ES250: Electrical Science. HW7: Energy Storage Elements

ES250:

Electrical Science

HW7: Energy Storage Elements

• This chapter introduces two more circuit elements, the

Introduction

This chapter introduces two more circuit elements, the capacitor and the inductor whose elements laws involve integration or differentiation; consequently, circuits

containing capacitors and/or inductors:

– are represented by differential equations, i.e., called

d i i it d t t ti i it

dynamic circuits, as opposed to static circuits – able to store energy

– have memory meaning that the voltages and currents athave memory, meaning that the voltages and currents at a particular time depend past values

• In addition we will see that:

Introduction

• In addition, we will see that:

– capacitor voltages and inductor currents are continuous functions of time assuming bounded currents/voltages, g / g , e.g., the voltage across a capacitor or the current through an inductor cannot change instantaneously

– in a dc circuit, capacitors act like open circuits and inductors act like short circuits

a set of series or parallel capacitors can be reduced to an – a set of series or parallel capacitors can be reduced to an

equivalent capacitor and a set of series or parallel inductors can be reduced to an equivalent inductor inductors can be reduced to an equivalent inductor

– an op amp and a capacitor can be used to make circuits that perform the mathematical operations of integration or differentiation

• A capacitor is a two‐terminal element that can be modeled

Capacitors

by two conducting plates separated by a nonconducting material, as shown below:

Capacitor law:

⇒

where

• The circuit symbol for a capacitor is shown with the passive

Capacitors

sign convention assumed:

Capacitor law:

⇒

where

• Find the current for a capacitor C = 1 mF when the voltage

Example 7.2‐1: Capacitor Current & Voltage

Find the current for a capacitor C 1 mF when the voltage across the capacitor is represented by the signal below:

Example 7.2‐1: Capacitor Current & Voltage

• We can solve for the voltage v(t) across a capacitor in terms

Capacitors

of the current i(t) by integrating the element law:

Capacitor law:

⇒

∴

• The time tThe time t₀₀ is called the initial time, and the capacitoris called the initial time, and the capacitor

voltage v(t₀) is called the initial condition; it is convenient to select t₀ = 0 as the initial time

• Find the voltage v for a capacitor C = 1/2 F when the current

Example 7.2‐2: Capacitor Current & Voltage

Find the voltage v for a capacitor C 1/2 F when the current is represented by the signal below and v(0) = 0:

It is important to note that the It is important to note that the voltage waveform cannot change instantaneously, but the current

f d

waveform may do so.

Example 7.2‐2: Capacitor Current & Voltage

• The resulting voltage waveform, as shown below, changes

Example 7.2‐2: Capacitor Current & Voltage

The resulting voltage waveform, as shown below, changes with t ² during the first 1 s, linearly with t during the period from 1 to 2 s, and stays constant equal to 3 V after t = 2 s:

• Determine the value of the capacitance associated with the

Example 7.2‐3 Capacitor Current & Voltage

Determine the value of the capacitance associated with the current and voltage waveforms for the circuit below:

Example 7.2‐3: Capacitor Current & Voltage

Note interpreting the equation graphically:

Note, interpreting the equation graphically:

Example 7.2‐3 Capacitor Current & Voltage

∴

∴

⇒

⇒

• Determine the values the constants, a and b, used to label

Example 7.2‐4: Capacitor Current & Voltage

Determine the values the constants, a and b, used to label the plot of the capacitor current associated with the current and voltage waveforms for the circuit below:

Example 7.2‐3: Capacitor Current & Voltage

Note interpreting the equation graphically:

Note, interpreting the equation graphically:

( )

24 1 5 2 10 600 0.04

5 a a a

µ

  −

=   − × = ⇒ =

 5

µ

 

( )

24  1 b 7 5 10⁻ 400b b 0 06

− =²⁴  

(

)

5 b b b

= 

µ

 

Questions?

Questions?

• The energy stored in a capacitor is:

Energy Storage in a Capacitor

The energy stored in a capacitor is:

• Since we have:

• Assuming the capacitor was uncharged at t = −∞, set

( ) 0 th f

v(−∞) = 0; therefore:

• For example, consider a 100‐mF capacitor that has a voltage

Energy Storage in a Capacitor

For example, consider a 100 mF capacitor that has a voltage of 100 V across it. The energy stored is:

• As long as the capacitor is not connected to any otherAs long as the capacitor is not connected to any other element, the energy of 500 J remains stored

• If we connect the capacitor to the terminals of a resistor weIf we connect the capacitor to the terminals of a resistor, we expect a current to flow until all the energy is dissipated as heat by the resistor

