Fundamentals of Electrical Engineering I

Lecture 4
 Capacitor
Objective:
Understand the relationships between charge,
current and voltage for capacitors

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 Application: Supercapacitors or
Ultracapacitors
• Two large-area metal foils with molecularly
thin separation, tightly packed, to make large
capacitance values in small packages
• Protects rechargeable batteries from current
spikes
• Found in many applications, from electronics
to electronics to electric vehicles
• Traditional capacitors have smaller values, e.g.
0.3 𝜇𝐹
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Application

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 Capacitor
Builds on:
• Passive convention:

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Charge and Voltage on a Capacitor
Capacitance (C) in Farads relates to the charge
(q) in Coulombs and the voltage (v) in volts,
according to
𝑞 = 𝐶𝑣

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 Test
A 5V voltage is placed across a 200 mF capacitor.
How much charge, in Coulombs, is stored on the
plates?

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 Capacitance
𝜀𝐴
• In a plate capacitor, 𝐶 =
𝑑

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 Test
You create a capacitor by placing two sheets of
metal of width w=20mm and length l=30mm
separated by a 0.1mm thick dielectric material
𝑝𝐹
with permittivity 𝜀 = 60 . What is capacitance
𝑚
of this capacitor?

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 Capacitors in Schematics
• The curved version indicates a polarized
capacitor, which must always have the positive
polarity of voltage on the flat side
– Appropriate for DC applications, only

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 Current and Voltage in a Capacitor
𝑑𝑞
• Current is the time rate of flow of charge, 𝑖 = , so
𝑑𝑡
we differentiate 𝑞 = 𝐶𝑣
𝑑𝑞 𝑑𝑣
=𝐶
𝑑𝑡 𝑑𝑡
𝑑𝑣
𝑖=𝐶
𝑑𝑡

• If 𝑣 is in Volts and 𝐶 is in Farads, then 𝑖 will be in Amps

• Current is a linear function of the voltage

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 Capacitor
𝑑𝑖
• From 𝑖 = 𝐶 , we can solve for the voltage
𝑑𝑡

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 Voltage from Current - Example
• Suppose the current through a 2 F capacitor is given
below. Find 𝑣 𝑡 for 𝑡 ≥ 0, given 𝑣 0 = −2𝑉.

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 Test
A 500𝜇𝐹 capacitor is measured and found to
have a voltage across the plates of 𝑣𝑐 =
2−2𝑡 𝑉. What is the current 𝑖 𝑡 ?

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 Key concepts
• Q V characteristics: 𝑞 = 𝐶𝑣
• In the passive convention,
𝑑𝑣
I V characteristic: 𝑖 = 𝐶
𝑑𝑡
• When using the integral relationship, must
add the initial voltage

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 Series and Parallel Capacitors
Builds on:
• Definition of series and parallel elements
𝜀𝐴
• Formula of capacitance 𝐶 =
𝑑
• Capacitor I V characteristics:
𝑑𝑣
𝑖=𝐶
𝑑𝑡
𝑡
1
𝑣 𝑡 = 𝑖 𝜏 𝑑𝜏 + 𝑣(𝑡0 )
𝐶
𝑡0
• KVL and KCL

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 Intuitive Parallel Example
𝜀𝐴
• Capacitance of one capacitor: 𝐶 =
𝑑

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 Intuitive Parallel Example
𝜀𝐴
• Capacitance of one capacitor: 𝐶 =
𝑑
• Consider an identical capacitor in parallel

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 Intuitive Parallel Example
𝜀𝐴
• Capacitance of one capacitor: 𝐶 =
𝑑
• Consider an identical capacitor in parallel

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 Intuitive Parallel Example
𝜀𝐴
• Capacitance of one capacitor: 𝐶 =
𝑑
• Consider an identical capacitor in parallel
𝜀(2𝐴)
𝐶𝑛𝑒𝑤 = = 2𝐶
𝑑

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 Intuitive Parallel Example
𝜀𝐴
• Capacitance of one capacitor: 𝐶 =
𝑑
• Consider an identical capacitor in parallel
𝜀(2𝐴)
𝐶𝑛𝑒𝑤 = = 2𝐶
𝑑
• For capacitors in parallel: add capacitances

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 Derive Parallel Formula
• Consider 𝑛 capacitors in parallel, each with
possibly distinct currents and capacitances

