Capacitor Integral Law and Inductor Behavior, Slides of Electrical Circuit Analysis

Download Capacitor Integral Law and Inductor Behavior and more Slides Electrical Circuit Analysis in PDF only on Docsity!

Capacitors &

Inductors

Capacitance & Inductance

• Introduce Two Energy STORING Devices
  • Capacitors
  • Inductors
• Outline
  • Capacitors
    • Store energy in their ELECTRIC field (electroStatic energy)
    • Model as circuit element
  • Inductors
    • Store energy in their MAGNETIC field (electroMagnetic energy)
    • Model as circuit element
  • Capacitor And Inductor Combinations
    • Series/Parallel Combinations Of Elements

Capacitance Defined

• Consider the Basic Physical Model

 Where

• A ≡The Horizontal Plate-Area, m 2
• d ≡ The Vertical Plate Separation Distance, m
• ε 0 ≡ “Permittivity” of Free Space; i.e., a vacuum - A Physical CONSTANT - Value = 8.85x10- Farad/m

 The Capacitance, C, of the Parallel-Plate Structure

w/o Dielectric d

A

C 0

ε

 Then What are the UNITS of Capacitance, C  Typical Cap Values → “micro” or “nano”

Capacitor Circuit Operation

• Recall the Circuit Representation

 LINEAR Caps Follow the Capacitance Law; in DC

 The Basic Circuit- Capacitance Equation

q ( )t^ = Cvc( )t

 Where

• Q ≡ The CHARGE STORED in the Cap, Coulombs
• C ≡ Capacitance, Farad
• Vc ≡ DC-Voltage Across the Capacitor  Discern the Base Units for Capacitance

Q = CV c

Volt

Coulomb Farad =

Forms of the Capacitor Law

• The time-Invariant Cap Law  If v (^) C at −∞ = 0, then the traditional statement of the Integral Law

 Leads to DIFFERENTIAL Cap Law ( )

( ) ( )

dt

dv t

C

dt

dq t

i t = = C

Q = CV c

 The Differential Suggests SEPARATING Variables

i ( )t dt^ = CdvC( )t

 Leads to The INTEGRAL form of the Capacitance Law

( ) (^) ( )

( ) ∫− (^) ∞ =^ ∫ −∞

v t v

t (^) C C

i y dy C dz

t

C i y dy

C

v t

 If at t 0 , vC = vC(t 0 ) (a KNOWN value), then the Integral Law becomes ( ) ( ) ( )

( ) ( ) (^) ∫ ( )

∫ ∫

= +

= (^) −∞ +

t C C t

t t

t C

i y dy C

v t v t

i y dy C

i y dy C

v t

0

0

0

1

1 1

0

Capacitor Integral Law

• Express the VOLTAGE Across the Cap Using the INTEGRAL Law

 Thus a Major Mathematical Implication of the Integral law

 If i(t) has NO Gaps in its i(t) curve then

 Even if i(y) has VERTICAL Jumps:

 The Voltage Across a Capacitor MUST be Continuous  An Alternative View

• The Differential Reln

( )

( ) = = ∫− (^) ∞ ( )

t C (^) C C iC y dy

q t v t

( ) (^) ∫ ( )

+∆ ∆ → +^ ∆ =∆→ −∞

t t t vC^ t t t C i y dy lim lim^1 0 0

v (^) C (t t) vC( )t t

∆ → 0

lim

 If vC is NOT Continous then dvC/dt → ∞, and So iC → ∞. This is NOT PHYSICALLY possible

( )

( ) dt

dv t iC t =C C

Capacitor Current

• Charges do NOT flow THRU a Cap
• Charge ENTER or EXITS The Capacitor in

Response to the Voltage Across it

• That is, the Voltage-Change DISPLACES the Charge Stored in The Cap - This displaced Charge is, to the REST OF THE CKT, Indistinguishable from conduction (Resistor) Current
• Thus The Capacitor Current is Called the

