Kirchhoff s Laws in Dynamic Circuits

Transcription

1 Kirchhoff s Laws in Dynamic Circuits Dynamic circuits are circuits that contain capacitors and inductors. Later we will learn to analyze some dynamic circuits by writing and soling differential equations. In these notes, we consider some simpler examples that can be soled using only Kirchhoff s laws and the element equations of the capacitor and the inductor. Example : Additionally, we are gien the following representations of the oltage source oltage and one of the resistor oltages: 0 V for t < 0 V for t < 0 = and = 0 V for > V for t > 0 s 5t t e We wish to express the capacitor current, i Plan: First, apply Kirchhoff s oltage law (KVL) to the loop consisting of the source, resistor R and the capacitor to determine the capacitor oltage, Next, use the element equation of the capacitor to determine the capacitor current as a function of time, t. Solution: Apply Kirchhoff s oltage law (KVL) to the loop consisting of the source, resistor R and the capacitor to write 8 V for t < 0 + s = 0 = s = 5t 6 8 e V for t > 0 Use the element equation of the capacitor to write 0 A for t < 0 d d 0 A for t < 0 i = C = 0.05 = d 5t = 5t dt dt 0.05 ( 6 8e ) for t > 0 e A for t > 0 dt

2 Example : where the resistor currents are gien by i 0.8 A for t < 0 0 A for t < 0 = and i = 0.8e 0.8 A for t > e A for t > 0 t 3 t Express the inductor oltage, Plan: First, apply Kirchhoff s current law (KCL) at the top node of the inductor to determine the inductor current, Next, use the element equation of the inductor to i determine the inductor oltage as a function of time, t. Solution: Apply Kirchhoff s current law (KCL) at the top node of the inductor to write 0.8 V for t < 0 i = i + i3 = 0 i = i i3 = t.6e 0.8 A for t > 0 Use the element equation of the inductor to write 0 V for t < 0 di di 0 V for t < 0 = L = = d t = t dt dt (.6e 0.8) for t > e V for t > 0 dt

3 Example 3: where ( ) ( ) t t s = cos 5 V and = 8.653cos V Express the capacitor current, i Plan: This example is ery similar to Example. In this case, the source oltage is a sinusoidal function of time. Apparently, that causes to also be a sinusoidal function of time. We will see that and i are both sinusoidal functions of time. Further, all of these sinusoidal functions hae the same frequency as the source oltage. As in Example, we first apply Kirchhoff s oltage law (KVL) to the loop consisting of the source, resistor R and capacitor to determine the capacitor oltage, Next, use the element equation of the capacitor to determine the capacitor current as a function of time, t. Solution: Apply KVL to the loop consisting of the source, resistor R and capacitor to write s s ( ) ( ) + = 0 = = cos 5t 8.653cos 5t This subtraction requires a couple of trig identities and some effort: ( ) ( ) ( ) ( ( ) ( ) ( ) ( t) ) = cos( 5t) ( 7.99cos( 5t) sin ( 5t) ) = 4.80cos( 5t) 4.80sin ( 5t) = cos 5t 8.653cos 5t = cos 5t cos 33.7 cos 5t + sin 33.7 sin 5 Next, use the element equation of the capacitor to write 4.80 = cos 5t + tan 4.80 = 6.790cos 5 45 V ( t ) 3

4 d d i = C = cos 5t 45 = sin 5t 45 dt dt = sin 5 45 ( ) ( )( ( )) ( ( t )) ( ( t )) ( t ) = cos = 0.849cos A The circuit in Example 3 is an example of an ac circuit. Let s talk about ac circuits. AC Circuits AC circuits are electric circuits in which The oltages of all independent oltage sources are sinusoidal functions of time. The currents of all independent current sources are sinusoidal functions of time. These oltage source oltages and current source currents all hae the same frequency. The oltages of the independent oltage sources and currents of the independent current sources are sometimes called the inputs to the circuit. Using this terminology we can say that: AC circuits are electric circuits in which all of the inputs are sinusoids haing the same frequency. Here s an important property of AC circuits: All the element oltages and currents of an AC circuit are sinusoids haing the same frequency as the inputs. The electric power system in the United States can be modeled as an AC circuit haing the frequency 60 Hz or 377 rad/s. Consequently, all of the oltages and currents in this circuit hae the form Acos( 377t + θ ). A particular current or oltage can be specified by proiding the alues of the amplitude, A, and the phase angle, θ. The circuit in Example 3 is an AC circuit haing the frequency 5 rad/s. Sinusoids and Complex Numbers It s useful to associate a complex number A θ with the sinusoid Acos( 377t+ θ ). The complex number A θ is gien in polar form so A represents the amplitude of the complex number and θ represents the phase angle of the complex number. Now, the addition of sinusoids corresponds to the addition of complex numbers: 4

