Exercise EE1-02.02 – Electric potentials and voltage (Solution)

Prof. Dr. Thorsten Jungmann (last updated 2025-10-21)

Relates to EE1-02 Electrical quantities and units, terms, symbols and models

Task

Three NiMH cells with a rated voltage of $U=1.2\mathrm{V}$ are connected in series in a rechargeable battery pack. The terminal labelled ‘0/COM’ of a voltmeter is connected to terminal B.

a) What are the measured potential differences at terminals A to D in the arrangement shown?

b) What would the potential differences be if A were chosen as the reference node?

c) What does the voltmeter display if point A is connected to terminal (+) and point D to terminal 0/COM?

d) How was the voltmeter connected if $U=3.6\mathrm{V}$ is displayed?

◀️ Back to the task

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Solution

If you connect the voltmeter between A (to +) and B (to 0/COM), it displays the voltage $U_{AB}$, i.e. the potential difference between the reference potential at 0/COM (here: ${\phi }_B$) and the electric potential at the + terminal (e.g. ${\phi }_A$): $U_{AB}={\phi }_A-{\phi }_B$

a)

$U_{AB}={\phi }_A-{\phi }_B=-U-0=-U=-1.2\mathrm{V}=\underline{-1.2\mathrm{V}}$
$U_{BB}={\phi }_B-{\phi }_B=\underline{0}$
$U_{CB}={\phi }_C-{\phi }_B=U-0=U=\underline{1.2\mathrm{V}}$
$U_{DB}={\phi }_D-{\phi }_B=2U-0=2U=2\cdot 1.2\mathrm{V}=\underline{2.4\mathrm{V}}$

b)

with ${\phi }_A=0$:
$U_{AA}={\phi }_A-{\phi }_A=\underline{0}$
$U_{BA}={\phi }_B-{\phi }_A=\underline{1.2\mathrm{V}}$
$U_{CA}={\phi }_C-{\phi }_A=\underline{2.4\mathrm{V}}$
$U_{DA}={\phi }_D-{\phi }_A=\underline{3.6\mathrm{V}}$

c)

$U_{AD}={\phi }_A-{\phi }_D=-3U=\underline{-3.6\mathrm{V}}$

d)

$U=3.6\mathrm{V}=-U_{AD}=U_{DA}={\phi }_D-{\phi }_A$ with ${\phi }_D=3U$ and ${\phi }_A=0$

The voltage determined corresponds to the potential difference between the measuring point D and the reference potential A. The voltmeter was therefore connected with + to D and 0/COM to A.