The resistance of the transmission line is 0.5 ohm, and the power station provides 105kW of electric power through the line with the voltage of 102kw, then the current and heat loss power in the transmission line will be reduced Isn't it possible to reduce the voltage if there is resistance on the transmission line?

The resistance of the transmission line is 0.5 ohm, and the power station provides 105kW of electric power through the line with the voltage of 102kw, then the current and heat loss power in the transmission line will be reduced Isn't it possible to reduce the voltage if there is resistance on the transmission line?

The higher the current, the lower the voltage
You said 102kw voltage, should be 102kv?
102kv should be line voltage. Load power factor is assumed to be 1
The current is about 105 / 1.732/102 ≈ 0.6A
The loss on the line is about I ^ 2 * r = 0.6 * 0.6 * 0.5 = 0.18w, which can be ignored compared with 105kW

The output AC voltage of the generator in a small power station is 500V and the output power is 100kW. If a transmission line with resistance of 3 Ω is used to send power to the remote users, what is the voltage and power that the users get? If the power loss on the transmission line is required to be 1.2% of the transmitted power, a step-up transformer should be installed in the power station. Before reaching the user, the step-down transformer should be changed into 220 V for the user's use. Regardless of the energy loss of the transformer, what are the turns ratios of the primary and secondary coils of the two transformers

(1) When the voltage is 500V, the current: I = Pu = 100 × 103500 = 200A, the lost voltage: u loss = IR = 200 × 3 = 600V ＞ 500V, so the voltage and power obtained by the user are 0. (2) when the voltage is 500V, the schematic diagram is as shown in the figure below. P loss = 1.2% P, i.e. P loss = 100 × 103 × 1.2% & nbsp; w = 1200 & nbsp; W. transmission current I1 = P loss r = 12003a = 20 & nbsp; A. Therefore, the current on the transmission line is 20A. According to the relationship between the current ratio and the turn ratio of the transformer, it can be seen that: NN1 = i1i = 20200 = 110, the transmission voltage after the step-up of the power station U1 = Pi1 = 100 × 10320 = 5000 & nbsp; V, the voltage lost on the transmission line u ′ = I · r = 20 × 3 & nbsp; v = 60 & nbsp; v. the voltage reaching the user's input transformer U2 = U-U ′ = (5 & nbsp; 000-60) v = 4 & nbsp; 940 & nbsp; 5. The turn ratio of step-down transformer is n3n4 = u2u3 = 4940220 = 24711 A: when the output AC voltage is 500V, the voltage and power obtained by users are 0; the turn ratio of step-up transformer and step-down transformer are 1:10247:11 respectively