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 Derive Series Formula
By KVL,
Consider 𝑛 capacitors in series, 𝑛

each with possibly distinct 𝑣 𝑡 = 𝑣𝑘 𝑡 =

𝑘=1
voltages and capacitances 𝑛 𝑡
1
= 𝑖𝑘 𝜏 𝑑𝜏 + 𝑣𝑘 (𝑡0 )
𝐶𝑘
𝑘=1 𝑡0
𝑛 𝑡 𝑛
1
= 𝑖𝑘 𝜏 𝑑𝜏 + 𝑣𝑘 (𝑡)
𝐶𝑘
𝑘=1 𝑡0 𝑘=1
𝑡
1
𝑡 = 𝑖𝑘 𝜏 𝑑𝜏 + 𝑣𝑘 (𝑡0 )
𝐶𝑒𝑓𝑓
1 𝑡0
𝑣𝑘 𝑡 = 𝑖𝑘 𝜏 𝑑𝜏 + 𝑣𝑘 (𝑡0 ) 𝟏
𝐶𝑘 𝑪𝒆𝒇𝒇 =
𝑡0 𝒏 𝟏
𝒌=𝟏 𝑪
𝒏

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 Example
Find the effective capacitance of the
combination below.

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 Test
By combining capacitance in parallel and series,
find the capacitance between in left terminals.

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 Key Concepts
𝑁
• Parallel: 𝐶𝑒𝑓𝑓 = 𝑛=1 𝐶𝑛
1
• Series: 𝐶𝑒𝑓𝑓 = 𝑁 1
𝑛=1𝐶𝑛

• The formulas for capacitors combinations are

OPPOSITE to those of resistors

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 Introduction to Inductors
Objective:
• Understand the relationship between current
and voltage for inductors

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 Inductance
𝜇𝑁2 𝐴
• Inductance in Henrys (H) is 𝐿 =
𝑙
• Where,
– 𝑙 is length and 𝐴 is the core area
– 𝑁 is the number of turns of the wire
– 𝜇 is the permeability of the core material

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Inductors in Schematics

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Current and Voltage in an Inductor

𝑑𝑖
𝑣=𝐿
𝑑𝑡
If 𝑖 is in Amps and 𝐿 is
in Henrys, then 𝑣 is in
Volts

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 Example
Suppose the current through a
0.5H inductor is given below.
Determine the voltage, 𝑣(𝑡)

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 𝑑𝑖
• From 𝑣 = we𝐿 ,
𝑑𝑡
can solve the current:
𝑡
1
𝑖 𝑡 = 𝑣 𝜏 𝑑𝜏 + 𝑖(𝑡0 )
𝐿
𝑡0
• Example: Suppose the
voltage across a 2H inductor
is as shown, and the current
at t=0 is 0.5 A

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 Test
The current through a 2mH inductor is given by
𝑖 𝑡 = −𝑡 2 𝐴
What is the voltage across the inductor at time
t=2?

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 Test
The voltage across a 3mH inductor is given by
𝑣 𝑡 = 2𝑡 + 1 𝑉
What is the current through inductor at time
t=3, assuming 𝑖 0 = 0?

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 DC Steady State
• A circuit “in steady state” an ongoing persistent behavior
• DC steady state means all voltages and currents are constant
𝑑𝑖
𝑣=𝐿 =0
𝑑𝑡
• Inductors behave like short circuits in DC steady state

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 Key Concepts
• In passive convention, the I V characteristic is:
𝑑𝑖
𝑣=𝐿
𝑑𝑡
• Current in an inductor is continuous:
𝑖 𝑡 − = 𝑖(𝑡 + )
• An inductor behaves as a short circuit to a DC
current

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 Series and Parallel Inductors
• Definitions of series and parallel elements
𝜇𝑁2 𝐴
• Formula for inductance: 𝐿 =
𝑙
• Inductor I V characteristics:
𝑑𝑖
𝑣=𝐿
𝑑𝑡
𝑡
1
𝑖 𝑡 = 𝑣 𝜏 𝑑𝜏 + 𝑖(𝑡0 )
𝐿
𝑡0

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 Intuitive Series Example
𝜇𝑁2 𝐴
• Inductance of one inductor: 𝐿 =
𝑙

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 Intuitive Series Example
𝜇𝑁2 𝐴
• Inductance of one inductor: 𝐿 =
𝑙
• Consider a second identical inductor in series
with it

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 Intuitive Series Example
𝜇𝑁2 𝐴
• Inductance of one inductor: 𝐿 =
𝑙
• Consider a second identical inductor in series
with it

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 Intuitive Series Example
𝜇𝑁2 𝐴
• Inductance of one inductor: 𝐿 =
𝑙
• Consider a second identical inductor in series
with it
𝜇(2𝑁)2 𝐴
𝐿𝑛𝑒𝑤 = = 2𝐿
2𝑙

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 Derive the General Formula
• Consider three inductors in series, each with
possibly distinct voltages and inductors
• Series elements share the same current:
𝑖1 = 𝑖2 = 𝑖3 = 𝑖
• KVL implies 𝑣1 + 𝑣2 + 𝑣3 = 𝑣

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 Derive the Parallel Formula
• Consider three inductors in • For the nth inductor,
parallel, each with possibly 𝑡
distinct currents and 1
𝑖𝑛 𝑡 = 𝑣 𝜏 𝑑𝜏 + 𝑖𝑛 (𝑡0 )
inductances 𝐿𝑛
𝑡0
• Parallel elements share the
same voltage, 𝑣
• KCL implies 𝑖 = 𝑖1 + 𝑖2 + 𝑖3

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 Example
Determine the voltage, v(t), in the circuit below.

