a) • z 2
1
+ z
2
· z
3
= − 46 .
83 + j 40 .
17 = 61 .
7 ∠ 139 .
4 ◦
• z
1
· z ∗
2
· z
3
= 53 .
30 − j 7 .
68 = 53 .
85 ∠ − 8 .
2 ◦
• z −
1
1 · ℑ ( z
2
) = 0 .
052 − j 0 .
121 = 0 .
1313 ∠ − 66 .
8 ◦
Please note that ℑ ( z
2
) = 1.
b)
R = | Z | cos ϕ = 500 Ω · 0 .
85 = 425 Ω
ωL = | Z | sin ϕ = 500 Ω · sin (arccos 0 .
85) = 263 .
4 Ω
ω = 2 πf = 100 π
1 s
→ L = 0 .
838 H
a) b = 325 .
27; T = 0 .
02 s → ω = 2 π
T
= 100 π s −
1 ; φ = 90 ◦ b) X = 0; X = b
√
2
= 230 c) X = 230 ( ℑ ( X ) = 0) d) The frequency has no influence on the RMS value.
e) When superimposing a positive offset, the RMS value increases.
(1)
(2)
(3)
a) Z s
= R + j
¡
ωL − 1
ωC
¢
= 140 Ω − j 96 .
1 Ω = 169 .
8 Ω ∠ − 34 .
5 ◦ b) Y p
=
1
Z p
=
1
R
+
1 jωL
+ jωC = 7 .
1 · 10 −
3 S − j 2 .
64 S = 2 .
64 S ∠ − 89 .
8 ◦
1

a) The impedance Z ab is purely ohmic if ℑ ( Z ab
) = 0. At first, we calculate the impedance between the terminals a and b :
Z ab
=
(
R
R +
+ j jωL
µ
)
¡
− j 1
¢
ωC
1
¶
ωL −
ωC
| {z
N
}
(4)
After multiplying the numerators, we obtain an expression of which we have to calculate the imaginary part and set this to zero. In order to do so, we have to reduce to higher terms with the complex conjugate: We multiply the fraction in (4) by N ∗
N ∗
, which gives us a real denominator
( N N ∗ ∈ R ):
Z ab
=
−
R
ωC
¡
ωL − 1
ωC
¢
R 2
³
+
RL
+
C
¡
− j
ωL −
R 2
ωC
1
ωC
¢
+
2
L
C
¡
ωL − 1
ωC
¢´
(5)
Now we can calculate the imaginary part of this expression and set it to zero. From that we obtain an equation for the capacitance:
C =
L
R 2 + ω 2 L 2
=
0 .
05
100 2 + (100 π ) 2 · 0 .
05 2
= 4 .
88 µ F (6)
The same result can be obtained by setting the imaginary part of the admittance to zero.
b) When one increases the frequency, the resistance of the inductance( ωL ) increases while the resistance of the capacitance (
1
ωC
) decreases. At the same voltage, a higher current will then flow through C and a lower current through L . Therefore, the impedance between a and b will have a capacitive character. Figure 1 shows the imaginary part of
Z ab as a function of the frequency.
a) At first, we calculate the voltage at the inductance: u ( t ) = L di ( t ) dt
= ωL I b cos ( ωt ) (7)
The instantaneous power results from the product of current and voltage; the amplitude of the oscillating instantaneous power is b =
ωL I b 2
2
=
100 π · 0 .
0009 · 247 2
2
= 8 .
62 kW (8)
2

0.5
0
−0.5
−1
−1.5
−2
−2.5
10 20 30 40 50 f in Hz
60 70 80 90 100
Figure 1: Behaviour of the imaginary part of Z ab quency f . For f < 50 Hz: ℑ ( Z ab
) > 0, i.e.
Z ab as a function of the freis inductive. For frequencies above 50 Hz: ℑ ( Z ab
) < 0, Z ab capacitive. At 50 Hz, the capacitance and inductance compensate each other.
b) Q
C
= U 2 /X
C
= U 2 ωC . The reactive power generated by the capacitance increases by 20%.
c) After an approximate graphical integration of the area under the power curve and division by about 7.5 million inhabitants, we obtain the average energy consumption per capita on this day:
E pk
≈ (8 .
5 GW · 24 h ) / 7 .
5 million = 27 .
2 kWh.
3