Exercise EE1-03.01 – Voltage drop across the internal resistance (Solution)

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

Relates to EE1-03 Sources

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Task

On a real voltage source, you measure a terminal voltage of $U_0=10\mathrm{V}$ without a load. You now connect a resistive load and find that the terminal voltage dips by $2\%$. You look at the load and see the marking $U_{\mathrm{rated}}=20\mathrm{V}$, $P_{\mathrm{rated}}=4.0\mathrm{W}$.

a) Calculate the resistance of the connected load!
b) Determine the terminal voltage under load!
c) Determine the value of the internal resistance of the voltage source!
d) What power is converted in the load?

Solution

a) $100\Omega$
b) $9.8\mathrm{V}$
c) $2.0\Omega$
d) $0.96\mathrm{W}$

▶️ Full worked solution

◀️ Back to the task

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Solution

Given

Given explicitly:

• Open-circuit voltage of the source: $U_0=10\mathrm{V}$
• Voltage dip under load: $2\%$
• Ratings of the load: $U_{\mathrm{rated}}=20\mathrm{V}$, $P_{\mathrm{rated}}=4.0\mathrm{W}$

Known:

• Linear voltage source with internal resistance: $U=U_0-R_i\cdot I$
• Ohm’s law: $U=R\cdot I$
• Power: $P=U\cdot I=\frac{U^2}{R}$

Required

a) Load resistance $R_L$ in $\Omega$
b) Terminal voltage under load $U_t$ in $\mathrm{V}$
c) Internal resistance $R_i$ in $\Omega$
d) Power in the load $P_L$ in $\mathrm{W}$

a) Load resistance

The resistance of the load follows from the ratings:

$$R_L=\frac{U_{\mathrm{rated}}^2}{P_{\mathrm{rated}}}=\frac{(20\mathrm{V}{)}^2}{4.0\mathrm{W}}=\frac{400{\mathrm{V}}^2}{4.0\mathrm{W}}=\underline{100\Omega }$$

b) Terminal voltage under load

With a voltage dip of $2\%$, the terminal voltage is:

$$U_t=U_0\cdot (1-0.02)=10\mathrm{V}\cdot 0.98=\underline{9.8\mathrm{V}}$$

c) Internal resistance of the source

The current through the load at the terminal voltage:

$$I=\frac{U_t}{R_L}=\frac{9.8\mathrm{V}}{100\Omega }=0.098\mathrm{A}$$

The voltage drop across the internal resistance:

$$\Delta U=U_0-U_t=10\mathrm{V}-9.8\mathrm{V}=0.2\mathrm{V}$$

The internal resistance according to Ohm’s law:

$$R_i=\frac{\Delta U}{I}=\frac{0.2\mathrm{V}}{0.098\mathrm{A}}=2.041\Omega \approx \underline{2.0\Omega }$$

Check via the voltage divider:

$$U_t=U_0\cdot \frac{R_L}{R_L+R_i}=10\mathrm{V}\cdot \frac{100\Omega }{100\Omega +2.04\Omega }=10\mathrm{V}\cdot 0.98=9.8\mathrm{V}$$

✓

d) Power in the load

The actual power in the load at the terminal voltage:

$$P_L=\frac{U_t^2}{R_L}=\frac{(9.8\mathrm{V}{)}^2}{100\Omega }=\frac{96.04{\mathrm{V}}^2}{100\Omega }=\underline{0.96\mathrm{W}}$$

Alternatively via current and voltage:

$$P_L=U_t\cdot I=9.8\mathrm{V}\cdot 0.098\mathrm{A}=0.96\mathrm{W}$$

✓

The load absorbs $0.96\mathrm{W}$, although it is designed for $4.0\mathrm{W}$ at $20\mathrm{V}$.

Internal resistance and maximum power transfer

The internal resistance of a real source limits the power it can deliver. In the present case, with $R_i=2.0\Omega$ and $R_L=100\Omega$, there is no maximum power transfer (this would be the case for $R_L=R_i$).

The load operates far below its rated power, since it is operated at only half its rated voltage. For resistive loads, the power falls with the square of the voltage: $P\sim U^2$.