– After all the energy dissipates, the current is zero and the voltage across the capacitor is zero.g p

• As noted in the previous section, the requirement of

Energy Storage in a Capacitor

As noted in the previous section, the requirement of conservation of charge implies that the voltage on a capacitor is continuous

– Thus, the voltage and charge on a capacitor cannot change instantaneously

– This statement is summarized by the equation

• For the circuit shown below assume the switch was closed

Energy Storage in a Capacitor

For the circuit shown below assume the switch was closed for a long time and the capacitor voltage had become

v_c = 10 V; then at time t = 0 we open the switch, as shown below

• Since the voltage on the capacitor is continuous:

• A 10‐mF capacitor is charged to 100 V, as shown in the

Ex 7.3‐1: Energy Stored by a Capacitor

A 10 mF capacitor is charged to 100 V, as shown in the circuit below; find the energy stored by the capacitor and the voltage of the capacitor at t = 0⁺ after the switch is

opened:

• Solution ‐ The voltage of the capacitor is v = 100 V at t = 0⁻; since the voltage at t = 0⁺ is the same at t = 0⁻, we have:

• The voltage across a 5‐mF capacitor varies as shown below;

Ex 7.3‐2: Power and Energy for a Capacitor

The voltage across a 5 mF capacitor varies as shown below;

determine and plot the capacitor current, power, and energy associated with the capacitor:

• The current is determined from i_c = C dv/dt as shown:

Solution

The current is determined from i_c C dv/dt as shown:

• The power is v(t)i(t), i.e., rhe product of the current curve

Solution

The power is v(t)i(t), i.e., rhe product of the current curve and the voltage curve as shown below

– note, the capacitor receives energy during the first two , p gy g seconds and then delivers energy for the period 2 < t < 3:

• The energy is ω = ∫p dt and can be found as the area under

Solution

The energy is ω ∫p dt and can be found as the area under the p(t) curve, as shown below:

Questions?

Questions?

• The equivalent circuit for N parallel capacitors:

Series and Parallel Capacitors

The equivalent circuit for N parallel capacitors:

≡

• The equivalent circuit for N series capacitors:

Series and Parallel Capacitors

≡≡

• Find the equivalent capacitance for the circuit below when

Ex: 7.4‐1 Parallel & Series Capacitors

C₁ = C₂ = C₃ = 2 mF, v₁(0) = 10 V, and v₂(0) = v₃(0) = 20 V:

• Since CSince C₂₂ and Cand C₃₃ are in parallel, we replace them with C ,are in parallel, we replace them with C_p, where:

• Circuit resulting from replacing C

Ex: 7.4‐1 Parallel & Series Capacitors

2 3 p

• Next, replace the two series capacitors C₁ and C_p with equivalent capacitor C where:

equivalent capacitor C_s where:

• To obtain the Equivalent circuit:

Ex: 7.4‐1 Parallel & Series Capacitors

• Note the initial voltage across the equivalent capacitor CNote, the initial voltage across the equivalent capacitor C_s can be computed as:

• Find the equivalent capacitance of the circuit below:

Exercise 7.4‐1

( )

(

)

Ceq =

( (

)

)

( )

( )

3mF 9mF 6mF

12 F 6 F

q

= + &

&

(

)

4mF

=

=

&

Questions?

Questions?

• An inductor can be formed

Inductors

from a wire shaped as a multi‐

turn coil with the voltage across

h d l

the inductor proportional to the rate of change of the

current through the inductor current through the inductor, where L is the constant of

proportionality called

p p y

inductance measured in henrys (H), expressed as shown:

• Inductance is a measure of the ability of a device to store energy in the form of a magnetic field

• A magnetic flux φ(t) is

Inductors

associated with a current i(t) in a coil

• In the case of an N‐turn coil where each line of flux passes th h ll il t th t t l through all coil turns; the total flux is given as Nφ = iL

• Faraday found that a changing

• Faraday found that a changing flux creates an induced voltage in each coil turn equal to the in each coil turn equal to the derivative of the flux φ, so the total voltage v across N turns is:

Note: Flux describes the strength and extent of an object's interaction with a magnetic field.

