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If two resistors R1 and R2 are connected to a 3V power supply, calculate the indication of the ammeter, and calculate the electric energy consumed when the 30s resistance is 3V
Ammeter: I = u / R1 (or R2) is used to measure each branch, and the resistance of main circuit can be calculated by the above method or the reciprocal of 1 / R1 + 1 / R2
Electric energy: w = uit = P & # 8198; t, P = I * 2R (* 2 instead of square) = u * 2 / R
In this way, understand the truth and then algebra
The electromotive force of the power supply is 3V, and the given constant resistance R1 = 10V, R2 = 5V. When the key s is connected to R1, the indication of the ammeter is 0.20a
The electromotive force of the power supply is 3V, and the constant resistance R1 = 10V, R2 = 5V is known. When the key s is connected to R1, the indication of the ammeter is 0.20a. Then when the key s is connected to R2, what is the indication of the current?
Let the electromotive force of the battery be ε, the internal resistance of the battery be r, the external voltage of the battery be u, and the current be I,
When the key is set at 1, 3V = 0.2A * (10 Ω + R), r = 5 Ω
When the key is set at 2, 3 = I * (5 Ω + 5 Ω), I = 0.3A can be obtained
Resistance R1 = 5 Ω, R2 = 4 Ω. When the switch is closed, the current indication is 0.9 a, then the ampere indication of ammeter is 1
At point B, the whole parallel connection U / [(12 + x) / 12x] = 0.5A, at point C, the front x is replaced by X / 3, 0.5 by 1, then u and X are contacted simultaneously, and then u is brought in so that the first equation is equal to 3, and the contact x is the minimum resistance
a. When resistance R1 is connected between a and B, the indication of ammeter is 0.2 A. if another resistance R2 is connected in series, the indication of ammeter is 0.16 a, then R1=
a. When resistance R1 is connected between B, the indication of ammeter is 0.2 a,
U=0.2R1
If another resistor R2 is connected in series, the indication of the ammeter is 0.16
U=0.16R1+0.16R2
Then: 0.04r1 = 0.16r2
R1=4R2
RELATED INFORMATIONS
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