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In the circuit shown in the figure, the voltage U remains unchanged, the resistance R1 = R2, when the switch S is open, the indication of the ammeter is I1; when the switch S is closed, the indication of the ammeter is I2, then I1: I2=______ ．
When switch S is off, the circuit is a simple circuit of R2; when switch S is closed, R1 and R2 are connected in parallel, because the voltage at both ends of R2 remains unchanged, so according to Ohm's law, the current through R2 remains unchanged, I1 ′ = I1; the current through R1 is I1 ″ = ur1, and the ammeter's indication is I2 = I1 ′ + I1 ″ = UR2 +
After R1 = 2 ohm, R2 = 4 ohm switch is closed, the indication of ammeter is 3a, find 1: the indication of voltmeter 2: what is the electric power consumed by R1 R2?
(1) Series connection: u = (R1 + R2) * I = (2 + 4) * 3 = 24 V
Parallel connection: u = U1 = 3 * 2 = 6V
Or = U2 = 3 * 4 = 12V
(2) Parallel: P = UI = 12 * 3 = 36V
Or P = UI = 6 * 3 = 18V
Series connection: P1 = u1i1 = 6 * 3 = 18W
P2=U2I2=12*3=36W
R1 = 40 ohm, R2 = 60 ohm, through R1 current I1 = 0.3A to calculate the power supply voltage, R2 electric power in parallel
U = i1r1 = 40 Ω times 0.3A = 12V I2 = u divided by R2 = 12V divided by 60 Ω = 0.2A P = UI = 12V times 0.2A = 2.4W
The voltage at both ends of a and B remains unchanged, the resistance in the circuit is r, the indication number of ammeter is I, if you want the indication number of ammeter is 4I, it should be in the circuit
How to write the formula
It is necessary to change the resistance into R / 4 in the circuit, or connect four RS in parallel
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