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In the circuit, the resistance R1 is 10 Ω. Close the switch s, the indication of ammeter A1 is 0.3A, and the indication of ammeter A2 is 0.5A Ask: (1) current through resistance R2: (2) power supply voltage please write the process of solving the problem clearly
Because the two meters show different numbers, it is a parallel circuit
The voltage is 10 * 0.3 = 3V
So the current of R2 is 3 / 0.5 = 6 Ω
I haven't studied physics for three or four years, right
As shown in the figure, the resistance R1 = 12 Ω. When the key SA is off, the current passing through is 0.3 A; when the key SA is closed, the ammeter shows 0.5 A. Q: what is the power supply voltage? What is the resistance value of resistance R2?
When the key SA is closed, the circuit is a simple circuit of R1, and the voltage of the power supply is u = i1r1 = 0.5A × 12 Ω = 6V; when the key SA is disconnected, the two resistors are connected in series, and the total resistance of the circuit is r = UI = 6v0.3a = 20 Ω, R2 = r-r1 = 20 Ω - 12 Ω = 8 Ω. Answer: the voltage of the power supply is 6V, and the resistance of R2 is 8 Ω
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