In the circuit shown in the figure, the resistance value of resistance R1 is 15 Ω, the resistance value of resistance R2 is 10 Ω, and the current through resistance R1 is 0.2 a

In the circuit shown in the figure, the resistance value of resistance R1 is 15 Ω, the resistance value of resistance R2 is 10 Ω, and the current through resistance R1 is 0.2 a

(1) ∵ parallel circuit, ∵ the voltage at both ends of each branch is equal to the power supply voltage, and ∵ I = ur, ∵ the indication of voltmeter: u = i1r1 = 0.2A × 15 Ω = 3V. (2) the current of resistance R2: I2 = UR2 = 3v10 Ω = 0.3A, ∵ the main circuit current in parallel circuit is equal to the sum of all branch currents, ∵ the indication of ammeter: I = I1 + I2 = 0.2A + 0.3A = 0.5A 5 a

Connect a 10 ohm resistor R1 in series with another resistor R2 into the circuit. The power supply voltage is known to be 12V, and the voltage at both ends of R2 is 9V. Calculate the R1 current?

Set the current to I
Because two resistors are connected in series, the current is equal everywhere
According to Ohm's Law:
I=U2/R2
I=E/R1+R2
The two equations are simultaneous
I=0.3A

Two resistors are connected in series in the circuit. It is known that R1 = 10 Ω. The ratio of the voltage at both ends of R1 to the supply voltage is 2:7. Calculate the resistance of R2
Can we use known: ask: answer:

Let the voltage at both ends of R1 be 2V and the power supply voltage be 7V
Then the voltage at both ends of R2 is 7-2 = 5V and the total current is 2 / 10 = 0.2A
So R2 = 5 / 0.2 = 25 Ω

Resistor R1 and R2 are connected in series to a 9V power supply. The current intensity through R1 is 3a, and the resistance value of R2 is 10 Ω. Calculate the voltage and resistance at both ends of R1

R = u / I, so the total resistance is 9 / 3 = 3 ohm! But the problem says that it is connected in series, and one of the resistances is 10 ohm. So your problem is wrong! Unless it is connected in parallel, it can be established!