“Displacement” Current

Capacitor Summary

• The Circuit Symbol (^)  From Calculus, Recall an Integral Property

 Compare Ohm’s Law and Capactitance Law Cap Ohm

 Now Recall the Long Form of the Integral Relation

v C

i C  Note The Passive Sign Convention

( ) (t ) dt

i t C dvc C =

∫ −∞

=

t vC t C iC(x)dx ( )^1 R R

R R

v Ri

v

R

i

∫ ( )^ = ∫ ( )^ +∫ ( ) −∞ −∞

t t

t t f x dx f x dx f x dx 0

0

∫ ∫ −∞

= +

0

0

( )^1 ( )^1 ( )

t (^) t

t

vC t C iC x dx C iC x dx

 The DEFINITE Integral is just a no.; call it vC(t 0 ) so

= + ∫

t

t

vC t vC t C iC x dx 0

( ) ( 0 )^1 ( )

iC Defined by

Differential

• The fact that the Cap Current is defined through a DIFFERENTIAL has important implications...
• Consider the Example at Left - Shows vC(t) - C = 5 μF - Find iC(t)

C= 5 μ F

i ( t )= Cdvdt ( t )

i F Vs 20 mA 6 10

5 10 [ ]^24 3

(^6) = 

 

 ×

= × − × −

− 60 mA

i (t )= 0 elsewhere

 Using the 1st^ Derivative (slopes) to find i(t)

Capacitor Energy Storage

• UN like an I-src or V-src a Cap Does NOT Produce Energy
• A Cap is a PASSIVE Device that Can STORE Energy
• Recall from Chp.1 The Relation for POWER

 For a Cap

 Then the INSTANTANEOUS Power

 Recall also

p = vi

p ( )t = v( )t ⋅i( )t

v C

i C

p (^) C ( )t = vC( )t ⋅iC( )t

( ) (t ) dt

dv i (^) C t = C C

 Subbing into Pwr Reln

dt

dv

p C (t ) =CvC(t) C

 By the Derivative CHAIN RULE  

( ) v^2 t dt

d p (^) C t C C

dt

dv dv

d dt

d (^) C C

= ⋅

Capacitor Energy Storage cont.

• Then Energy in Terms of Capacitor Stored-Charge

 The Total Energy Stored during t = 0-6 ms

 Short Example (^)  w C Units?

 Charge Stored at 3 mS

wC ( 0 , 6 )= CvC^2 − Cv C^2

V Coul V J V

Coul F ⋅V^2 = ⋅^2 = ⋅ =

( 1 , 2 )^22 q^2 t 1 C

q t C

wC t t = C − C

VC (t) C = 5 μF

5 * 10 [ ]*( 24 ) [ ] 2

w ( 0 , 6 )^16 F^2 V^2 C =^ −

qC ( 3 )=CvC( 3 ) qC( 3 )= 5 * 10 −^6 [F]* 12 [V]= 60 μ C

Some Questions About Example

• For t > 8 mS, What is the Total Stored CHARGED?

 For t > 8 mS, What is the Total Stored ENERGY?

qC( t > 8 mS) = 0

vC (t) C = 5 μF

wC( t > 8 mS) = 0

CHARGING Current DIScharging Current 2 2 (^ )^502 2

9 1 2

w ( −∞, t )=^1 Cv t = = ⋅ F⋅ C C^ μ

2 2 (^2 )^02

5

( , )^1111 F

q t C

wC t C μ

−∞ = = =

Capacitor Summary cont.

• Consider A Cap Driven

by A SINUSOIDAL V- Src

 At 135° = (3/4)π

 Electric power absorbed by Cap at a given time

C = 2 μ F

v ( t )= 130 sin( 120 π t )

v ( t ) i(t)

p (^) C ( t)=vC(t)iC(t )

p ( ) V ( mA) mW

v V

i mA

i

C

C

C

C

1160 91. 9 69. 3 6371

1160 130 sin( 0. 75 ) 91. 9

1160 69. 3

1 160 2 * 10 6 * 130 * 120 cos( 0. 75 )

= ∗ − = −

= =

=−

= −

π

π π

 The Cap is SUPPLYING Power at At 135° = (3/4)π = 6.25 mS

• That is, The Cap is RELEASING (previously) STORED Energy at Rate of 6.371 J/s

WhiteBoard Work

vc(t)(V)

-12^0 1 2 3 4 5 t(s) Figure P5.

See ENGR-43_Lec-06- 1_Capacitors_WhtBd.ppt

 Let’s Work this Problem

• The VOLTAGE across a 0.1-F capacitor is given by the waveform in the Figure Below. Find the WaveForm Eqn for the Capacitor CURRENT

iC ( )t = [ 2 A] cos( 50000 t) + v C(t)^ -

ANOTHER PROB 0.5 μF