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 Test
Using parallel and series calculations of equivalent
inductance, find the inductance between point a and
point b. Enter the value in mH in the box without units.

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 Key Concepts
• Series:
𝑁
𝐿𝑒𝑓𝑓 = 𝑛=1 𝐿𝑛
• Parallel:
1
𝐿𝑒𝑓𝑓 = 𝑁
𝑛=1 𝐿𝑛
• The series and parallel inductor formulas are
similar to those of resistors

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 Energy in Reactive Elements
Objective:
• Learn how to compute the energy stored and
the power supplied or absorbed in inductors
and capacitors

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 Energy in Reactive Elements
Builds on:
• The passive convention, power absorbed is
𝑝𝑎𝑏𝑠 = 𝑣𝑖
• 𝑝𝑠𝑢𝑝 = −𝑝𝑎𝑏𝑠
𝑡
• Energy stored is 𝑤 𝑡 = 𝑝
−∞ 𝑎𝑏𝑠
𝜏 𝑑𝜏
𝑑𝑣
• 𝑖=𝐶
𝑑𝑡
𝑑𝑖
• 𝑣=𝐿
𝑑𝑡
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 Example – Indirect Method
• The voltage below is across a 3F
capacitor. Determine the energy
stored, w(t).

• Procedure (indirect):
𝑡

𝑤 𝑡 = 𝑖 𝜏 𝑣 𝜏 𝑑𝜏
−∞
𝑑𝑢
Using 𝑖 = 𝐶
𝑑𝑡

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 Example – Direct method
• Given 𝑣 𝑡 across the capacitor, the stored
energy can be calculated directly, as
1
𝑤 𝑡 = 𝐶𝑣 2 (𝑡)
2

• In C is in Farads and 𝑣 is in Volts, then w is in

Joules (J)
• The current direction doesn’t matter

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Proof of the Direct Method
𝑡 𝑡
𝑑𝑣
𝑤 𝑡 = 𝑣𝑖 𝑑𝜏 = 𝑣 𝐶 𝑑𝜏 =
𝑑𝜏
−∞ −∞
𝑡
1 2
=𝐶 𝑣 𝑑𝑣 = 𝐶𝑣
2
−∞

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 Example – Direct Method
• Suppose the voltage below is
across a 3F capacitor. Calculate the
stored energy for t≥0.

• Procedure:
1
– Simply calculate: 𝑤 𝑡 = 𝐶𝑣 2
2

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 Test
2V is placed across a 500 𝜇𝐹 capacitor. How
much energy is being stored in the capacitor?

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 Energy stored in an Inductor
• Given the current 𝑖(𝑡) flowing through the inductor,
the stored energy can be calculated directly, as

1 2
𝑤 𝑡 = 𝐿𝑖 (𝑡)
2

• If 𝐿 is in Henrys, 𝑖 is in Amps, then 𝑤 is in Joules (J)

𝑡
• Indirect method: 𝑤 𝑡 = −∞
𝑖𝑣 𝑑𝜏, where
𝑑𝑖
𝑣=𝐿
𝑑𝑡
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 Inductor Energy Example - Direct
• Suppose the current shown flows through a
0.5H inductor. Determine the energy stored,
𝑤(𝑡), and the power supplied, 𝑝𝑠𝑢𝑏 (𝑡)

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 Test
For a L=2H inductor, and at a specified point in
time, the voltage is 𝑣 = 3𝑉 and the current is
𝑖 = 4𝐴. How much energy is stored in the
inductor at this moment in time

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 Compare Energy for Capacitor and
Inductors
1 1 2
𝑤 𝑡 = 𝐶𝑣 2 (𝑡) 𝑤 𝑡 = 𝐿𝑖 (𝑡)
2 2

Exchange 𝐿 for 𝐶 and 𝑖 for 𝑣

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 Key Concepts
• The direct method for calculating energy
stored in a reactive element:
1 2
–𝑤 𝑡 = 𝐿𝑖 (𝑡), for inductors
2
1
–𝑤 𝑡 = 𝐶𝑣 2 (𝑡), for capacitors
2
• Find the power absorbed using
𝑑𝑤
𝑝𝑎𝑏𝑠 =
𝑑𝑡

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© G. Berdibekov