VOLTAGE AND CURRENT DIVISION
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We have already shown how the
elementary methods of DC circuit analysis can be extended and used in AC
circuits to solve for the complex peak or effective values of voltage and
current and for complex impedance or admittance. In this chapter, we�ll solve
some examples of voltage and current division in AC circuits.
Example
1
Find
the voltages v1(t) and v2(t), given that vs(t)=110cos(2p50t).
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here to load or save this circuit
Let�s
first obtain this result by hand calculation using the voltage division formula.
The
problem can be considered as two complex impedances in series: the impedance of
the resistor R1, Z1=R1
ohms (which is a real number), and the equivalent impedance of R2 and
L2 in series, Z2
= R2 + j w
L2.
Substituting
the equivalent impedances, the circuit can be redrawn in TINA as follows:
Note
that we have used a new component, a complex impedance, now available in TINA
v6. You can define the frequency dependence of Z by means of a table that you
can reach by double clicking the impedance component. In the first row of the
table you can define either the DC impedance or a frequency independent complex
impedance (we have done the latter here, for the inductor and resistor in
series, at the given frequency).
Using
the formula for voltage division:
V1
= Vs*Z1 / (Z1
+ Z2)
V2
= Vs*Z2 / (Z1
+ Z2)
Numerically:
Z1
= R1 = 10 ohms
Z2
= R2 + j w
L = 15 + j 2*p*
50*0.04 =15 + j 12.56 ohms
V1=
110*10/ (25+j12.56) = 35.13-j17.65 V = 39.31 e �j26.7
°
V
V2=
110*(15+j12.56)/ (25+j12.56) = 74.86+j17.65 V
= 76.92 e j 13.3°
V
The
time function of the voltages:
v1(t)
= 39.31 cos (wt
- 26.7°)
V
v2(t)
= 76.9 cos (wt
+ 13.3°)
V
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V1 |
V2 |
Next let�s check these results with TINA�s Interpreter:
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{Solution
by TINA's Interpreter} |
Note
that when using the Interpreter we did not have to declare the values of the
passive components. This is because we are using the Interpreter in a work
session with TINA in which the schematic is in the schematic editor. TINA�s
Interpreter looks in this schematic for the definition of the passive component
symbols entered into the Interpreter program.
The
diagram shows that Vs is
the sum of the phasors V1 and V2,
Vs
= V1 + V2.
By moving the phasors we can also demonstrate that V2 is the difference between Vs and V1, V2 = Vs � V1.
This
figure also demonstrates the subtraction of vectors. The resultant vector should
start from the tip of the second vector, V1.
In a similar way we can demonstrate that V1 = Vs � V2. Again, the resultant vector should start from the tip of the second vector, V1.
Of
course, both phasor diagrams can be considered as a simple triangle rule diagram
for Vs =
V1 + V2 .
The
phasor diagrams above also demonstrate Kirchhoff�s voltage law (KVL).
As we have learned in our study of DC circuits, the applied voltage of a series circuit equals the sum of the voltage drops across the series elements. The phasor diagrams demonstrate that KVL is also true for AC circuits, but only if we use complex phasors!
Example
2
In
this circuit, R1 represents the DC resistance of the coil L; together
they model a real world inductor with its loss component. Find the voltage
across the capacitor and the voltage across the real world coil.
L = 1.32 h, R1 = 2 kohms, R2 = 4 kohms, C = 0.1 mF, vS(t) = 20 cos (wt) V, f = 300Hz.
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here to load or save this circuit
|
V1 |
V2 |
Solving
by hand using voltage division:
=
13.91 e j 44.1°
V
and
v1(t)
= 13.9 cos (w×t
+ 44°)
V
=
13.93 e �j 44.1°
V
and
v2(t)
= 13.9 cos(w×t
� 44.1°)
V
Notice
that at this frequency, with these component values, the magnitudes of the two
voltages are nearly the same, but the phases are of opposite sign.
Once
again, let�s have TINA do the tedious work by solving for V1 and V2
with the Interpreter:
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{Solution
by TINA's Interpreter!} |
Example
3
The
current source iS(t) = 5 cos (wt) A, the resistor R = 250 mohm, the inductor L = 53
uH, and the frequency f = 1 kHz.
Find the current in the inductor and the current in the resistor.
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here to load or save this circuit
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IR |
IL |
Using
the formula for current division:
iR(t)
= 4 cos (w×t
+ 37.2°)
A
Similarly:
iL(t)
= 3 cos(w×t
� 53.1°)
And using the Interpreter in
TINA:
|
{Solution
by TINA's Interpreter} |
The phasor diagram shows that the generator current IS is the resultant vector of the complex currents IL and IR. It also demonstrates Kirchhoff�s current law (KCL), showing that the current IS entering the upper node of the circuit equals the sum of IL and IR, the complex currents leaving the node.
Example
4
Determine i0(t), i1(t) and i2(t). The component values and the source voltage, frequency, and phase are given on the schematic below.
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here to load or save this circuit
i0
i1
i2
In our solution, we will use the principle of current division. First we
find the expression for the total current i0:
I0M = 0.315 e j
83.2°
A
and i0(t)
= 0.315 cos (w×t
+ 83.2°)
A
Then using current division, we find the current in the capacitor C:
I1M = 0.524 e j
91.4°
A
and i1(t) = 0.524 cos(w×t
+ 91.4°)
A
And the current in the inductor:
I2M = 0.216 e-j
76.6°
A
and i2(t)
= 0.216 cos(w×t
� 76.6°)
A
With anticipation, we seek confirmation of our hand calculations using TINA�s Interpreter.
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{Solution by TINA's Interpreter} |
VRLCM = 8.34 e j 1.42°
V
and
IC
=
I1= VRLCM*jwC
= 0.524 e j 91.42°
A
and hence
iC
(t) = 0.524 cos (w×t
+ 91.4°)
A.
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