Power Distribution in a DC Generator, Line, and Load: 04-BS-9, May 2016, Q7

Question 7: Power Distribution in a DC Generator, Line, and Load (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.

Quantity	Value
EMF, $\mathcal{E}$	$12$ V
Internal resistance $r_{\text{int}}$	$0.1\ \Omega$
Transmission line resistance $R_{\text{line}}$	$1\ \Omega$
Load resistance $R_{\text{load}}$	$10\ \Omega$

Find. $P_{\text{load}}$, $P_{\text{line}}$, and $P_{\text{int}}$.

Series DC circuit: generator EMF and internal resistance, transmission-line resistance, resistive load.

Approach. All three resistances form one series loop, so a single current flows through all of them; find it from the total resistance, then compute each element's power as $I^2R$.

Series current. $$I=\frac{\mathcal{E}}{r_{\text{int}}+R_{\text{line}}+R_{\text{load}}}=\frac{12}{0.1+1+10}$$ $$I=\boxed{1.081\ \text{A}}$$
Power in each element, $P=I^2R$. $$P_{\text{load}}=(1.081)^2(10)=\boxed{11.69\ \text{W}}\qquad P_{\text{line}}=(1.081)^2(1)=\boxed{1.169\ \text{W}}\qquad P_{\text{int}}=(1.081)^2(0.1)=\boxed{0.1169\ \text{W}}$$
Check against total EMF power. $P_{\text{total}}=\mathcal{E}I=(12)(1.081)=12.97$ W, which equals $P_{\text{load}}+P_{\text{line}}+P_{\text{int}}=11.69+1.169+0.1169=12.97$ W — energy balance confirmed.

Quantity	Result
Series current $I$	$1.081$ A
$P_{\text{load}}$	$11.69$ W
$P_{\text{line}}$	$1.169$ W
$P_{\text{int}}$	$0.1169$ W

Concept & Theory

Topic. Series DC circuit power distribution among source, line, and load resistances.

Key relations. Series current $I=\mathcal{E}/R_{\text{total}}$; power dissipated in any resistor, $P=I^2R$; conservation, $P_{\text{total}}=\mathcal{E}I=\sum P_i$.

Why this works. With only one loop and no branching, the same current flows through the source, line, and load in series, so Ohm's law fixes $I$ from the total resistance and each element's dissipated power follows independently from that single shared current — a useful energy-conservation check is that the three powers must sum to the total power delivered by the EMF.

Common pitfall. Computing terminal voltage or load voltage first and using $P=V^2/R$ inconsistently across elements (easy to mix up which voltage belongs to which resistor); the $I^2R$ route avoids that entirely since $I$ is common to all three.

Source. Sadiku, Fundamentals of Electric Circuits, Ch. 2–4 (Ohm's law, series circuits, power).

Question 7 of 8: Power Distribution in a DC Generator, Line, and Load

Question 7: Power Distribution in a DC Generator, Line, and Load (20 marks)