5 ( ) ( ) ( ) Acos 377t+ θ + Bcos 377t+ φ = Ccos 377t+ γ A θ + B φ = C γ Suppose we need to find the sum of sinusoid, Acos( 377t θ ) Bcos( 377t φ ) Instead, we can choose to find the sum of complex numbers, A θ + B φ. The sum of the complex numbers, C γ, proides the sum of the sinusoids, Ccos( 377t+ γ ). It s customary to conert complex numbers from polar form to rectangular form before adding them. That is A θ = a+ jb where A θ is the complex number in polar form and a+ jb is the complex number in rectangular form. The conersion from polar form to rectangular form and isa ersa is described by a= the real part of A θ = Acos( θ ) b= A = A the imaginary part of θ sin ( θ ) A= the magnitude of a+ jb= a + b b tan when a > 0 a θ = the angle of a+ jb= b 80 tan when a < 0 a Two complex numbers in rectangular form, a+ jb and c+ jd, are added as ( a+ jb) + ( c+ jd) = ( a+ c) + j( b+ d) = e+ j f (The real part of e+ j f is equal to the sum of the real parts of a+ jb and c+ jd. Also, the imaginary part of e+ j f is equal to the sum of the imaginary parts of a+ jb and c + jd.) To summarize, to add two complex numbers in polar form, A θ and B φ, we first conert both complex numbers to rectangular form: ( cos sin ) ( cos sin ) ( ) ( ) A θ + B φ = A θ + ja θ + B φ+ jb φ = a+ jb + c+ jd Next, we add the rectangular forms of the complex numbers: ( a+ jb) + ( c+ jd) = ( a+ c) + j( b+ d) = e+ j f 5

6 Finally, we conert the sum from rectangular form to polar form: f e + f tan when e> 0 e e+ j f = C γ = f e + f 80 tan when e< 0 e In Example 3 we needed to calculate the sum of two sinusoids haing at the frequency 5 rad/s: The corresponding sum of complex numbers is ( t) ( t+ ) cos cos ( ( ) j ( )) ( ) j ( ) ( j0) ( 7.99 j4.80) ( 7.99) j ( ) ( ) = cos 0 + sin cos sin 33.7 = + + = + = 4.80 j 4.80 ( ) 4.80 = tan 4.80 = The sinusoid at the frequency 5 rad/s corresponding to is cos 5t 45. Consequently Example 4: ( t) ( t+ ) = ( t ) cos cos cos 5 45 ( ) where the resistor currents are gien by ( ) ( ) i =.040 cos 3t 9.4 A and i = cos 3t A 3 Express the inductor oltage, 6

7 Plan: This example is ery similar to Example. First, apply KCL at the top node of the inductor to determine the inductor current, Next, use the element equation of the i inductor to determine the inductor oltage as a function of time, t. Solution: First, apply Kirchhoff s current law (KCL) at the top node of the inductor to write 3 3 ( ) ( ( t ) i = i i = 0 i = i + i =.040cos 3t cos 3t = 0.83cos A Next, use the element equation of the inductor to write Example 5: di d = L 0.83cos( 3t 56.3 ) 4.993cos( 3t 33.7 V dt = dt = + ) ) where Gien 0 V for t < 0 V for t < 0 = and = 0 V for > V for t > 0 s 5t t e R = 0 Ω, determine the alue of R. Plan: This example is similar to Example. As in Example we determine 8 V for t < 0 0 A for t < 0 = and i = 6 8 e V for t > 0 e A for t > 0 5t 5t Next, apply Ohm s law to resistor R to determine determine the current in R. Noticing that determine the alue of R. i. Apply KCL at the top node of R to R, apply Ohm s law to is the oltage across 7

8 Solution: Apply Ohm s law to resistor R to write i 0. A for t < 0 = = = 5t R 0 0.8e A for t > 0 Apply KCL at the top node of R to write R 0. A for t < 0 = i i = 5t e A for t > 0 Comparing this equation to our equation for when t < 0, 8 R = = i i = 0. R = 40 Ω.) R, we conclude that R = 40 Ω. (For example, 8