Exercise EE1-04.05 – Conductor resistance and voltage drop of a cable reel (Solution)

Prof. Dr. Thorsten Jungmann (last updated 2025-10-26)

Relates to EE1-04 Resistance

Task

From a large online department store, you order a $50\mathrm{m}$ cable reel of copper electric line with a cross-section of $1.5\mathrm{m}{\mathrm{m}}^2$ to run the $2000\mathrm{W}$ electric heater in the garden shed. You notice that the heater does not heat the garden shed as well as when you run it from a socket-outlet in the house.

What technical cause could explain why the actual thermal power differs from the value stated on the packaging?

a) Sketch the electric circuit and locate the potential cause.

b) What conductor resistance does the cable reel have at 20 °C?

c) Calculate the voltage drop across the electric line while the heater is operating. What voltage is actually present at the heater?

d) What power does the heater draw, and how much does it emit as heat ($\eta =1$)?

e) Check your assumption from a).

f) Why should you always fully unwind a cable reel before use?

Solution

$R_{\mathrm{line}}\approx 1.2\Omega$
$\Delta U_{\mathrm{line}}\approx 9.9\mathrm{V}$ → $U_{\mathrm{heater}}\approx 220.1\mathrm{V}$
$P_{\mathrm{heater}}\approx 1.83\mathrm{kW}$ (≈ 8.4 % below rated value)
$P_{\mathrm{line}}\approx 82\mathrm{W}$ (conductor loss)

▶️ Full worked solution

◀️ Back to the task

---

Solution

Given

Given explicitly:

• Cable length (single): $L=50\mathrm{m}$ (outgoing and return conductors → $\ell =2L=100\mathrm{m}$)
• Conductor cross-section: $A=1.5\mathrm{m}{\mathrm{m}}^2$
• Material: copper
• Rated power of the heater: $P_{\mathrm{rated}}=2000\mathrm{W}$ (at $U_{\mathrm{rated}}=230\mathrm{V}$)

Known:

• Resistivity of copper (20 °C): ${\rho }_{\mathrm{Cu}}\approx 0.0178\Omega \mathrm{m}{\mathrm{m}}^2/m$
• Conductor resistance: $R=\rho \frac{\ell }{A}$
• Ohm’s law/power: $U=IR,P=UI=I^{2}R=\frac{U^2}{R}$
• Electric circuit: source $230\mathrm{V}$ – electric line $R_{\mathrm{line}}$ – heater $R_{\mathrm{heater}}$

Required

a) Sketch of the electric circuit & cause
b) Conductor resistance at $20^{\circ }\mathrm{C}$
c) Voltage drop and terminal voltage at the heater
d) Power drawn by the heater and thermal power emitted
e) Plausibility check of the assumption from a)
f) Why fully unwind the cable reel?

---

a) Electric circuit & cause

• Cause: voltage drop across the finite conductor resistance $R_{\mathrm{line}}$ → lower terminal voltage at the heater → lower power.

b) Conductor resistance (20 °C)

Double length (outgoing and return conductors): $\ell =2\cdot 50\mathrm{m}=100\mathrm{m}$

$$R_{\mathrm{line}}={\rho }_{\mathrm{Cu}}\frac{\ell }{A}=0.0178\Omega \mathrm{m}{\mathrm{m}}^2/m\cdot \frac{100\mathrm{m}}{1.5\mathrm{m}{\mathrm{m}}^2}=1.186\dots \Omega \approx \underline{1.2\Omega }$$

c) Voltage drop and terminal voltage

Heater resistance from the rated values (at $230\mathrm{V}$, $2\mathrm{kW}$):

$$R_{\mathrm{heater}}=\frac{U_{\mathrm{rated}}^2}{P_{\mathrm{rated}}}=\frac{230^2{\mathrm{V}}^2}{2000\mathrm{W}}=26.45\Omega$$

Current during operation (series connection of $R_{\mathrm{line}}$ and $R_{\mathrm{heater}}$):

$$I=\frac{U_{\mathrm{rated}}}{R_{\mathrm{line}}+R_{\mathrm{heater}}}=\frac{230\mathrm{V}}{(1.19+26.45)\Omega }=8.32\mathrm{A}$$

Voltage drop across the electric line:

$$\Delta U_{\mathrm{line}}=I{R}_{\mathrm{line}}=8.32\mathrm{A}\cdot 1.19\Omega \approx \underline{9.9\mathrm{V}}$$

Terminal voltage at the heater:

$$U_{\mathrm{heater}}=230\mathrm{V}-\Delta U_{\mathrm{line}}=\underline{220.1\mathrm{V}}$$

d) Power drawn by the heater

$$P_{\mathrm{heater}}=I^2{R}_{\mathrm{heater}}=(8.32\mathrm{A}{)}^2\cdot 26.45\Omega =\underline{1.83\mathrm{kW}}(\approx 1830\mathrm{W})$$

(Relative deviation: $\approx 8.4\%$ below rated power.)

e) Checking the assumption

• Voltage drop: $\frac{\Delta U_{\mathrm{line}}}{230\mathrm{V}}\approx \underline{4.29\%}$
• Power reduction (resistive load, $P\propto U^2$): $\approx \underline{8.4\%}$

→ Confirmed: the lower heating power is caused by the voltage drop across the long electric line.

Side calculation (loss in the electric line):

$$P_{\mathrm{line}}=I^2{R}_{\mathrm{line}}=8.32^2{\mathrm{A}}^2\cdot 1.19\Omega =\underline{82\mathrm{W}}$$

The electric line is a middling heater, of the order of an electric heating pad.

f) Why fully unwind?

• Wound up on the reel, the electric line is hardly cooled → the losses ($P_{\mathrm{line}}\approx 82\mathrm{W}$) cause heat to build up → risk of overheating/fire.
• With the outgoing and return conductors close together, inductive effects are largely compensated; the main problem is the thermal load (not the magnetic field).

→ Therefore: always fully unwind the cable reel.