# Series connection of three conductors. The voltage drop across a conductor with resistance

Series connection of three conductors. The voltage drop across a conductor with resistance R1 = 36 ohms is equal to U1 = 9V. Determine the voltage on the conductor with R2 = 64 Ohm and the resistance of the conductor R3, if the voltage at its ends is 120 V.

R1 = 36 ohms.

U1 = 9 V.

R2 = 64 ohms.

U3 = 120 V.

U2 -?

R3 -?

With a series connection of conductors, the current strength at all resistances will be the same: I = I1 = I2 = I3.

Let’s write Ohm’s law for a section of a circuit for each resistance: I1 = U1 / R1, I2 = U2 / R2, I3 = U3 / R3.

U1 / R1 = U2 / R2.

The formula for determining the voltage across the second resistance will be: U2 = U1 * R2 / R1.

U2 = 9 V * 64 Ohm / 36 Ohm = 16 V.

U1 / R1 = U3 / R3.

The resistance of the third conductor R3 is expressed by the formula: R3 = U3 * R1 / U1.

R3 = 120V * 36 ohms / 9V = 480 ohms.

Answer: U2 = 16 V, R3 = 480 ohms.

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