Three conductors with resistances of 12 ohms, 9 ohms and 3 ohms are connected in series.

Three conductors with resistances of 12 ohms, 9 ohms and 3 ohms are connected in series. The voltage at the ends of the circuit is 120 V. Find the voltage on the second wire.

Circuit resistance with series connection of conductors:
R = R1 + R2 + R3, where R1, R2, R3 are the resistances of the conductors (R1 = 10 ohms, R2 = 9 ohms, R3 = 3 ohms).
R = R1 + R2 + R3 = 12 + 9 + 3 = 24 ohms.
According to Ohm’s law, the current in the circuit is:
I = U / R, where U is the voltage at the ends of the circuit (U = 120 V).
I = U / R = 120/24 = 5 A.
The voltage on the second conductor is calculated by the formula:
U2 = I * R2 = 5 * 9 = 45 V.
Answer: The voltage on the second conductor is 45 V.