Chapter 6

EE 201 ELECTRIC CIRCUITS
LECTURE 20
The material covered in this lecture will be as follows:
The Inductor.
Current-Voltage Relations of Inductor.
Power and Energy in the Inductor.
At the end of this lecture you should be able to:
Understand and apply the inductor’s current-voltage differential relation.
Derive and apply the inductor’s current-voltage integral relation.
Understand that zero inductor’s voltage implies D.C. inductor’s current and vice versa.
Derive the inductor’s power and energy relations.
Apply the inductor’s energy relation.
Energy Storage Elements:
Some electric circuit elements can store electric energy. These include the capacitor and the
inductor.
A resistor clearly is not an energy storage element, because the electric energy it absorbs is
converted to heat energy and thus it is lost.
Ideal voltage sources and ideal current sources are not considered energy storage elements.
Energy storage elements (ESE) can absorb electric energy. This energy remains in electric form
(stored).
ESE can also deliver electric energy, but they can only deliver the energy that has been previously
absorbed (stored energy).
The Inductor
Symbol of the inductor is shown in the diagram.
The unit of the inductance L is the Henry (H).
1

Current-Voltage Relationship in the Inductor:
1- Differential Relation:
The current-voltage relation in an inductor is:
v L t = L di L(t)
dt
( eq.1)
This differential relation is valid only if i L enters the (+) side of v L .
A minus sign must be inserted in the above relation if i L enters the (-) side of v L [According to the
passive sign convention].
[The voltage across an inductor is proportional to the rate of change of the current through the
inductor].
Example 6.1
The independent current source in the circuit generates zero current for t < 0 and a pulse of 10te -5t A
for t > 0.
i t =
0 t < 0
10te −5t A t > 0
a) Sketch the current waveform.
b) Express the voltage across the inductor as a function of time.
c) Sketch the voltage waveform.
v L t = L di L(t)
dt
di
dt = 10e−5t + 10t(−5e −5t )
di
= (10 − 50t)e−5t
dt
v L t = (1 − 5t)e −5t V
2

2- Integral Relation
v L t = L di L(t)
dt
(eq. 2)
v L dt = Ldi L (eq. 3)
Integrate both sides
i L t = 1 L
t
t 0
v L d + i(t 0 )
(eq.4)
Example 6.2
The voltage pulse applied to the 100 mH inductor shown in Figure is given by
v t =
0 t < 0
20te −10t V t > 0
Assume i = 0 for t ≤ 0.
a) Sketch the voltage waveform.
b) Express the inductor current as a function of time.
c) Sketch the current waveform.
3

i L t = 1
0.1
t
0
20e −10 d + i(0)
i L t = 200
−e −10
100 (10 + 1) t 0
= 2 − 20te −10t − 2e −10t A
Power and Energy Relations in the Inductor:
The instantaneous power p(t) absorbed by an inductor is given by:
p t = v L t i L t
p = Li di
dt
(eq.5)
(eq.6)
Or expressing the current in terms of the voltage
p = v 1 L
t
t 0
v d + i(t 0 )
(eq.7)
Also, power is the time rate of expending energy, so
p = dw
dt
pdt = dw
dw = pdt
dw = Li di
(eq.8)
(eq.9)
(eq.10)
(eq.11)
Integrate both sides
w = 1 2 Li2
(eq.12)
4

Example 6.3
The independent current source in the circuit generates zero current for t < 0 and a pulse of 10te -5t A
for t > 0.
i t =
0 t < 0
10te −5t A t > 0
a) Derive the expressions for the inductor voltage, power, and energy.
b) Sketch the current, voltage, power, and energy as functions of time.
c) Specify the interval of time when energy is being stored in the inductor.
d) Specify the interval of time when energy is being extracted from the inductor.
e) Evaluate the integrals
0.2
p dt
0
and
∞
0.2
p dt
Solution
a)
v L t = (1 − 5t)e −5t V
p t = vi = 10te −10 − 50t 2 e −10t W
w t = 1 2 Li2 = 1 2 × 100 × 10−3 100t 2 e −10t
b)
= 5t 2 e −10t J
5

c) Energy is being stored in the capacitor whenever the power is positive. In the interval (0, 0.2) s.
d) Energy is being extracted whenever the power is negative. In the interval ( 0.2, ∞) s.
e)
0.2
p dt
0
= 10 e−10t
0.2
100 −10t − 1 − 50 t2 e −10t
−10 + 2
10
0
e −10t
100 −10t − 1 0
0.2
= 0.2e −2
= 27.07 mJ.
∞
p dt
0.2
= 10 e−10t
∞
100 −10t − 1 − 50 t2 e −10t
−10 + 2
10
0.2
e −10t
∞
100 −10t − 1 0.2
= −0.2e −2
= −27.07 mJ.
6

