Exercise EE1-03.03 – Operating point of a source (Solution)

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

Relates to EE1-03 Sources

Task

Determine the operating point of a linear source ($U_0=9.0\mathrm{V}$, $R_{\mathrm{i}}=1.0\Omega$) connected to a load with $R_{\mathrm{L}}=10\Omega$ graphically and by calculation.

Solution

Operating point: $I_A\approx 0.82\mathrm{A}$, $U_{\mathrm{op}}\approx 8.2\mathrm{V}$.

▶️ Full worked solution

◀️ Back to the task

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Solution

Given

Given explicitly:

• Open-circuit voltage: $U_0=9.0\mathrm{V}$
• Internal resistance: $R_i=1.0\Omega$
• Load resistance: $R_L=10\Omega$

Known:

• Thevenin equivalent: voltage source $U_0$ in series with $R_i$
• Characteristic curves:
  • Source: $U(I)=U_0-R_{i}I$
  • Load: $U(I)=I{R}_L$

Required

Operating point $(I_A,U_{\mathrm{op}})$ of the linear source at $R_L$ – graphically and by calculation.

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Characteristic curves

Parameters of the linear source:

$$U_0=9\mathrm{V}$$ $$I_{\mathrm{sc}}=\frac{U_0}{R_i}=\frac{9\mathrm{V}}{1\Omega }=9\mathrm{A},U(I)=9\mathrm{V}-1\Omega \cdot I$$

Load characteristic curve:

$$U(I)=I\cdot R_L=10\Omega \cdot I$$

Reference points (for the graph):

• Source characteristic curve: $(I=0,U=9\mathrm{V})$ and $(I=9\mathrm{A},U=0\mathrm{V})$
• Load curve: $(I=0,U=0)$ and $(I=0.9\mathrm{A},U=9\mathrm{V})$

c) Operating point by calculation

Intersection of the lines: $U_0-R_{i}I=I{R}_L$

$$9-1\cdot I=10\cdot I\Rightarrow 9=11I\Rightarrow I_A=\frac{9}{11}=0.818\mathrm{A}\approx \underline{0.82\mathrm{A}}$$ $$U_{\mathrm{op}}=I_A{R}_L=0.818\mathrm{A}\cdot 10\Omega =8.182\mathrm{V}\approx \underline{8.18\mathrm{V}}$$

(Check: $U_{\mathrm{op}}=U_0-I_A{R}_i=9-0.818\cdot 1=8.182\mathrm{V}$ ✓)

d) Operating point graphically

The intersection of the two lines ($U=9-1\cdot I$ and $U=10\cdot I$) in the $I$–$U$ diagram gives:

$$(I_A,U_{\mathrm{op}})\approx (0.8\mathrm{A},8.2\mathrm{V})$$