As the circuit shown in the figure, the power supply voltage is 12V, the resistance value of R1 is 10 Ω, the number of a meter and A1 is 1a and 0.4A respectively, then what is the resistance value of R2?

As the circuit shown in the figure, the power supply voltage is 12V, the resistance value of R1 is 10 Ω, the number of a meter and A1 is 1a and 0.4A respectively, then what is the resistance value of R2?


It can be seen from the circuit diagram that R1 and R2 are connected in series with sliding rheostat after parallel connection, ammeter a measures the current of main circuit, ammeter A1 measures the current of R1 branch; ∵ the current of main circuit in parallel circuit is equal to the sum of the currents of each branch, ∵ the current through R2 branch: I2 = i-i1 = 1a-0.4a = 0.6A, ∵ the voltage at both ends of each branch in parallel circuit is equal, ∵ according to Ohm's law, u = i1r1 = i2r2, that is 0.4A × 10 Ω = 0.4A 6A × R2, R2 ≈ 6.7 Ω. Answer: the resistance of R2 is 6.7 Ω



The reading of ammeter A1 is 0.5A, and that of ammeter A2 is 1.25A. If the resistance of R2 is 8 Ω, what is the resistance of R1


Parallel, so the voltage is equal
That is, the current and resistance are equal
So R1 = 20 Ω



In the circuit as shown in the figure, the resistance R1 = 12 ohm, R2 = R3 = R4 = 6 ohm. When the key K is open, the indication of the voltmeter is 12V, and the electric power consumed by the whole circuit is 13W. Calculate: (1) the electromotive force E and internal resistance r of the power supply. (2) after the key K is closed, what are the indication of the voltmeter and the ammeter? (the internal resistance of the ammeter is negligible, and the internal resistance of the voltmeter is very large)


(1) When the key s is disconnected, the total resistance of the external circuit is r outer = R4 + R1 (R2 + R3) R1 + R2 + R3 = 6 Ω + 6 Ω = 12 Ω, the total current is I = u outer r outer = 1212a = 1a. The electric power consumed by the whole circuit is p = EI, then the electromotive force of the power supply is e = pi = 13V. The internal resistance of the power supply is r = e − UI = 13 − 121 Ω = 1 Ω (2)



In the circuit shown in the figure, R1 = R2, when switch S is off, the indication of ammeter is 1a, when switch S is closed, the indication of ammeter will ()
A unchanged, B increased to 2a, C decreased to 0.5A, D increased to 4a
The picture shows that R1 and R2 are connected in parallel, and the ammeter is on the branch where R1 is located


A parallel circuit does not affect each other