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 for t < 0
vs = ⎨
⎩20 V for t > 0
and
2 V for t < 0
⎧
v1 = ⎨ −5t
⎩8 e + 4 V for t > 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 for t < 0
⎧
v1 + v 2 − v s = 0 ⇒ v 2 = v s − v1 = ⎨
−5 t
⎩16 − 8 e V for t > 0
Use the element equation of the capacitor to write
0 A for t < 0
⎧
⎧ 0 A for t < 0
⎪
i2 = C
= 0.025
=⎨
= ⎨ −5t
d
−5 t
dt
d t ⎪0.025 (16 − 8 e ) for t > 0 ⎩1e A for t > 0
dt
⎩
d v2
d v2
1
Example 2:
Consider this circuit
where the resistor currents are given by
0.8 A for t < 0
⎧
i1 = ⎨
−2 t
⎩0.8 e − 0.8 A for t > 0
and
0 A for t < 0
⎧
i3 = ⎨
−2 t
⎩−0.8 e A for t > 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 for t < 0
⎧
i 1 = i 2 + i 3 = 0 ⇒ i 2 = i1 − i 3 = ⎨
−2 t
⎩1.6 e − 0.8 A for t > 0
Use the element equation of the inductor to write
0 V for t < 0
⎧
0 V for t < 0
⎧
⎪
v2 = L
=2
=⎨ d
=⎨
−2 t
1.6 e − 0.8 for t > 0 ⎩−6.4 e −2 t V for t > 0
dt
d t ⎪2
d
t
⎩
d i2
d i2
(
)
2
Example 3:
Consider this circuit
where
v s = 12 cos ( 5 t ) 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
v1 + v 2 − v s = 0 ⇒ v 2 = v s − v1 = 12cos ( 5 t ) − 8.653cos ( 5 t + 33.7° )
This subtraction requires a couple of trig identities and some effort:
v 2 = 12 cos ( 5 t ) − 8.653cos ( 5 t + 33.7° ) = 12 cos ( 5 t ) − 8.653 ( cos ( 33.7° ) cos ( 5 t ) + sin ( 33.7° ) sin ( 5 t ) )
= 12 cos ( 5 t ) − ( 7.199 cos ( 5 t ) + 4.801sin ( 5 t ) )
= 4.801cos ( 5 t ) − 4.801sin ( 5 t )
⎛
⎛ −4.801 ⎞ ⎞
= 4.801 2 + 4.801 2 cos ⎜ 5 t + tan −1 ⎜
⎟⎟
⎝ 4.801 ⎠ ⎠
⎝
= 6.790 cos ( 5 t − 45° ) V
Next, use the element equation of the capacitor to write
3
i2 = C
d v2
dt
= 0.025
d
6.790 cos ( 5 t − 45° ) = 0.025 ( 6.790 ) ( −5sin ( 5 t − 45° ) )
dt
= 0.849 ( − sin ( 5 t − 45° ) )
= 0.849 ( − cos ( 5 t − 45° − 90° ) )
= 0.849 cos ( 5 t + 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 A cos ( 377 t + 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 A cos ( 377 t + 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:
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A cos ( 377 t + θ ) + B cos ( 377 t + φ ) = C cos ( 377 t + γ ) ⇔
A∠θ + B∠φ = C∠γ
Suppose we need to find the sum of sinusoid, A cos ( 377 t + θ ) + B cos ( 377 t + φ ) . Instead, we
can choose to find the sum of complex numbers, A∠θ + B∠φ . The sum of the complex
numbers, C ∠γ , provides the sum of the sinusoids, C cos ( 377 t + γ ) .
It’s customary to convert complex numbers from polar form to rectangular form before adding
them. That is
A∠θ = a + j b
where A∠θ is the complex number in polar form and a + j b is the complex number in
rectangular form. The conversion from polar form to rectangular form and visa versa is described
by
a = the real part of A∠θ = A cos (θ )
b = the imaginary part of A∠θ = A sin (θ )
A = the magnitude of a + j b = a 2 + b 2
⎧
⎛b⎞
tan −1 ⎜ ⎟ when a > 0
⎪
⎪
⎝a⎠
θ = the angle of a + j b = ⎨
⎪180° − tan −1 ⎛ b ⎞ when a < 0
⎜
⎟
⎪⎩
⎝ −a ⎠
Two complex numbers in rectangular form, a + j b and c + j d , are added as
( a + j b) + (c + j d ) = ( 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 + j b and c + j d . Also, the
imaginary part of e + j f is equal to the sum of the imaginary parts of a + j b and c + j d .)
