Problem + Hard 33-1 Source:
Question 1 - 1st test -
Electric Circuit Course I - School of Engineering - UFRGS - 1976.
In the circuit below determine the parameters "h" and find the
values of K and R, knowing that:
a) If I1 = 35A and E2 = 10V so I2 = 1A
b) If I1 = - 30A and E2 = 24V then I = 1A.
Solution of the Problem + Hard 33-1
Analyzing the 1st Test
a) If I1 = 35A and
E2 = 10V so I2 = 1A.
For the first test, the given conditions are shown in the figure below, with
the indication of the currents in the circuit. The resistors that
were in parallel, were replaced by their respective equivalents.
Making the C mesh, indicated by the green arrow, the following equation is established:
- 4 I - 2 + 10 = 0
Carrying out the calculation, the value of I is found, or:
I = 2 A
And making the mesh A, indicated by the green arrow:
V1 = 32 (2 R) / (2 + R) = (64 R) / (2 + R) ⓵
And finally, making the mesh A + B, we find the following equation:
V1 = 2 K + 12 + 8 = 2 K + 20 ⓶
Considering the two equations for V1 and equating ⓵ a
⓶, we find the following relationship:
(64 R) / (2 + R) = 2 K + 20 ⓷
Analyzing the 2nd Test
a) If I1 = - 30A and
E2 = 24V so I = 1A.
The circuit shown in the figure below, expresses
the conditions of the second test, including the flow and values of the currents in the circuit.
The value of I2 is found by doing the mesh equation C. Note
that, according to test data, the value of I is 1A. Thus, in the resistance of
2 ohms flows a current of 2A, and this causes a potential difference of
4 volts on it. This way:
- 4 - 2 I2 + 24 = 0 or I2 = 10 A
Making the mesh A, indicated by the green arrow, the following relationship is established:
V1 =-22 (2 R)/(2 + R) = -(44 R) / (2 + R) ⓸
Using the same reasoning for the B mesh, indicated by the green arrow, we obtain the following
relationship:
V1 = K - 32 + 4 = K - 28 ⓹
Repeating the process done previously, and equating ⓸ to
⓹, you get:
- (44 R) / (2 + R) = K - 28 ⓺
Repeating the equation ⓷, found earlier,
a system of two equations with two unknowns is achieved, when used in conjunction with ⓺.
(64 R) / (2 + R) = 2 K + 20 ⓷
Solving the system we easily find the values of K and R. Soon:
R = 2
K = 6
With the values of K and R determined, from this moment on, you can
calculate the parameters "h" of the circuit. These parameters have the following equations:
V1 = h11 I1 + h12 V2
I2 = h21 I1 + h22 V2
h11 and h21 are calculated by doing
E2 = 0. Therefore, if E2 = 0, that is,
short circuit at the output, the resistance of
2 ohms where I2 circulates, it is in parallel with the other
resistance of 2 ohms, through which a current of 2 I flows.
It follows that I2 = - 2 I, as they are under the same
potential difference. Based on this conclusion, the current arriving at point c
is 4 I. See the figure below for the information indicated in the circuit.
Setting V1 = 26 volts, calculate the current I, as:
- 26 + 6 I + 16 I + 4 I = 0
So the value of I is:
I = 1 A
From this data, it is concluded that:
I1 = 30 A
I2 = - 2 A
At this moment, one can calculate h11 and
h21. Then:
h11 = V1 / I1 = 26 / 30 = 13 / 15 Ω
h21 = I2 / I1 = - 2 / 30 = - 1 / 15
To determine the parameters h12 and h22,
you must make the current I1 = 0.
A current source of 12A is placed at the output. Like this,
I2 = 12A. Making the mesh in the direction indicated by the green arrow,
the following relationship is established:
-I2 + 2 I + 6 I - 4 I2 + 8 I + 4 I = 0
Remembering that I2 = 12A and performing the calculation determines the value
from I. Soon:
I = 3 A
Now you can calculate the values of V1
and V2.
V1 = 1 (I2 - 2 I) = 6 V
In turn, V2 holds:
V2 = 2 I2 + 4 I = 36 V
All data needed to calculate h11 and
h21 were calculated. Then:
h12 = V1 / V2 = 6 / 36 = 1 / 6
h22 = I2 / V2 = 12 / 36 = 1 / 3 siemens
Therefore, the quadripole equations are as follows:
