Exercise 2.7.9 - Variation: Simplifying Circuits II (written exam task, approx 8% of a 60-minute written exam, WS2020)
Given is the adjoining circuit with
$R_1=10 ~\Omega$
$R_2=20 ~\Omega$
$R_3=5 ~\Omega$
1. Determine the equivalent resistance $R_{\rm eq}$ between A and B by summing the resistances.
\begin{align*}
R_{ges} &= \frac{20 ~\Omega \cdot 5 ~\Omega}{\frac{3}{2}\cdot 20 ~\Omega + 5 ~\Omega} = 2.858 ~\Omega \rightarrow 2.9 ~\Omega
\end{align*}
2. Now let the voltage from A to B be: $U_{\rm AB}=U_0= 10 ~\rm V$. What is the current $I$?
\begin{align*}
I =\frac{10~\rm V}{2 \cdot 5 ~\Omega} = 1 ~\rm A
\end{align*}