ELEC1200 Problem Sheet 6
Kier Davis
December 3, 2015
Question 1
As given in the question, we know:
R = 50 Ω
L = 7 × 10−7 H
ω = 2π × 1 × 108 = 6.28 × 108 rad s−1
Therefore the impedance of the resistor is:
ZR (jω) = R = 50 Ω
and that of the inductor is:
ZL (jω) = jωL
= j × 6.28 × 108 × 7 × 10−7
= 440j Ω
Combining the two impedances in parallel gives:
1
1
+
Ztotal (jω)
ZR (jω) ZL (jω)
1
1
+
=
50 440j
= 0.02 − 0.00227j
1
=
Ztotal (jω) = (0.02 − 0.00227j)
−1
= 49.4 + 5.61j Ω
Question 2
The voltage across the supply vsupply (t) can be modelled using a phasor as follows:
vsupply (t) = 2 sin (1000t)
= Im 2ej×1000t
= Im ejωt · Vsupply (jω)
where ω = 1000 rad s−1 and the supply voltage phasor Vsupply (jω) = 2 V.
The voltage vR (t) across the resistor and vC (t) across the capacitor can be modelled with similar phasors:
vR (t) = Im ejωt · VR (jω)
vC (t) = Im ejωt · VC (jω)
The impedance of the resistor is:
ZR (jω) = R = 1000 Ω
and the impedance of the capacitor is:
ZC (jω) = −
1
j = −500j Ω
ωC
1
It is easy to see from Kirchoff’s current law that a single current with phasor I(jω) flows in the circuit, and
from Kirchoff’s voltage law we know that:
Vsupply (jω) = VR (jω) + VC (jω)
Applying the general impedance law V (jω) = Z(jω)I(jω) gives us:
Zsupply (jω)I(jω) = ZR (jω)I(jω) + ZC (jω)I(jω)
Zsupply (jω) = ZR (jω) + ZC (jω)
= 1000 − 500j Ω
Since we know that Vsupply (jω) = 2 V, we can find the current phasor I(jω) as follows:
I(jω) =
Vsupply (jω)
2
=
= 0.0016 + 0.0008j A
Zsupply (jω)
1000 − 500j
Then the capacitor volage phasor VC (jω) can be found by re-applying the impedance law:
VC (jω) = ZC (jω)I(jω)
= (−500j) (0.0016 + 0.0008j)
= 0.4 − 0.8j V
and so:
vC (t) = Im ejωt · (0.4 − 0.8j)
= Im ejωt · 0.894e−1.11j
n
o
= Im 0.894ej(ωt−1.11)
= 0.894 sin (ωt − 1.11) V
Question 3
The given voltage v(t) across the load can be written using a phasor as follows:
v(t) = 5 sin 105 t
n
o
5
= Im 5ej (10 t)
n
o
5
= Im ej (10 t) · V (jω)
where ω = 105 rad s−1 and the voltage phasor V (jω) = 5 V.
The phasor I(jω) of the current through the load is found similarly:
π
i(t) = 0.5 sin 105 t +
6 o
n
5
π
= Im 0.5ej (10 t+ 6 )
n
o
5
= Im ej (10 t) · I(jω)
π
where I(jω) = 0.5ej 6 A.
Hence the impedance of the load Z(jω) is given by:
V (jω)
I(jω)
5
=
π
0.5ej 6
π
= 10ej (− 6 )
Z(jω) =
= 8.66 − 5.00j Ω
2
An example of a circuit with this impedance would be a 8.66 Ω resistor in series with a capacitor of capacitance
1
C. C is found by equating the imaginary part of Z(jω) with the impedance of the capacitor − ωC
j as follows:
−
1
j = −5.00j
105 C
1
= 5.00 × 105
C
C = 2 × 10−6 F
= 2 µF
Question 4
Given the voltage v(t), we first find the voltage phasor V (jω):
v(t) = 10 cos (ωt + 0.436)
o
n
= Re 10ej(ωt+0.436)
= Re ejωt · V (jω)
where V (jω) = 10ej0.436 V.
