V
1
:
=
10
V
{\displaystyle {\mathit {V1}}\mathrm {\colon } =10V}
R
1
:
=
1000
Ω
{\displaystyle {\mathit {R1}}\mathrm {\colon } =1000\Omega }
I
:
=
V
1
R
1
=
0.01000
A
{\displaystyle I\mathrm {\colon } ={\frac {\mathit {V1}}{\mathit {R1}}}={\text{0.01000}}A}
m
A
:
=
0.001
A
{\displaystyle {\mathit {mA}}\mathrm {\colon } =0.001A}
I
:
=
V
1
R
1
=
10.00
m
A
{\displaystyle I\mathrm {\colon } ={\frac {\mathit {V1}}{\mathit {R1}}}={\text{10.00}}{\mathit {mA}}}
V
1
:
=
10
V
{\displaystyle {\mathit {V1}}\mathrm {\colon } =10V}
R
g
e
s
:
=
R
1
+
R
2
=
3 300
Ω
{\displaystyle {\mathit {Rges}}\mathrm {\colon } ={\mathit {R1}}+{\mathit {R2}}={\text{3 300}}\Omega }
I
:
=
V
1
R
g
e
s
=
3.030
m
A
{\displaystyle I\mathrm {\colon } ={\frac {\mathit {V1}}{\mathit {Rges}}}={\text{3.030}}{\mathit {mA}}}
U
R
1
:
=
I
⋅
R
1
=
3.030
V
{\displaystyle {U}_{\mathit {R1}}\mathrm {\colon } =I\cdot {\mathit {R1}}={\text{3.030}}V}
U
R
2
:
=
I
⋅
R
2
=
6.970
V
{\displaystyle {U}_{\mathit {R2}}\mathrm {\colon } =I\cdot {\mathit {R2}}={\text{6.970}}V}
Zur grafischen Lösung kann man die Widerstandsgerade von R2 in das Diagramm von R1 spiegelverkehrt eintragen und am Schnittpunkt die Werte ablesen. Der Schnittpunkt wird Arbeitspunkt genannt.
