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1 Pure Resistor Network Simplification
(written test, approx. 12 % of a 60-minute written test, SS2023)

The circuit below shall be given.

The values in the circuit are

  • $R_1 = 60 ~\Omega$
  • $R_2 = 40 ~\Omega$
  • $R_3 = 40 ~\Omega$
  • $R_4 = 100 ~\Omega$
  • $U_{\rm AB} = 10 ~\rm V$

electrical_engineering_1:cgeyprm6oboukccicircuit.svg

1. Calculate the voltage at node $\rm K$, when switch $\rm S$ is open.
It might be beneficial to redraw the circuit first.

Solution steps

Rearranging the circuit one can get: electrical_engineering_1:cgeyprm6oboukccicircuitsolve1.svg

Once the switch $\rm S$ is opened, the upper part is a parallel circuit. Therefore, $R_{\rm eq}$ is given as: \begin{align*} R_{\rm eq} &= (R_1+R_2)||(R_1+R_2)+R_4 \\ &= {{1}\over{2}}\cdot(R_1+R_2)+R_4 \\ &= {{1}\over{2}}\cdot(60~\Omega + 40~\Omega) + 100~\Omega \\ \end{align*}

Result

\begin{align*} R_{\rm eq} &= 150~\Omega \end{align*}

2. Calculate the voltage at node $\rm K$, when switch $\rm S$ is closed.

Solution steps

The voltage divider for node $\rm K$ has the same proportionality as the voltage divider for node $\rm K'$. Therefore, the potential of $\rm K$ is the same as for $\rm K'$. There will be no current flow through $R_3$. The resistance does not create a voltage drop and therefore does not interfere with the circuit.

Result

The equivalent resistance is similar to the circuit with opened switch. \begin{align*} R_{\rm eq} &= 150~\Omega \end{align*}