Problem + Hard 53-2
Source: exercise 8-21 - page 407 - THOMAS, Roland E.,
ROSA, Albert J. , THOUSSAINT, Gregory J. - Book: The Analysis & Design of Linear Circuits
- 6th Edition - Ed. John Willey & Sons, Inc. - 2009.
Using superposition find the voltage Vo(t) in the circuit
shown in the figure below, knowing that:
i1(t) = 100 cos(10,000t) mA and
v1(t) = 20 cos(20,000t - 45°) V.
Solution of the Problem + Hard 53-2
This problem has the particularity of having two sources of energy that work
on different frequencies. In this way, to solve the problem we will use
superposition, analyzing the circuit independently for each source and,
at the end, we add up the answers.
Due to this particularity, the capacitor will have two different reactances,
one for each font. Let's start with the current source you have
ω = 10,000 rad/s.
Xc = 1 / ω C = 100 Ω
Considering only the current source, we must eliminate the voltage source. we do this
short-circuiting this source. So we are left with a circuit where the capacitor and the resistor
are in parallel. Calculating the equivalent impedance of this parallel, we have:
Zeq = 100 (- j 100) / (100 - j 100) = 70.70 ∠-45°
We can easily calculate the voltage vo(t) because we just need to multiply the impedance
electric current flowing through it. Like this:
vo(t) = Zeq. i1(t) = 70.70 ∠-45° x 0.1∠0°
Note that the value of the current source was in mA and we changed it to ampere.
Carrying out the calculation, we get:
vo(t) = 7.07 ∠-45° = 7.07 cos(10,000t - 45°)
Now we must calculate vo(t) devclass the voltage source. Therefore,
we must eliminate the current source. This is achieved by eliminating it from the circuit because,
in this case, current source is an open circuit. So our circuit boils down to
to a capacitor in series with the resistor.
Note that the voltage source has a frequency of ω = 20,000 rad/s.
With this new frequency, let's calculate the new reactance of the capacitor.
Xc = 1 / ω C = 50 Ω
So the impedance of the circuit is
Zeq = 100 - j 50 = 111.8 ∠-26.56° Ω.
To calculate vo(t), just apply a voltage divisor, that is:
Veja na figura abaixo, a forma de onda que se desenvolve sobre o capacitor quando
temos duas fontes de energia atuando sobre o circuito com frequências diferentes.