A1 measures the main circuit current, ammeter A2 measures the branch current of R2, and voltmeter measures the voltage of power supply. (1) ∵ in parallel circuit, the main circuit current is equal to the sum of the branch currents, ∵ the current passing through R1: I1 = i-i2 = 0.5a-0.3a = 0.2A, ∵ the voltage at both ends of each branch in parallel circuit is equal, ∵ according to Ohm's law, the voltage of power supply is electricity Indication of voltmeter: u = i1r1 = 0.2A × 6 Ω = 1.2V; (2) resistance of R2: R2 = ui2 = 1.2v0.3a = 4 Ω. Answer: (1) reading of voltmeter is 1.2V; (2) resistance of R2 is 4 Ω



R1 and R2 are connected in series. The voltmeter measures the voltage of R2. R1 is the constant resistance, R2 is the sliding rheostat, and the indication of the ammeter is 0.15A
When slide P of sliding rheostat runs all resistance wires, the indication of voltmeter is 9V; when slide P runs half resistance wires, the indication of voltmeter is 6V;
1. Maximum resistance of sliding rheostat R2 (answer: 60 Ω)
2. Power supply voltage and R1 resistance (answer: 18V; 60ohm)


(1) If you know R2 current and voltage at both ends of R2, you can get the maximum resistance value: R2 = u / I = 9V / 0.15A = 60 Ω (2) first find out the current in the circuit when slide P runs half resistance wire is I = u / r2 '= 6V / 30 Ω = 0.2A, then use the voltage equation: 0.2A (R1 + R2) = 0.15A (R1 + R2'), 0.2A (R1 + 30 Ω) = 0.15A (



As shown in the figure, the ammeter shows 1.5 A, the resistance of R1 is 6 Ω, the power supply voltage is 6 V and remains unchanged. Calculate the resistance R2


12 ohm, right
6v/6=1A
1.5A-1A=0.5A
6V/0.5A=12



As shown in Figure 2, R1 = 2 ohm, R2 = 3 ohm, the indication of voltmeter is 1V, calculate the voltage of R2 two sections and power supply voltage


What's the picture? R1 and R2 are connected in series, and the voltmeter measures the voltage at both ends of R1
Right? It should be
The current through R1 is 0.5A, the two resistors are in series, the current is equal, the voltage at both ends of R2 is U2 = r2xi = 3x0.5 = 1.5V, and the power supply voltage is 2.5V