Therefore,
1 + 2j
× 10ej0.436
0.2 + 0.6j
= (3.5 − 0.5j) × 10ej0.436
I(jω) =
= 3.54ej(−0.142) × 10ej0.436
= 35.4ej0.294 A
Translating back to the time domain gives a result of:
i(t) = Re ejωt · I(jω)
= Re ejωt · 35.4ej0.294
n
o
= Re 35.4ej(ωt+0.294)
= 35.4 cos (ωt + 0.294)
Question 5
As shown in the figure above, the circuit is equivalent to a single load consisting of a combination of
fundamental components, with a voltage v(t) = 100 cos (314t) applied across it.
The voltage phasor can be derived as follows:
v(t) = 100 cos (314t)
n
o
= Re 100ej(314t)
n
o
= Re ej(314t) · V (jω)
3
giving ω = 314 rad s−1 and the voltage phasor V (jω) = 100 V.
The impedances of the four components are:
ZR1 = R1
ZR2 = R2
1
ZC =
ωCj
ZL = ωLj
First, ZR2 and ZL are added to get the impedance of the inner series combination Zseries :
Zseries = ZR2 + ZL
= R2 + ωLj
Next, the impedances ZC and Zseries are combined in parallel to get Zparallel :
Zparallel =
=
=
1
1
ZC
+
1
Zseries
1
ωCj +
1
R2 +ωLj
R2 + ωLj
(1 − ω 2 LC) + ωR2 Cj
Finally, ZR1 and Zparallel are added to get the overall impedance Zload :
Zload = ZR1 + Zparallel
R2 + ωLj
= R1 +
(1 − ω 2 LC) + ωR2 Cj
Substituting in our component values and the value of ω = 314 rad s−1 gives Zload = 1500 − 0.126j Ω.
We can now find the phasor of the current through the load:
V (jω)
Zload
100
=
1500 − 0.126j
= 0.0667 + 5.58 × 10−6 j
I(jω) =
−5
= 0.0667ej (8.37×10 ) A
And so the current i(t) through the load is:
n
o
i(t) = Re ej(314t) · I(jω)
n
o
−5
= Re ej(314t) · 0.0667ej (8.37×10 )
o
n
−5
= Re 0.0667ej (314t+8.37×10 )
= 0.0667 cos 314t + 8.37 × 10−5
4
Question 6
As shown in the figure above, the circuit is equivalent to a single load consisting
of a combination of
fundamental components, with a voltage vsupply (t) = 10 cos 106 t + 0.349 applied across it.
The supply voltage phasor can be derived as follows:
vsupply (t) = 10 cos 106 t + 0.349
n
o
6
= Re 10ej (10 t+0.349)
n
o
6
= Re ej (10 t) · V (jω)
giving ω = 106 rad s−1 and the supply voltage phasor Vsupply (jω) = 10ej0.349 V.
The impedances of the four components are:
ZR1 = R1
ZR2 = R2
1
ZC =
ωCj
ZL = ωLj
The impedance Zparallel of the parallel combination of C and R2 is given by:
1
+ Z1R2
1
=
ωCj + R12
Zparallel =
=
1
ZC
R2
1 + ωR2 Cj
Thus the impedance Zload of the whole load is found by adding ZR1 , ZL and Zparallel :
Zload = ZR1 + ZL + Zparallel
R2
= R1 + ωLj +
1 + ωR2 Cj
Substituting in our component values gives us a value of Zload = 3000 + 1000j = 3160ej0.322 Ω.
We can now find the current phasor Iload (jω):
Vsupply (jω)
Zload
10ej0.349
=
3160ej0.322
= 0.00316ej0.0273 A
Iload (jω) =
As a result of Kirchoff’s current law, the current through the parallel combination Iparallel (jω) is equal to the
current through the overall load Iload (jω).
5
Additionally, Kirchoff’s voltage law tells us that the voltage VR2 (jω) across the resistor R2 is equal to the
voltage Vparallel (jω) across the parallel combination.
Therefore, we can find the phasor of the voltage across R2 as follows:
VR2 (jω) = Vparallel (jω)
= Zparallel Iparallel (jω)
R2
=
Iload (jω)
1 + ωR2 Cj
= 4470ej(−1.11) × 0.00316ej0.0273
= 14.1ej(−1.08) V
So the voltage v(t) across R2 is:
n
o
6
v(t) = Re ej (10 t) · VR2 (jω)
n
o
6
= Re ej (10 t) · 14.1ej(−1.08)
n
o
6
= Re 14.1ej (10 t−1.08)
= 14.1 cos 106 t − 1.08
6