Power in Electric Circuits — Question 5

Section 26-6 · Physics pages 4260–4261

Question 5 High transmission voltage reduces the loss for a fixed delivered power A load receives constant power PL at voltage VL > 0. The two supply wires together have resistance Rw. Use I = PL/VL and source voltage Vs = VL + IRw. source load total Rw Schematic; labels, not drawing lengths, specify the data. (a) Find wire loss, required source power and delivery efficiency. (b) For PL = 12.0 kW and Rw = 2.00 Ω, compare losses at VL = 240 V and 2400 V.
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Question 5 - Solution High transmission voltage reduces the loss for a fixed delivered power source load total Rw (a) Method and calculation Pw = I2Rw = P 2 LRw/V 2 L, Ps = VsI = PL + Pw. η = PL/Ps = 1 1 + PLRw/V 2 L . (b) Interpretation and further calculation I240 = 50.0 A, Pw,240 = 5000 W. I2400 = 5.00 A, Pw,2400 = 50.0 W, Pw,2400/Pw,240 = .0100. Physical check The comparison holds delivered power fixed; raising voltage across a fixed resistance would be a different constraint.

Question 5

High transmission voltage reduces the loss for a fixed delivered power

A load receives constant power PLP_L at voltage VL>0V_L>0. The two supply wires together have resistance RwR_w. Use I=PL/VLI=P_L/V_L and source voltage Vs=VL+IRwV_s=V_L+IR_w.

Diagram shown in the original SVG worksheet.

  1. Find wire loss, required source power and delivery efficiency.

  2. For PL=12.0P_L=12.0 kW and Rw=2.00ΩR_w=2.00\,\Omega, compare losses at VL=240V_L=240 V and 2400 V.

Question 5 - Solution

High transmission voltage reduces the loss for a fixed delivered power

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Pw=I2Rw=PL2Rw/VL2,Ps=VsI=PL+Pw.P_w=I^2R_w=P_L^2R_w/V_L^2,\quad P_s=V_sI=P_L+P_w.η=PL/Ps=11+PLRw/VL2.\eta=P_L/P_s=\frac1{1+P_LR_w/V_L^2}.

(b) Interpretation and further calculation

I240=50.0A,Pw,240=5000W.I_{240}=50.0\,\mathrm A,\quad P_{w,240}=5000\,\mathrm W.I2400=5.00A,Pw,2400=50.0W,Pw,2400/Pw,240=.0100.I_{2400}=5.00\,\mathrm A,\quad P_{w,2400}=50.0\,\mathrm W,\quad P_{w,2400}/P_{w,240}=.0100.

Physical check

The comparison holds delivered power fixed; raising voltage across a fixed resistance would be a different constraint.