Week 3 Capacitance & Inductance Transitions

Try your hand at graphing these functions Graph: y = e x Graph: y = e -x Graph: y = 1 - e -x

exy y = e x

exy y = e -x

exy y = 1 - e -x

First-order System Has the same equation

Time Constant

RC Circuit – Initial Conditions An RC circuit is one where you have a capacitor and resistor in the same circuit. Suppose we have the following circuit: Initially, the capacitor is UNCHARGED (q = 0) and the current through the resistor is zero. A switch (in red) then closes the circuit by moving upwards. The question is: What happens to the current and voltage across the resistor and capacitor as the capacitor begins to charge as a function of time? Time(s) VCVC Which path do you think it takes?

Voltage Across the Resistor - Initially V Resistor t (sec)  If we assume the battery has NO internal resistance, the voltage across the resistor will be the EMF. After a very long time, V cap =  as a result the potential difference between these two points will be ZERO. Therefore, there will be NO voltage drop across the resistor after the capacitor charges. Note: This is while the capacitor is CHARGING.

Current Across the Resistor - Initially t (sec) I max =  /R Since the voltage drop across the resistor decreases as the capacitor charges, the current across the resistor will reach ZERO after a very long time. Note: This is while the capacitor is CHARGING.

Voltage Across the Capacitor - Initially t (sec) V cap  As the capacitor charges it eventually reaches the same voltage as the battery or the EMF in this case after a very long time. This increase DOES NOT happen linearly. Note: This is while the capacitor is CHARGING.

Current Across the Capacitor - Initially t (sec) I max =  /R Since the capacitor is in SERIES with the resistor the current will decrease as the potential difference between it and the battery approaches zero. It is the potential difference which drives the value for the current. Note: This is while the capacitor is CHARGING.

Time Domain Behavior The graphs we have just seen show us that this process depends on the time. Let’s look then at the UNITS of both the resistance and capacitance. Unit for Resistance =  = Volts/Amps Unit for Capacitance = Farad = Coulombs/Volts

The “Time” Constant It is clear, that for a GIVEN value of "C”, for any value of “R” it effects the time rate at which the capacitor charges or discharges. Thus the PRODUCT of R and C produce what is called the CIRCUIT Capacitive TIME CONSTANT. We use the Greek letter, Tau, for this time constant. The question is: What exactly is the time constant? Another way to express R Another way to express Farads as Coulombs/Volt

The “Time” Constant The time constant is the time that it takes for the capacitor to reach 63% of the EMF value during charging.

Let’s test our function 1RC31RC2RC4RC  0.63   0.86   0.95   0.98   Applying each time constant produces the charging curve we see. For practical purposes the capacitor is considered fully charged after 4-5 time constants( steady state). Before that time, it is in a transient state. Steady State Transient State ε is the full voltage of the source

Charging Functions Likewise, the voltage function can be divided by another constant, in this case, “R”, to derive the current charging function. Now we have 3 functions that allows us to calculate the Charge, Voltage, or Current at any given time “t” while the capacitor is charging. Charge and voltage build up to a maximum… …while current fades to zero

Capacitor Discharge – Resistor’s Voltage Suppose now the switch moves downwards towards the other terminal. This prevents the original EMF source to be a part of the circuit. V Resistor t (sec)  At t =0, the resistor gets maximum voltage but as the capacitor cannot keep its charge, the voltage drop decreases.

Capacitor Discharge – Resistor’s Current I Resistor t (sec) I  R Similar to its charging graph, the current through the resistor must decrease as the voltage drop decreases due to the loss of charge on the capacitor.

Capacitor Discharge – Capacitor's Voltage The discharging graph for the capacitor is the same as that of the resistor. There WILL be a time delay due to the TIME CONSTANT of the circuit. In this case, the time constant is reached when the voltage of the capacitor is 37% of the EMF.

Capacitor Discharge – Capacitor’s Current I cap t (sec) I  R Similar to its charging graph, the current through the capacitor must decrease as the voltage drop decreases due to the loss of charge on the capacitor.