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∠θ + B∠φ = ( A cos θ + j A sin θ ) + ( B cos φ + j B sin φ ) = ( a + j b ) + ( c + j d )
Next, we add the rectangular forms of the complex numbers:
( a + j b) + (c + j d ) = ( a + c ) + j (b + d ) = e + j f
5
Finally, we convert the sum from rectangular form to polar form:
⎧
⎛ f ⎞
e 2 + f 2 ∠ tan −1 ⎜ ⎟
⎪
⎝e⎠
⎪
e + j f = C ∠γ = ⎨
⎪ e 2 + f 2 ∠ ⎛180° − tan −1 ⎛ f
⎜
⎜
⎪⎩
⎝e
⎝
when e > 0
⎞⎞
⎟ ⎟ when e < 0
⎠⎠
In Example 3 we needed to calculate the sum of two sinusoids having at the frequency 5 rad/s:
12 cos ( 5 t ) − 8.653cos ( 5 t + 33.7° )
The corresponding sum of complex numbers is
12∠0 − 8.653∠33.7° = (12 cos ( 0 ) + j 12sin ( 0 ) ) − ( 8.653cos ( 33.7° ) + j 8.653sin ( 33.7° ) )
= (12 + j 0 ) − ( 7.199 + j 4.801)
= (12 − 7.199 ) + j ( 0 − 4.801)
= 4.801 − j 4.801
2
⎛ −4.801 ⎞
= 4.8012 + ( −4.801) ∠ tan −1 ⎜
⎟
⎝ 4.801 ⎠
= 6.790∠ − 45°
The sinusoid at the frequency 5 rad/s corresponding to 6.790∠ − 45° is 6.790 cos ( 5 t − 45° ) .
Consequently
12 cos ( 5 t ) − 8.653cos ( 5 t + 33.7° ) = 6.790 cos ( 5 t − 45° )
Example 4:
Consider this circuit
where the resistor currents are given by
i1 = 1.040 cos ( 3 t − 19.4° ) A
and
i 3 = 0.6240 cos ( 3 t + 33.7° ) A
Express the inductor voltage, v 2 , as a function of time, t.
6
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
i1 = i 2 − i 3 = 0 ⇒ i 2 = i1 + i 3 = 1.040 cos ( 3 t − 19.4° ) + 0.6240 cos ( 3 t + 33.7° )
= 0.8321cos ( 3 t − 56.3° ) A
Next, use the element equation of the inductor to write
v2 = L
d i2
dt
=2
d
0.8321cos ( 3 t − 56.3° ) = 4.993cos ( 3 t + 33.7° ) V
dt
Example 5:
Consider this circuit
where
⎧10 V for t < 0
vs = ⎨
⎩20 V for t > 0
and
2 V for t < 0
⎧
v1 = ⎨ −5t
⎩8 e + 4 V for t > 0
Given R1 = 10 Ω , determine the value of R 2 .
Plan:
This example is similar to Example 1. As in Example 1 we determine
8 V for t < 0
⎧
v2 = ⎨
−5 t
⎩16 − 8 e V for t > 0
and
⎧ 0 A for t < 0
i 2 = ⎨ −5t
⎩1 e A for t > 0
Next, apply Ohm’s law to resistor R1 to determine i1 . Apply KCL at the top node of R 2 to
determine the current in R 2 . Noticing that v 2 is the voltage across R 2 , apply Ohm’s law to
determine the value of R 2 .
7
Solution:
Apply Ohm’s law to resistor R1 to write
i1 =
v1
R1
=
0.2 A for t < 0
⎧
=⎨
−5 t
10 ⎩0.8 e + 0.4 A for t > 0
v1
Apply KCL at the top node of R 2 to write
0.2 A for t < 0
⎧
= i1 − i 2 = ⎨
−5 t
R2
⎩0.4 − 0.2 e A for t > 0
v2
Comparing this equation to our equation for v 2 , we conclude that R 2 = 40 Ω . (For example,
when t < 0,
v2
8
=
= i1 − i 2 = 0.2 ⇒ R 2 = 40 Ω .)
R2 R2
8