• Inductors are wound in various forms with practical

Inductors

inductances ranging from 1 μH to 10 H

• The circuit symbol for an indictor is shown with the passive sign convention assumed:

Inductor law:

• Conservation of flux applies to the inductor and therefore pp the flux , and thus current, through the inductor cannot have discontinuities, i.e., cannot change instantaneously

• We can solve for the current i(t) across an inductor in terms

Inductors

of the voltage v(t) by integrating the element law:

Inductor law:

∴

• The time tThe time t₀₀ is called the initial time, and the inductoris called the initial time, and the inductor

current i(t₀) is called the initial condition; it is convenient to select t₀ = 0 as the initial time

• Consider the voltage waveform shown below for an inductor

Inductors

when L = 0.1 H and i(t₀) = 2 A:

• Since v(t) = 2 V between t = 0 and t = 2, we have:

Inductors

which results in the following current waveform:

• Find the voltage across a L = 0.1 H inductor when the

Ex. 7.5‐1: Inductor Current & Voltage

current in the inductor for t > 0 and i(0) = 0 is:

• The voltage for t > 0 is given by:

Solution

• The plots below represent the current and voltage of the

Ex. 7.5‐2: Inductor Current & Voltage

inductor in the circuit; determine the value of L:

• The current and voltage of the inductor are related by:

Solution

Note, interpreting the equation graphically:

• Picking convenient values for t and t₀, e.g., t₀ = 2 ms and t = 6 ms yields:

t 6 ms yields:

Solution

∴

⇒

• The input voltage and current to the inductor circuit shown

Ex. 7.5‐3: Inductor Current & Voltage

below are given by:

• Determine the values of the inductance, L, and resistance, R, if the initial inductor current is i_L(0) = −3.5 A

Apply KCL at either node to get:

• Substituting in the time domain expressions for the

Ex. 7.5‐3: Inductor Current & Voltage

specified inductor voltage and current yields:

⇒

• Equating coefficients gives:

Ex. 7.5‐3: Inductor Current & Voltage

Questions?

Questions?

• Similar to a capacitor, the ideal inductor only stores energy;

Energy Storage in an Inductor

Similar to a capacitor, the ideal inductor only stores energy;

it is thus referred to as a passive element meaning it cannot generate or dissipate energy

• The energy stored in an inductor can be calculated in a similar manner as we did for a capacitor; resulting in:

• Find the current in a L = 0.1 H inductor when the initial

Ex. 7.6‐1: Inductor Voltage & Current

Find the current in a L 0.1 H inductor when the initial current is zero and the voltage across the inductor is

as shown below:

• Given the voltage waveform, the current can be computed

Ex. 7.6‐1: Inductor Voltage & Current

Given the voltage waveform, the current can be computed as:

• Find the power and energy for an inductor of 0.1 H when

Ex. 7.6‐2: Power & Energy for an Inductor

Find the power and energy for an inductor of 0.1 H when the current and voltage are given as:

• Find the power and the energy stored in a 0.1‐H inductor

Ex. 7.6‐3: Power & Energy for an Inductor

Find the power and the energy stored in a 0.1 H inductor when i = 20te^−2t A and v = 2e^−2t(1 − 2t) V for t ≥ 0 and i = 0 for t < 0

Solution ‐

• The energy is then given by:

Questions?

Questions?

• A series connection of N inductors can be reduced to an

Series and Parallel Inductors

A series connection of N inductors can be reduced to an equivalent series inductor L_s, as shown:

≡

• A parallel connection of N inductors can be reduced to an

Series and Parallel Inductors

A parallel connection of N inductors can be reduced to an equivalent parallel inductor L_p, as shown:

≡

• Find the equivalent inductance for the circuit of below

Ex. 7.7‐1: Series & Parallel Inductors

Find the equivalent inductance for the circuit of below assuming all the inductor currents are zero at t₀:

• Solution ‐

mH and 20‐mH inductors in parallel as shown:

mH and 20‐mH inductors in parallel, as shown:

⇒

• This equivalent inductor is in series with the 2‐mH and 3‐mH

Ex. 7.7‐1: Series & Parallel Inductors

This equivalent inductor is in series with the 2 mH and 3 mH inductors as shown below:

≡≡

therefore, we obtain

• Find the equivalent inductance of the circuit shown below:

Exercise 7.7‐2

Find the equivalent inductance of the circuit shown below:

( )

(

)

L = & + &

≡

( )

(

)

4m 12m

2mH 20mH

4 +12

Leq +

 ×  

=   + 

& &

&

≡ ₍

₎

 

  

 

= + & = &

5m 20m 5m+20m 4mH

= × =

Questions?

Questions?

