When the voltage at both ends of a resistance wire is u, the power consumed is p. if the voltage at both ends of the resistance wire is increased by one fifth, what is the original voltage?

When the voltage at both ends of a resistance wire is u, the power consumed is p. if the voltage at both ends of the resistance wire is increased by one fifth, what is the original voltage?

When the voltage at both ends of a resistance wire is u, the consumed electric power is p. if the voltage at both ends of the resistance wire is increased by one fifth, what is the consumed electric power? Suppose the resistance of the resistance wire is r, 1. The original power of the resistance wire: P = u ^ 2 / R, r = u ^ 2 / P.1 formula 2. The consumed electric power of the resistance wire: P = ((1 + 1 / 5) U) ^

When the voltage at both ends of a resistance wire is u, the power consumed is p. if the voltage at both ends of the resistance wire is increased by one fifth, what is the voltage?

This question should be "when the voltage at both ends of a resistance wire is u, the consumed electric power is p. if the voltage at both ends of the resistance wire is increased by one fifth, what is the power of the resistance wire?" if it is a corrected question, then one of the power calculation formulas derived from Ohm's law is: P = V ^ 2 / r if

The output AC voltage of the generator of a small hydropower station is 500V, and the output power is 50KW. If a transmission line with resistance of 3 Ω is used to send power to remote users, what is the voltage and power that users get? If it is required that the power loss on the transmission line should not exceed 0.6% of the output power, a step-up transformer should be installed in the power station. After reaching the user, the step-down transformer should be reduced to 220 V for use by the user. Assuming that the transformer is an ideal transformer, what are the turns ratio of the primary and secondary coils of the step-up transformer and the step-down transformer?

According to P = UI, I = Pu = 50KW, 500V = 100A; P1 = I2R = (100) 2 × 3 = 30000w; 50kw-30kw = 20kW; IR = 100 × 3 = 300V; 500v-300v = 200V; P = UI, I = Pu = 50KW, 500V = 100A; P = UI, I = UI, I = Pu = 50KW, 500V = 100A, I = I2R = (100) 2 × 3 = 30000w

The output voltage of a generator is 500V, and it transmits 40A current to the factory. The resistance of the transmission line is 3 ohm. Ask the output power of the generator

The output power of this generator is p = UI = 500x40w = 20000w = 20kW
Note: 1. Can not use p = u & # 178; / r = (500 & # 178; / 3) W. because 500V voltage does not completely fall on the transmission line, the voltage on the transmission line is only 40x3 = 120V, more fall on factory users
2. We can't use p = I & # 178; r = 40 & # 178; x3w, because this is only the power lost by the heating of the transmission line