Exercise EE1-03.02 – D cell (Solution)

Prof. Dr. Thorsten Jungmann (last updated 2025-10-27)

Relates to EE1-03 Sources

Task

A D cell has the open-circuit voltage $U_0=1.5\mathrm{V}$ and the short-circuit current $I_{\mathrm{sc}}=1.9\mathrm{A}$. Which terminal voltage $U_{\mathrm{t}}$ results when the current $I=0.19\mathrm{A}$ is drawn from it?

a) Sketch the equivalent circuits for the linear voltage source and the linear current source.

b) Calculate the elements of the equivalent circuits.

c) Draw the I-U characteristic curve of the linear source.

d) Determine the terminal voltage by calculation as specified in the task.

e) Determine the terminal voltage graphically as specified in the task and compare the result with the result of the calculation in d).

Solution

a) Sketches
b) $0.79\Omega$ / $1.3\mathrm{S}$
c) Characteristic curve
d) $1.35\mathrm{V}$
e) approximately agrees

▶️ Full worked solution

◀️ Back to the task

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Solution

Given

Given explicitly:

• Open-circuit voltage: $U_0=1.5\mathrm{V}$
• Short-circuit current: $I_{\mathrm{sc}}=1.9\mathrm{A}$
• Current drawn: $I=0.19\mathrm{A}$

Known:

• Internal resistance (linear): $R_i=\frac{U_0}{I_{\mathrm{sc}}}$
• Thevenin/Norton equivalence: voltage source $U_0$ in series with $R_i$ $\iff$ current source $I_{\mathrm{sc}}$ in parallel with $R_i$
• Characteristic curve: $U(I)=U_0-R_{i}I$

Required

a) Sketches of the equivalent circuits (Thevenin/Norton)
b) Elements of the equivalent circuits
c) I-U characteristic curve of the source
d) Terminal voltage $U_{\mathrm{t}}$ by calculation for $I=0.19\mathrm{A}$
e) $U_{\mathrm{t}}$ graphically and comparison with d)

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a) Equivalent circuits

Thevenin (linear voltage source): $U_0$ in series with $R_i$

Norton (linear current source): $I_{\mathrm{sc}}$ in parallel with $G_i$

b) Calculating the elements

Internal resistance:

$$R_i=\frac{U_0}{I_{\mathrm{sc}}}=\frac{1.5\mathrm{V}}{1.9\mathrm{A}}=0.789\dots \Omega \approx \underline{0.79\Omega }$$

Internal conductance:

$$G_i=\frac{I_{\mathrm{sc}}}{U_0}=\frac{1.9\mathrm{A}}{1.5\mathrm{V}}=1.266\dots \mathrm{S}\approx \underline{1.3\mathrm{S}}$$

c) I-U characteristic curve

Equation of the straight line (voltage form):

$$U(I)=U_0-R_{i}I=1.5\mathrm{V}-0.789\Omega \cdot I$$

Reference points: $(I=0,U=1.5\mathrm{V})$ and $(I=1.9\mathrm{A},U=0\mathrm{V})$.

d) Terminal voltage by calculation

For $I=0.19\mathrm{A}$:

$$U_{\mathrm{t}}=U_0-R_{i}I=1.5\mathrm{V}-\frac{1.5\mathrm{V}}{1.9\mathrm{A}}\cdot 0.19\mathrm{A}=1.5\mathrm{V}-0.15\mathrm{V}=\underline{1.35\mathrm{V}}$$

(Check: $I$ is $10\%$ of $I_{\mathrm{sc}}$ ⇒ voltage drop $=10\%$ of $U_0$.)

e) Terminal voltage graphically and comparison

Read off from the straight line $U(I)$ at $I=0.19\mathrm{A}$:

$$U_{\mathrm{t}}\approx 1.3\mathrm{V}$$

Comparison: approximately agrees with d) (inaccuracy of the graphical reading)