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04-BS-9 · May 2013

Question 7 of 8: Transmission-Line Efficiency Comparison

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2013 — 04-BS-9 Basic Electromagnetics. Three-hour, closed-book exam (approved Casio/Sharp calculator only). Aids given: $\varepsilon_0=8.85\times10^{-12}$ F/m, $\mu_0=4\pi\times10^{-7}$ H/m, $e=1.6\times10^{-19}$ C. Format: eight questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All eight are solved below for completeness.

Reference texts: Sadiku, Elements of Electromagnetics / Hayt & Buck, Engineering Electromagnetics — electrostatics, Gauss's law, capacitance, magnetostatics, Faraday's law, Maxwell's equations and displacement current; Young & Freedman, University Physics with Modern Physics — dipole fields, torque on a current loop, electromagnetic wave propagation in dielectrics.

Question 7: Transmission-Line Efficiency Comparison (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

QuantityValue
Generator 11000 V, $R_{\text{load,1}}=1\ \Omega$
Generator 22000 V, $R_{\text{load,2}}=4\ \Omega$
Transmission line resistance (case B)0.1 $\Omega$
Generator internal impedancezero (ideal source)

Find. Efficiency $\eta=P_{\text{load}}/P_{\text{total}}$ for both generators, in cases A (direct) and B (through the line), and a comparison.

Vgenerator (zero Z_int)R_line = 0.1 Ω(case B only; 0 in case A)R_loadSystem (1000V, 1Ω) and (2000V, 4Ω); line 0.1Ω
Series circuit: ideal EMF source, line resistance (present only in case B), load resistance.

Approach. With zero internal impedance, case A has no resistive element other than the load itself, so all delivered power reaches the load. Case B is a simple series divider between the line and load resistances, so efficiency reduces to $\eta=R_{\text{load}}/(R_{\text{load}}+R_{\text{line}})$, independent of the EMF.

  1. Case A — direct connection. With no line resistance and an ideal source, every watt generated is dissipated in the load: $$\eta_{A,1}=\eta_{A,2}=\boxed{100\%}\quad\text{for both generators.}$$
  2. Case B — 1000 V / 1 Ω system. $$\eta_{B,1}=\frac{R_{\text{load,1}}}{R_{\text{load,1}}+R_{\text{line}}}=\frac{1}{1.1}$$ $$\eta_{B,1}=\boxed{90.9\%}$$ (Current $I=1000/1.1=909$ A; line loss $I^2R_{\text{line}}=82.6$ kW against a 1000 V, 909 A source.)
  3. Case B — 2000 V / 4 Ω system. $$\eta_{B,2}=\frac{R_{\text{load,2}}}{R_{\text{load,2}}+R_{\text{line}}}=\frac{4}{4.1}$$ $$\eta_{B,2}=\boxed{97.6\%}$$ (Current $I=2000/4.1=487.8$ A; line loss $=23.8$ kW.)
  4. Comparison and comment. Direct connection is loss-free for both (case A, trivially 100%), but once a line resistance is introduced (case B) the 2000 V/4 Ω system is markedly more efficient (97.6% vs 90.9%). This is because efficiency in case B depends only on the ratio $R_{\text{line}}/R_{\text{load}}$ — the higher-voltage system drives a smaller current (487.8 A vs 909 A) through the same line resistance for a comparable order of delivered power, so the line dissipates a smaller fraction of the total. This is the textbook justification for transmitting power at higher voltage (and correspondingly higher load impedance/lower current): it minimizes $I^2R_{\text{line}}$ losses as a share of the power delivered.
QuantityResult
$\eta_{A,1}$, $\eta_{A,2}$ (direct)100%, 100%
$\eta_{B,1}$ (1000V/1Ω via line)90.9%
$\eta_{B,2}$ (2000V/4Ω via line)97.6%
Conclusionhigher-voltage/higher-load-R system loses proportionally less over the line