EO204
Tutorial 3 and Tutorial 2 – Cramer’s rule
Q1.
Two mutually coupled coils are connected in series. Given the
following component values and properties, calculate the total
effective inductance when the induced voltages are (a) both in the
same direction and (b) in opposing directions. L1 = 200 mH, L2 = 400
mH, M = 100 mH.
Inductances are all summed if the induced voltages are in the same
direction, or M is subtracted if they are in opposing directions.
𝐿𝑡𝑜𝑡 = 𝐿1 + 𝐿2 ± 2𝑀
a)
𝐿𝑡𝑜𝑡 = 200 + 400 + 2 × 100 = 800𝑚𝐻
b)
𝐿𝑡𝑜𝑡 = 200 + 400 − 2 × 100 = 400𝑚𝐻
Q2, 3 and 4.
The circuit shown above contains an ideal transformer. The
vertical lines between coils indicate a magnetic core.
2) If the output current I2 is 5 mA (rms), find the turns ratio N2/N1.
Since the output current is 5𝑚𝐴, then the output voltage is:
𝑉𝑜 = 5𝑚𝐴 × 10𝑘Ω = 50𝑉𝑟𝑚𝑠
Since the input voltage is 10𝑉𝑟𝑚𝑠 and the transformer is ideal,
then the turns ratio, 𝑛 is given by:
𝑛=
𝑛2 50𝑉𝑟𝑚𝑠
=
=5
𝑛1 10𝑉𝑟𝑚𝑠
3) For this circuit, find also the value of input resistance, 𝑅𝑎𝑏 . To
do this you will need to use the expression for impedance
transformation.
𝑅𝑎𝑏
𝑛
𝑉𝑜 (𝑛1 )
𝑉1
𝑉𝑜 𝑛1 2
𝑛1 2
2
= =
= ( ) = 𝑅2 ( )
𝐼1 𝐼 (𝑛2 ) 𝐼2 𝑛2
𝑛2
2 𝑛
1
Therefore:
1
2
𝑅𝑎𝑏 = 10𝑘Ω × ( ) = 400Ω
5
4) Using the value of 𝑅𝑎𝑏 found in the previous question,
calculate the current supplied by the voltage source.
The current supplied by the voltage source is therefore:
𝐼1 =
10𝑉𝑟𝑚𝑠
= 25𝑚𝑉
400Ω
Tutorial 1 – Cramer’s rule Question 3)
(Week 2) Using mesh current analysis and a supermesh, find the
matrix equation to describe this circuit. Then use Cramer’s rule
to determine the current through the load impedance at 𝜔 = 𝜔0
using the following values:
𝑉𝑠 = 3 V, 𝐼𝑠 = 1 A, R = 3 Ω, ZL = 0.5(1+j3) Ω. Hint: eliminate C
1
from your matrix equation using 𝜔0 =
𝑅𝐶
We are given that:
𝜔0 =
1
𝑅𝐶
Using Kirchoff’s voltage law for each of the 3 meshes gives:
For mesh 1:
𝑗𝑉𝑠 = 𝐼1 𝑅 + (𝐼1 − 𝐼2 )
𝐼𝑠 = 𝐼2 − 𝐼3
1
𝑗𝜔𝐶
(1)
(2)
1
𝐼3 𝑍𝐿 + (𝐼2 − 𝐼1 )
𝑗𝜔𝐶
=0
(3)
Rearranging equations (1), (2) and (3) gives:
𝐼1 (𝑅 +
1
𝑗𝜔𝐶
) − 𝐼2 (
1
𝑗𝜔𝐶
) + 0𝐼3 = 𝑗𝑉𝑠
0𝐼1 + 𝐼2 − 𝐼3 = 𝐼𝑠
−𝐼1 (
1
𝑗𝜔𝐶
) + 𝐼2 (
(2)
1
𝑗𝜔𝐶
Substituting 𝜔0 =
) + 𝐼3 𝑍𝐿 = 0
1
𝑅𝐶
(3)
and arranging in matrix form gives:
𝑅 (1 − 𝑗
[
(1)
𝜔0
)
𝜔
0
𝜔0
𝑗
𝑅
𝜔
𝜔0
𝑅
𝜔
1
𝜔0
−𝑗
𝑅
𝜔
𝑗
0
𝐼1
𝑗𝑉𝑠
−1 [𝐼2 ] = [ 𝐼𝑠 ]
0
𝑍𝐿 ] 𝐼3
Substituting the values given of 𝑉𝑠 = 3𝑉, 𝐼𝑠 = 1𝐴, 𝑅 = 3Ω and 𝑍𝐿 =
1
2
(1 + 3𝑗) Ω, and setting 𝜔 = 𝜔0 gives the matrix:
[
3(1 − 𝑗)
0
3𝑗
1
3𝑗
−3𝑗
0
−1
𝐼1
3𝑗
] [𝐼2 ] = [ 1 ]
1
(1 + 3𝑗) 𝐼3
0
2
Using Cramer’s rule:
𝐼3 =
𝐷3
𝐷
𝐷3 is given by:
[
3(1 − 𝑗)
0
3𝑗
1
3𝑗
−3𝑗
0
−1
𝐼1
3𝑗
] [𝐼2 ] = [ 1 ]
1
(1 + 3𝑗) 𝐼3
0
2
3(1 − 𝑗)
𝐷3 = | 0
3𝑗
3𝑗
1
−3𝑗
3𝑗
1|
0
Therefore:
𝐷3 = 3(1 − 𝑗)(1 × 0 − (−3𝑗 × 1))
−3𝑗(0 × 0 − 1 × 3𝑗)
+3𝑗(0 × −3𝑗 − 1 × 3𝑗)
𝐷3 = 3(1 − 𝑗) × 3𝑗 + 3𝑗 × 3𝑗 − 3𝑗 × 3𝑗 = 𝟗(𝟏 + 𝒋)
𝐷 is given by:
3(1 − 𝑗)
0
𝐷=|
3𝑗
3𝑗
1
−3𝑗
0
−1
|
1
(1 + 3𝑗)
2
Therefore:
1
𝐷 = 3(1 − 𝑗) (1 × (1 + 3𝑗) − (−3𝑗 × −1))
2
1
−3𝑗 (0 × (1 + 3𝑗) − (−1 × 3𝑗))
2
+0(0 × −3𝑗 − 1 × 3𝑗)
1
𝐷 = 3(1 − 𝑗) × ( (1 + 3𝑗) − 3𝑗) − 3𝑗 × 3𝑗 + 0 = 𝟔(𝟏 − 𝒋)
2
Therefore:
𝐼3 =
𝐷3 9(1 + 𝑗)
=
= 1.5𝑗 = 𝟏. 𝟓 < 𝟗𝟎° 𝑨
𝐷
6(1 − 𝑗)