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Exercise E3 Equivalent Linear Source
(written test, approx. 10 % of a 60-minute written test, SS2023)

The circuit below has to be simplified. Use equivalent linear sources for simplification.
Calculate the internal resistance $R_\rm i$ and the source voltage $U_\rm s$ of an equivalent linear voltage source.

  • $R_1 = 5 ~\Omega$
  • $U_0 = 10 ~\rm V$
  • $R_2 = 7.5 ~\Omega$
  • $I_3 = 0.5 ~\rm A$
  • $R_4 = 10 ~\Omega$
  • $U_5 = 4 ~\rm V$

electrical_engineering_1:lefxcuaxiu8ewcr9circuit.svg

Solution

The principle idea here is to find parts of the circuit which are already a linear (voltage or current) source. Then this can be transformed into the equivalent other source, as shown in the next picture.

electrical_engineering_1:lefxcuaxiu8ewcr9circuitsolve1.svg

In order to get the currents one has to calculate it by $I_x = {{U_x}\over{R_x}}$ \begin{align*} I_0 &= {{U_0}\over{R_1}} = {{10~\rm V}\over{5 ~\Omega}} = 2 ~\rm A\\ I_5 &= {{U_5}\over{R_4}} = {{4 ~\rm V}\over{10~\Omega}} = 0.4 ~\rm A \\ \end{align*}

$I_3$ and $I_0$ can be combined to $I_{03}=I_0 - I_3$ facing upwards: \begin{align*} I_{03}=1.5~\rm A \end{align*}

Then, the linear current source $I_{03}$ with $R_1$ gets transformed into a linear voltage source with $U_{03}=R_1\cdot I_{03}$ facing down. \begin{align*} U_{03}=7.5~\rm V \end{align*}

Then, the resistors $R_1$ and $R_2$ can be combined to $R_{12}= R_1 + R_2$.

After this, the next step is to make a linear current source out of $U_{03}$ and $R_{12}$. The current will be $I_{0123}={{U_{03}}\over{R_{12}}}$, facing up again. \begin{align*} I_{0123}=0.6~\rm A \end{align*}

The second-last step is the sum up of the current sources $I_{0123}$ and $I_5$ as $I_{01235}=I_{0123}-I_5$ and the resistors as $R_{124}=R_{12}||R_4$. \begin{align*} I_{01235} &=0.2~\rm A \\ R_{124} &=5.55... ~\Omega \end{align*}

The final step is the back-transformation to a linear voltage source, with $U_{\rm AB} = R_{124} \cdot I_{01235}$.

The simplest and fastest (= for exams) is to work with interim results in the calculation.
Here, there there is also a full final formula given:

\begin{align*} U_\rm s &= U_{\rm AB} = I_{01235} \cdot R_{124} \\ &= (I_{0123}-I_5) \cdot (R_{12}||R_4) \\ &= \left({{U_{03}}\over{R_{12}}}-I_5\right) \cdot \left((R_1 + R_2)||R_4\right) \\ &= \left({{R_1\cdot I_{03}}\over{R_1+R_2}}-I_5\right) \cdot \left((R_1 + R_2)||R_4\right) \\ &= \left({{R_1\cdot \left({{U_0}\over{R_1}} - I_3\right)}\over{R_1+R_2}}-I_5\right) \cdot \left((R_1 + R_2)||R_4\right) \\ \end{align*}

Result

\begin{align*} U_\rm s &= U_{\rm AB} =1.11...~\rm V \\ R_{\rm i} &= 5.55... ~\Omega \end{align*}