Kirchhoff's Laws in Dynamic Circuits

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 solving differential equations. In these notes, we consider some simpler examples that can be solved using only Kirchhoff’s laws and the element equations of the capacitor and the inductor.

Example 1: Consider this circuit

Additionally, we are given the following representations of the voltage source voltage and one of the resistor voltages:

⎧⎧10 V fortt<< 0 2 V for 0 vvs ==⎨⎨and 1 −5t ⎩⎩20 V forte>+ 0 8 4 V fort> 0

We wish to express the capacitor current, i 2 , as a function of time, t.

Plan:

First, apply Kirchhoff’s voltage law (KVL) to the loop consisting of the source, resistor R1 and the capacitor to determine the capacitor voltage, v 2 , as a function of time, t. Next, use the element equation of the capacitor to determine the capacitor current as a function of time, t.

Solution:

Apply Kirchhoff’s voltage law (KVL) to the loop consisting of the source, resistor R1 and the capacitor to write ⎧ 8 V fort < 0 vvv12+−=ss0 ⇒ v2 =−= vv 1⎨ −5t ⎩16− 8et V for> 0

Use the element equation of the capacitor to write

⎧ 0 A fort < 0 dv22 dv ⎪ ⎧ 0 A fort < 0 iC2 ==0.025 =⎨⎨d −5t =−5t dt dt ⎪0.025() 16−> 8et for 0 ⎩1et A for> 0 ⎩ dt

1 Example 2: Consider this circuit

where the resistor currents are given by

⎧⎧0.8 A fortt<< 0 0 A for 0 ii13==⎨⎨−−22ttand ⎩⎩0.8et−> 0.8 A for 0 −0.8 e A fort> 0

Express the inductor voltage, v 2 , as a function of time, t.

Plan: First, apply Kirchhoff’s current law (KCL) at the top node of the inductor to determine the

inductor current, i 2 , as a function of time, t. Next, use the element equation of the inductor to determine the inductor voltage 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 fort < 0 iii123=+=0 ⇒ iii2 =−=1 3⎨ −2t ⎩1.6et− 0.8 A for> 0

Use the element equation of the inductor to write

⎧ 0 V fort < 0 di22 di ⎪ ⎧ 0 V fort < 0 vL2 ===2 ⎨⎨d −2t = −2t dt dt ⎪2() 1.6et−> 0.8 for 0 ⎩−6.4et V for> 0 ⎩ dt

2 Example 3: Consider this circuit

where

vts ==12cos() 5 V and v1 8.653cos( 5 t+ 33.7°) V

Express the capacitor current, i 2 , as a function of time, t.

Plan: This example is very similar to Example 1. In this case, the source voltage is a sinusoidal

function of time. Apparently, that causes v1 to also be a sinusoidal function of time. We will see

that v 2 and i 2 are both sinusoidal functions of time. Further, all of these sinusoidal functions have the same frequency as the source voltage.

As in Example 1, we first apply Kirchhoff’s voltage law (KVL) to the loop consisting of the

source, resistor R1 and capacitor to determine the capacitor voltage, v 2 , as a function of time, t. 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 R1 and capacitor to write

vvv12+−=ss0 ⇒ v2 =−= vv 112cos( 5 t) − 8.653cos( 5 t + 33.7 °)

This subtraction requires a couple of trig identities and some effort:

vt2 =−12cos() 5 8.653cos ( 5 t +°=−°+° 33.7 ) 12cos( 5 t) 8.653( cos( 33.7)() cos 5t sin ( 33.7 ) sin( 5t)) =−12cos() 5ttt() 7.199cos () 5 + 4.801sin () 5 =−4.801cos() 5tt 4.801sin () 5

22⎛−⎞− 1⎛⎞4.801 =+4.801 4.801 cos⎜⎟ 5t + tan ⎜⎟ ⎝⎠⎝⎠4.801 =−°6.790cos() 5t 45 V

Next, use the element equation of the capacitor to write

3 dv2 d iC2 = =0.025 6.790cos()()() 5t−°= 45 0.025 6.790() − 5sin 5t −° 45 dt dt =−−°0.849() sin() 5t 45 =−−°−°0.849() cos() 5t 45 90 =+°0.849cos() 5t 45 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 voltages of all independent voltage sources are sinusoidal functions of time. • The currents of all independent current sources are sinusoidal functions of time. • These voltage source voltages and current source currents all have the same frequency.

The voltages of the independent voltage 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 having the same frequency.

Here’s an important property of AC circuits:

All the element voltages and currents of an AC circuit are sinusoids having the same frequency as the inputs.

The electric power system in the United States can be modeled as an AC circuit having the frequency 60 Hz or 377 rad/s. Consequently, all of the voltages and currents in this circuit have the form Atcos( 377 +θ ). A particular current or voltage can be specified by providing the values of the amplitude, A, and the phase angle, θ .

The circuit in Example 3 is an AC circuit having the frequency 5 rad/s.

Sinusoids and Complex Numbers It’s useful to associate a complex number A∠θ with the sinusoid Atcos( 377 +θ ). The complex number A∠θ is given 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:

AtBtCtcos()() 377++θ cos 377 +=φγθφ cos( 377 +) ⇔ ABC ∠+∠=∠γ

Suppose we need to find the sum of sinusoid, AtBtcos( 377+θ ) + cos( 377 +φ ). Instead, we can choose to find the sum of complex numbers, AB∠θ +∠φ . The sum of the complex numbers, C∠γ , provides the sum of the sinusoids, Ctcos( 377 +γ ) .

