The resistor R1 and R2 are connected in series on the power supply with a voltage of 4.5V. The resistance of R1 is known to be 5 Ω, and the voltage at both ends of R2 is 3V. The total resistance in the circuit and the total resistance through R2 are calculated The resistance of an electric bell is 10 Ω. When it works normally, the voltage at both ends is 4V. Now it is connected to 9V power supply. In order to make it work normally, how much resistance should it be connected in series?

The resistor R1 and R2 are connected in series on the power supply with a voltage of 4.5V. The resistance of R1 is known to be 5 Ω, and the voltage at both ends of R2 is 3V. The total resistance in the circuit and the total resistance through R2 are calculated The resistance of an electric bell is 10 Ω. When it works normally, the voltage at both ends is 4V. Now it is connected to 9V power supply. In order to make it work normally, how much resistance should it be connected in series?

1、15Ω,0.3A
The voltage of R1 is 4.5v-3v = 1.5V, and the current of R1 is I = 1.5v/5 Ω = 0.3A
Resistor R1 and R2 are connected in series, so the current of R2 is equal to 0.3A of R1
R2=3V/0.3A=10Ω
The total resistance is r = R1 + R2 = 5 Ω + 10 Ω = 15 Ω
2. When the bell works normally, the current is 4V / 10 Ω = 0.4A
To make it work normally, a resistor should be connected in series to divide the voltage: 9v-4v = 5V
In series, both have the same current of 0.4A
The resistance value of the resistance is: 5V / 0.4A = 12.5 Ω
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The resistance R1 and R2 are in series, and the resistance R2 is twice of R1. The power supply voltage is 3V. How much is the voltage of r1r2

Since the resistance of R2 is twice that of R1, the voltage of R2 is twice that of R1
R1 voltage 1V R2 voltage 2V