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.