It’s customary to convert complex numbers from polar form to rectangular form before adding them. That is A∠θ =+ajb

where A∠θ is the complex number in polar form and ajb+ is the complex number in rectangular form. The conversion from polar form to rectangular form and visa versa is described by aAA=∠the real part of θ = cos(θ )

bA=∠the imaginary part of θ =A sin (θ )

Aa=+the magnitude of jb=a22+b

⎧ −1 ⎛⎞b ⎪ tan⎜⎟ whena > 0 ⎪⎝⎠a θ =+=the angle of ajb⎨ ⎪ −1 ⎛⎞b 180°− tan⎜⎟ whena < 0 ⎩⎪ ⎝⎠−a

Two complex numbers in rectangular form, ajb+ and c+ jd, are added as

()()ajbcjdac+++=+++=+( ) jbd( ) ejf

(The real part of ej+ f is equal to the sum of the real parts of ajb+ and cjd+ . Also, the imaginary part of ej+ f is equal to the sum of the imaginary parts of ajb+ and c+ jd.)

To summarize, to add two complex numbers in polar form, A∠θ and B∠φ , we first convert both complex numbers to rectangular form:

A∠+∠=θφBA()cos θ + jAB sin θ +( cos φ + jBajbcj sin φ) =( +)( + +d )

Next, we add the rectangular forms of the complex numbers:

()()ajbcjdac+++=+++=+( ) jbd( ) ejf

Finally, we convert the sum from rectangular form to polar form:

⎧ 22− 1⎛⎞f ⎪ ef+∠tan⎜⎟ whene> 0 ⎪ ⎝⎠e ejfC+=∠=γ ⎨ ⎪ 22⎛⎞−1⎛⎞f ef+ ∠°−⎜⎟180 tan⎜⎟ whene< 0 ⎩⎪ ⎝⎠⎝⎠e

In Example 3 we needed to calculate the sum of two sinusoids having at the frequency 5 rad/s:

12cos( 5tt) − 8.653cos( 5+° 33.7 ) The corresponding sum of complex numbers is

12∠− 0 8.653 ∠ 33.7 °=() 12cos() 0 +jj 12sin( 0) −( 8.653cos( 33.7°+)( 8.653sin 33.7°)) =+()(12jj 0 − 7.199 + 4.801 ) =−()()12 7.199 +j 0 − 4.801 =−4.801j 4.801

212 − ⎛⎞−4.801 =+−∠4.801() 4.801 tan ⎜⎟ ⎝⎠4.801 =∠−°6.790 45

The sinusoid at the frequency 5 rad/s corresponding to 6.790∠ −°45 is 6.790cos() 5t −° 45 . Consequently 12cos() 5tt−+°= 8.653cos( 5 33.7) 6.790cos( 5 t−° 45 )

Example 4: Consider this circuit

where the resistor currents are given by

it13=−°=1.040cos() 3 19.4 A and i 0.6240cos( 3 t + 33.7°) A

Express the inductor voltage, v 2 , as a function of time, t.

Plan: This example is very similar to Example 2. First, apply KCL at the top node of the inductor to determine the inductor current, i 2 , as a function of time, t. Next, use the element equation of the inductor to determine the inductor voltage as a function of time, t.

Solution: First, apply Kirchhoff’s current law (KCL) at the top node of the inductor to write

iii=−=0 ⇒ iii =+=1.040cos( 3 t − 19.4 °+)( 0.6240cos 3t + 33.7 °) 123 21 3 =−°0.8321cos() 3t 56.3 A

Next, use the element equation of the inductor to write

di d vL==2 2 0.8321cos() 3t−°= 56.3 4.993cos ( 3t +° 33.7) V 2 dt dt

Example 5: Consider this circuit

where ⎧⎧10 V fortt<< 0 2 V for 0 vvs ==⎨⎨and 1 −5t ⎩⎩20 V forte>+ 0 8 4 V fort> 0

Given R1 =Ω10 , determine the value of R2 .

Plan: This example is similar to Example 1. As in Example 1 we determine

⎧⎧8 V fortt<< 0 0 A for 0 vi22==⎨⎨−−55ttand ⎩⎩16−> 8et V for 0 1 et A for> 0

Next, apply Ohm’s law to resistor R1 to determine i1 . Apply KCL at the top node of R2 to determine the current in R2 . Noticing that v 2 is the voltage across R2 , apply Ohm’s law to determine the value of R2 .

Solution:

Apply Ohm’s law to resistor R1 to write

vv11⎧ 0.2 A fort < 0 i1 ===⎨ −5t R1 10 ⎩0.8et+ 0.4 A for> 0

Apply KCL at the top node of R2 to write

v 2 ⎧ 0.2 A fort < 0 =−=ii12⎨ −5t R2 ⎩0.4− 0.2et A for> 0

Comparing this equation to our equation for v 2 , we conclude that R2 = 40 Ω . (For example,

8 v 2 when t < 0, ==−=⇒ii120.2 R2 = 40 Ω.) RR22