The circuit below shall be given.
The values in the circuit are
$R_1 = 60 ~\Omega$
$R_2 = 40 ~\Omega$
$R_3 = 40 ~\Omega$
$R_4 = 100 ~\Omega$
$U_{\rm AB} = 10 ~\rm V$
1. Calculate the voltage at node $\rm K$, when switch $\rm S$ is open.
It might be beneficial to redraw the circuit first.
Solution
Rearranging the circuit one can get:
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
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*}