Four resistors of 1 Ohm each are connected in series and connected to a current source

Four resistors of 1 Ohm each are connected in series and connected to a current source, EMF of which is 4.5 V and an internal resistance of 0.5 Ohm. Find the voltage across one resistor.

R1 = R2 = R3 = R4 = 1 ohm.

EMF = 4.5 V.

r = 0.5 ohm.

U1 -?

When connecting the conductors in series:

the voltage at the terminals of the current source will be the sum of the voltages at each resistance U = U1 + U2 + U3 + U4;

the current in the circuit and in each conductor is the same I = I1 = I2 = I3 = I4;

the resistance of the entire circuit is the sum of all resistances R = R1 + R2 + R3 + R4.

We express the current in the circuit I according to Ohm’s law for a closed loop: I = EMF / (R + r).

R = 4 * R1.

I = 4.5 V / (4 * 1 Ohm + 0.5 Ohm) = 1 A.

We find the voltage across the resistor according to Ohm’s law for a section of the circuit: U1 = I1 * R1 = I * R1.

U1 = 1 A * 1 Ohm = 1 V.

Answer: the voltage across the resistors is U1 = U2 = U3 = U4 = 1 V.