EE 201 ELECTRIC CIRCUITS
LECTURE 21
The material covered in this lecture will be as follows:
Energy Storage Elements.
The Capacitor.
Capacitor’s Current-Voltage Relations.
Power and Energy in the Capacitor.
At the end of this lecture you should be able to:
Differentiate between energy storage elements and non-energy storage elements.
Derive and apply the differential relation between the capacitor’s current and voltage.
Derive and apply the integral relation between the capacitor’s current and voltage.
Apply the passive sign convention to the differential and integral relations.
Recognize that zero capacitor’s current implies D.C. capacitor’s voltage and vice versa.
Derive and apply the power and energy relations for the capacitor.
Understand the relationship between the energy and power for an energy storage element.
The Capacitor:
Symbol of the capacitor is shown in the diagram
The unit of the capacitance C is in Farads (F)
The charge q stored in the capacitor is proportional to the voltage across the capacitor. The constant of
proportionality is the capacitance C.
q = Cv C
7

Voltage-Current Relationship in the Capacitor:
1- Differential Relation:
dq
dt = C dv C
dt
i C = C dv C
dt
[Explicit time dependence is not shown for simplicity]
This is a differential relation between the voltage and current in a capacitor. It is valid only if i C enters the
(+) side of v C . [Passive sign convention]
2- Integral Relation:
t
v C t = 1 i
C C d + v t 0
t 0
Power and Energy Relations in the Capacitor:
p t = v C t i C t = Cv dv
dt
w = 1 2 Cv2
8

Example 6.5
An uncharged 0.2 µF capacitor is driven by a triangular current pulse. The current pulse is described
by
i t =
0 t ≤ 0
5000t A 0 ≤ t ≤ 20 × 10 −6 s
0.2 − 5000t A 20 ≤ 2 ≤ 40 × 10 −6 s
0 t ≥ 40 × 10 −6 s
a) Derive the expressions for the capacitor voltage, power, and energy.
b) Sketch the voltage, current, power, and energy as functions of time.
Solution
For 0 ≤ t ≤ 20 µs
t
v = 5 × 10 6 5000 d + 0
0
= 12.5 × 10 9 t 2 V
10

p = vi = 62.5 × 10 12 t 3 W
w = 1 2 Cv2 = 15.625 × 10 12 t 4 J
For 20 ≤ t ≤ 40 µs
t
v = 5 × 10 6 (0.2 − 5000) d + 5
20µs
= (10 6 t − 12.5 × 10 9 t 2 − 10) V
p = vi = (62.5 × 10 12 t 3 − 7.5 × 10 9 t 2 + 2.5 × 10 5 t − 2) W
w = 1 2 Cv2 = (15.625 × 10 12 t 4 − 2.5 × 10 9 t 3 + 0.125 × 10 6 t 2 − 2t + 10 −5 ) J
For t ≥ 40 µs
v = 10 V
p = vi = 0
w = 1 2 Cv2 = 10 µJ
11

EE 201 ELECTRIC CIRCUITS
LECTURE 22
The material covered in this lecture will be as follows:
⇒ Inductors and Capacitors in Series and in Parallel
⇒ Inductors and Capacitors in D.C. Circuits
⇒ Continuity Relations for the Inductor and the Capacitor
At the end of this lecture you should be able to:
⇒Combine inductors and capacitors in series and in parallel
⇒ Understand the meaning of a D.C. circuit
⇒ Deal with inductors and capacitors in D.C. circuits
⇒ Understand that the inductor’s current is continuous
⇒ Understand that the capacitor’s voltage is continuous
Inductors and Capacitors in Series and in Parallel:
Inductors in series and in parallel are combined in the same way as resistance.
12

Capacitors in series and in parallel are combined in the same way as conductance.
13

AP6.4
i(t)
+
v(t)
-
i 1 (t)
60 mH
i 2 (t)
240 mH
i 1 0 = +3 A
i 2 0 = −5 A
v t = −30e −5t mV for t ≥ 0
(a) L eq =
60 240
60+240
= 48 mH.
(b) i 0 = +3 + −5 = −2 A.
(c) i = 1
0.048
(d) i 1 = 1
0.06
t
0
−0.03e −5τ d + (−2)
= 0.125e −5t − 2.125 A
t
0
= 0.1e −5t + 2.9 A
i 2 = 1 t
0.24
−0.03e −5τ d + (3)
−0.03e −5τ d + (−5)
0
= 0.025e −5t − 5.025 A
i 1 + i 2 = i
15

AP6.5
i(t) + v 1 -
2 mF
+
v 2
-
8 mF
i t = 240e −10t μA for t ≥ 0
v 1 0 = −10 V
v 2 0 = −5 V
t
1
v 1 =
2 × 10 −6 240e−10τ × 10 −6 d + (−10)
0
= −12e −10t + 2 V
v 1 ∞ = 2 V
v 2 ∞ = −2 V
t
1
v 2 =
8 × 10 −6 240e−10τ × 10 −6 d + (−5)
0
= −3e −10t − 2 V
W = 1 2 2 2 2 + 1 2 8 −2 2 × 10 −6
= 20μJ
See P6.6 and P6.11
16