A small electric bell with a resistance of 10 ohm is connected in series with a fixed resistance of 20 ohm and connected to a 6V power supply. If the small electric bell just works normally, what is the voltage at both ends of the electric bell?

According to the resistance ratio, 10: (10 + 20) = 2:6

For the electric bell marked "6V & nbsp; 10 Ω", please Join one Ω resistance can make it work normally under 9V voltage

When the bell works normally, the voltage U bell = 6V, the resistance R bell = 10 Ω. According to the partial voltage characteristics of the series circuit, a resistor partial voltage should be connected in series to make the bell work normally at 9V. Because the total voltage in the series circuit is equal to the sum of the partial voltages, the voltage at both ends of the series resistance: ur = u bell = 9v-6v = 3

A resistance of 100 ohm Ω, working voltage of 3V bell, now need to be connected to 12V power supply, ask must be connected in series with a voltage divider!

Bell current = V / r = 3V / 100 ohm = 0.03A = circuit current = total voltage / total resistance = 12V / (100 + unknown resistance) ohm, the solution is 400 = 100 ohm + unknown resistance, the unknown resistance is 300 ohm

The bell with "6V 10 ohm" on the name plate needs to be connected in series with a resistor of several ohm to make it work under 9V voltage

In series, the current is equal
I1 = I2, that is U1 / R1 = U2 / R2
6V/10Ω=(9V-6V)/R2
R2=5Ω
A 5 Ω resistor should be connected in series to make it work at 9V

A small electric bell with a resistance of 10 ohm is connected in series with a fixed resistance of 20 ohm and connected to a 6V power supply. If the small electric bell just works normally, what is the voltage at both ends of the electric bell?