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The power supply voltage is 6 V, the indication of ammeter A is 2.5 A, R1 = 4 Ω, calculate the indication of ammeter a1.a2 and the resistance of resistance R2 Well
R1 and R2 should be in parallel, A1 = 6V △ 4eu = 1.5A
So A2 = a-a1 = 2.5a-1.5 = 1a
So R2 = 6V △ 1A = 6ohm
Connect the circuit, the power supply voltage is 12V, keep unchanged, the resistance R2 = 40 Ω, when he turns off the switch, the ammeter's indication is 0.4A
Resistance. (2) when he closes the switch, what is the indication of the ammeter?
Your topic is incomplete
I assume that resistance R1 and R2 are in parallel
R1=U/I1=12v/0.4a=30Ω
The current of single R2 is I2 = u / r2 = 12V / 40 Ω = 0.3A
Total current = I1 + I2 = 0.4 + 0.3 = 0.7A
As shown in the figure, the circuit, resistance R1 = 9 Ω, R2 = 15 Ω, power EMF e = 12V, internal resistance R = 1 Ω, ammeter degree I = 0.4A, calculate the resistance value of resistance R3 and other consumed electric power, R1, R2, ammeter series as a way, R3 and its three parallel
(9 + 15) × 0.4 = 9.6V, so the total current of the circuit is 2.4 / 1 = 2.4a
So R3 flows through 2A
24 Ω: R3 = 2:0.4, so R3 = 4.8 Ω
The power is UI = 9.6 × 2 = 19.2a
As shown in the figure, in the circuit, the power supply voltage remains unchanged, and the ratio of resistance values of R1 and R2 is 3:2. When the switch S is closed, the ratio of indication values of ammeter A1 and A2 is ()
A. 2：3B. 3：2C. 3：5D. 5：3
According to the circuit diagram, R1 and R2 are connected in parallel. Ammeter A1 measures the current of branch R2 and ammeter A2 measures the current of main circuit. The voltage at both ends of each branch in parallel circuit is equal. According to Ohm's law, the current ratio of two branches: i1i2 = ur1ur2 = r2r1 = 23. The current of main circuit in parallel circuit is equal to the current of each branch
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