Question 8 of 8: Power Delivered and Power Lost — Two Parallel Transmission Lines
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 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 — induced EMF in a rotating loop, plane-wave impedance.
Question 8: Power Delivered and Power Lost — Two Parallel Transmission Lines (20 marks)
Check: the paper's steel resistivity "$20\times20^{-8}$" is read as a misprint for $20\times10^{-8}\ \Omega\cdot\text{m}$ (about 12× copper's, physically typical for steel). Each 1 km line's resistance is taken as $\rho L/A$ using the stated 1 km length directly (the length given is treated as the line's own electrical length, not doubled for a separate return conductor).
Given.
Quantity
Value
Source
$115$ V, zero internal impedance
Load
$10\ \Omega$ resistive
Line length $L$ (each)
$1$ km $=1000$ m
Conductor area $A$ (each)
$1\ \text{mm}^2=1\times10^{-6}\ \text{m}^2$
Copper resistivity $\rho_{\text{Cu}}$
$1.7\times10^{-8}\ \Omega\cdot\text{m}$
Steel resistivity $\rho_{\text{st}}$
$20\times10^{-8}\ \Omega\cdot\text{m}$ (see check note)
Find. (i) power delivered to the $10\ \Omega$ load; (ii) power lost in the steel line.
Copper and steel lines in parallel from an ideal 115 V source to a 10 Ω load.
Approach. Compute each line's resistance from $R=\rho L/A$, combine the two lines in parallel to get the line-to-load path resistance, solve the resulting series circuit (parallel lines $+$ load) for the total current, then split that current between the two lines by their own resistances to get the steel line's loss.
Individual line resistances.
$$R_{\text{Cu}}=\frac{\rho_{\text{Cu}}L}{A}=\frac{(1.7\times10^{-8})(1000)}{1\times10^{-6}}=17.0\ \Omega$$
$$R_{\text{st}}=\frac{\rho_{\text{st}}L}{A}=\frac{(20\times10^{-8})(1000)}{1\times10^{-6}}=200.0\ \Omega$$
Parallel combination and total current. The two lines connect the source to the load node in parallel:
$$R_{\text{par}}=\frac{R_{\text{Cu}}R_{\text{st}}}{R_{\text{Cu}}+R_{\text{st}}}=\frac{(17.0)(200.0)}{217.0}=15.67\ \Omega$$
The source, $R_{\text{par}}$, and the load form a series loop:
$$I_{\text{total}}=\frac{V}{R_{\text{par}}+R_{\text{load}}}=\frac{115}{15.67+10}=\boxed{4.480\ \text{A}}$$
Part (i): Power delivered to the load.
$$P_{\text{load}}=I_{\text{total}}^2 R_{\text{load}}=(4.480)^2(10)$$
$$P_{\text{load}}=\boxed{200.7\ \text{W}}$$
Part (ii): Power lost in the steel line. The two lines share the same terminal voltage $V_{\text{line}}=I_{\text{total}}R_{\text{par}}=(4.480)(15.67)=70.20$ V, which drives current through each line independently:
$$I_{\text{st}}=\frac{V_{\text{line}}}{R_{\text{st}}}=\frac{70.20}{200.0}=0.3510\ \text{A}$$
$$P_{\text{st}}=I_{\text{st}}^2 R_{\text{st}}=(0.3510)^2(200.0)$$
$$P_{\text{st}}=\boxed{24.6\ \text{W}}$$