The bottom line to take away…. When charging a capacitor Charge and voltage build up to a maximum… …while current fades to zero. When discharging a capacitor All three fade away during discharge. Time to charge 63% = time constant “tau” = τ = RC

Capacitor Circuit Operation

Recall the Circuit Representation  LINEAR Caps Follow the Capacitance Law; in DC  The Basic Circuit-Capacitance Equation Where Q  The CHARGE STORED in the Cap, Coulombs C  Capacitance, Farad V c  DC-Voltage Across the Capacitor Discern the Base Units for Capacitance

“Feel” for Capacitance Pick a Cap, Say 12 µF  Recall Capacitor Law Now Assume That The Cap is Charged to hold 15 mC Find V c  Solving for V c Caps can RETAIN Charge for a Long Time after Removing the Charging Voltage Caps can Be DANGEROUS!

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  The Differential Suggests SEPARATING Variables  Leads to The INTEGRAL form of the Capacitance Law  If at t 0, v C = v C (t 0 ) (a KNOWN value), then the Integral Law becomes

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 Application  If v C is NOT Continuous then dv C /dt → , and So i C → . This is NOT PHYSICALLY possible

Capacitor Differential Law Express the CURRENT “Thru” the Cap Using the Differential Law  Thus a Major Mathematical Implication of the Differential Law  If v C = Constant Then  This is the DC Steady-State Behavior for a Capacitor  A Cap with CONSTANT Voltage Across it Behaves as an OPEN Circuit  Cap Current Charges do NOT flow THRU a Cap –Charge ENTER or EXITS The Cap in Response to Voltage CHANGES

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 Capacitance Law CapacitorOhm Now Recall the Long Form of the Integral Relation Note The Passive Sign Convention The DEFINITE Integral is just a number; call it v C (t 0 ) so

Capacitor Summary cont Consider Finally the Differential Application Some Implications For small Displacement Current dv C /dt is small; i.e, v C changes only a little Obtaining Large i C requires HUGE Voltage Gradients if C is small Conclusion: A Capacitor RESISTS CHANGES in VOLTAGE ACROSS It

CONCLUSION: Capacitance

The Inductor Second of the Energy-Storage Devices Basic Physical Model: Circuit Symbol

Physical Inductor Inductors are Typically Fabricated by Winding Around a Magnetic (e.g., Iron) Core a LOW Resistance Wire –Applying to the Terminals a TIME VARYING Current Results in a “Back EMF” voltage at the connection terminals Some Real Inductors

Inductance Defined From Physics, recall that a time varying magnetic flux, , Induces a voltage Thru the Induction Law  Where the Constant of Proportionality, L, is called the INDUCTANCE  L is Measured in Units of “Henrys”, H 1H = 1 Vs/Amp  Inductors STORE electromagnetic energy  They May Supply Stored Energy Back To The Circuit, But They CANNOT CREATE Energy For a Linear Inductor The Flux Is Proportional To The Current Thru it

Inductance Sign Convention Inductors Cannot Create Energy; They are PASSIVE Devices All Passive Devices Must Obey the Passive Sign Convention

Inductor Circuit Operation Recall the Circuit Representation Separating the Variables and Integrating Yields the INTEGRAL form Previously Defined the Differential Form of the Induction Law In a development Similar to that used with caps, Integrate −  to t 0 for an Alternative integral Law

Drill Problem 4-13, pp The 0.05F in P4-13 is initially charged to 8v. At t = 0, a 20v source is connected. Determine the expressions for: I(t) and v c (t) for t > 0 I(t)=1.2e -2t v c (t) = 20-12e -2t

Summary

Reactive Element Initial Conditions t = 0 + Final Condition t = ∞ Stored Quantity? Source? NoYes DCAC Capacitor Short Circuit Voltage Source Open CircuitShort Circuit InductorOpen CircuitCurrent SourceShort CircuitOpen Circuit Circuit Behavior of Reactive Components

CONCLUSION: Inductance