• In this section we concentrate on finding the change in

Initial Conditions of Switched Circuits

In this section we concentrate on finding the change in

selected variables in a circuit when a switch is thrown from open to closed or vice versa

• The time of throwing the switch is considered to be t = 0,

and we want to determine the value of the variable at t = 0−

and t = 0+, immediately before and after throwing the

switch; thus, a switched circuit is a circuit with one or more switches that open or close at time t0 = 0 as shown below:

switches that open or close at time t0 = 0, as shown below:

• We are particularly interested in the change in the current

Initial Conditions of Switched Circuits

We are particularly interested in the change in the current and voltage of energy storage elements after the switch is thrown since these variables along with the sources will dictate the behavior of the circuit for t > 0

• We summarize the important characteristics of the behavior of an inductor and a capacitor on the next slide:

– note that we assume t0 = 0

– recall that an instantaneous change in the inductor current or the capacitor voltage is not permitted

– however, it is possible to instantaneously change an inductor's voltage or a capacitor's current

• Consider the switched inductor circuit shown previously:

– the switch is open at t = 0⁻ and closes at t = 0

– the circuit has attained steady‐state conditions before the switch is thrown (closed), so the inductor is a short circuit – let R₁ = R₂ = 1, so that the source current at t = 0⁻ divides

equally between i₁ and i_L, with

• Since the current cannot change instantaneously for the

Initial Conditions of Switched Circuits

Since the current cannot change instantaneously for the inductor, we have

• However, the current in the resistor can change , g instantaneously; note prior to t = 0, we have

• After the switch is thrown, we require that the voltage across R₁ be equal to zero because of the switched short circuit, and therefore

• Thus, the resistor current changes abruptly from 1 to 0

• Now consider the switched capacitor circuit shown below:

Initial Conditions of Switched Circuits

Now consider the switched capacitor circuit shown below:

• prior to t = 0, the switch has been closed for a long time

• since the source is a constant the current in the capacitor issince the source is a constant, the current in the capacitor is zero for t < 0 because the capacitor appears as an open

circuit in a steady‐state condition

• therefore, the voltage across the capacitor for t < 0 can be obtained using voltage divider as shown

obtained using voltage divider as shown

• Therefore at t 0⁻ when v 10 and R R 1 Ω we

Initial Conditions of Switched Circuits

• Therefore at t = 0 when v_s = 10 and R₁ = R₂ = 1 Ω, we obtain:

• However, since the voltage across a capacitor cannot change instantaneously, we have

• Thus, when the switch is opened at t = 0, the source is removed from the circuit, but v_c(0⁺) remains equal to 5 V

• To analyze a circuit with both a capacitor and an inductor, as

Initial Conditions of Switched Circuits

To analyze a circuit with both a capacitor and an inductor, as shown below, we must find v_c(0⁻) and i_L(0⁻) in steady‐state:

0 t<

≡

( )

( )

(0 ) 3 (2 3) 10

6V (0 )

c

c

v

v

−

+

= +

= =

(0 ) (0 ) 3

2A (0 )

L c

L

i v

i

− −

+

=

= =

• Find i_L(0⁺), v_c(0⁺), dv_c(0⁺)/dt, and di_L(0⁺)/dt for the circuit

Ex. 7.8‐1: Initial Conditions in a Switched Circuit

Find i_L(0 ), v_c(0 ), dv_c(0 )/dt, and di_L(0 )/dt for the circuit below:

• Redraw the circuit for t = 0⁻ by replacing the inductor with a

Solution

Redraw the circuit for t 0 by replacing the inductor with a short circuit and the capacitor with an open circuit, as

shown below:

Note that:

• In order to find dv_c(0⁺)/dt and di_L(0⁺)/dt, we throw the

Solution

In order to find dv_c(0 )/dt and di_L(0 )/dt, we throw the switches at t = 0 and redraw the circuit, as shown below:

• Since we wish to find dv_c(0⁺)/dt, recall that

⇒

• To find i_c we write KCL at node a to obtain

Solution

To find i_c we write KCL at node a to obtain

• Therefore at t = 0⁺

• Therefore, at t = 0

• Hence, we obtain

• Since we wish to find di_L(0⁺)/dt, recall that

Solution

Since we wish to find di_L(0 )/dt, recall that

T fi d it KVL f th i ht h d h t bt i

⇒

• To find v_L we write KVL for the right‐hand mesh to obtain

• Therefore, at t = 0⁺

• Hence, we obtain

• Thus, we find that at the switching time t = 0, the current in

Solution

Thus, we find that at the switching time t 0, the current in the inductor and the voltage of the capacitor remain

constant

• However, the inductor voltage changes instantaneously from v_L(0⁻) = 0 to v_L(0⁺) = −2 V, and we determined that di_L(0⁺)/dt = −2 A/s

• Also, the capacitor current changed instantaneously from p g y i_c(0⁻) = 0 to i_c(0⁺) = 6 A and we found that dv_c(0⁺)/dt = 12 V/s

Questions?

Questions?