Exercise 1 : Pure Resistor Network Simplification
(written test, approx. 13% of a 60-minute written test, WS2022)

The following circuit with $R_1=200 \Omega$, $R_2=R_3=100 \Omega$ and the switch $S$ is given.

1. The switch shall now be open. Calculate the equivalent resistance $R_{eq}$ between $A$ and $B$.

Solution

With the switch open the resistor $R_3$ dies not take part into the resulting resistor. The equivalent resistor is given by a parallel configuation of resistors in series: \begin{align*} R_{eq} &= (R_2 + R_1 + R_1)||(R_2 + R_2)\\ R_{eq} &= (100 \Omega + 200 \Omega + 200 \Omega )&&||(100 \Omega + 100 \Omega ) &&\\ R_{eq} &= (500 \Omega )&&||(200 \Omega )&&\\ R_{eq} &= {{500 \Omega \cdot 200 \Omega }\over{500 \Omega + 200 \Omega}}&&\\ \end{align*}

Final result

2. The switch shall now be closed. Calculate the equivalent resistance $R_{eq}$ between $A$ and $B$.

Solution

Now a wye-delta transformation is necessary. Since $R_2=R_3$ and based on the equations for the transformation, the transformed $R_Y$ is given as: \begin{align*} R_{Y} &= {{R_2 \cdot R_2}\over{R_2 + R_2 + R_2}} \\ &= {{(100 \Omega)^2}\over{3 \cdot 100 \Omega}} \\ &= {{1}\over{3}} \cdot 100 \Omega = 33.33 \Omega \end{align*}

The equivalent resistor is given by a parallel configuation of resistors in series: \begin{align*} R_{eq} &= R_Y + (R_Y + R_1 + R_1)||(R_Y + R_2)\\ R_{eq} &= 33.33 \Omega + (33.33 \Omega + 400 \Omega)||(33.33 \Omega + 100 \Omega)\\ \end{align*}

Final result