rc circuits

RC CircuitsAP Physics C

Montwood High SchoolR. Casao

RC Circuits To date, we have studied steady-state

direct current circuits in which the current is constant.

In circuits containing a capacitor, the current varies over time.

When a potential difference is applied across a capacitor, the rate at which the capacitor charges depends on the capacitance and on the resistance in the circuit.

An RC circuit can store charge and release it at a later time.

Charging a Capacitor Consider a series circuit

containing a resistor and a capacitor that is initially uncharged.

With switch S open, there is no current in the circuit.

When switch S is closed at t = 0 s, charges begin to flow and a current is present in the circuit and the capacitor begins to charge.

Charging a Capacitor The gap between the capacitor plates

represents an open circuit and charge does not pass from the positive plate to the negative plate.

Charge is transferred from one plate to the other plate through the resistor, switch, and battery until the capacitor is fully charged.

The value of the maximum charge depends on the EMF of the battery.

Once the maximum charge is reached, the current in the circuit is zero.

Charging a Capacitor Applying Kirchhoff’s loop rule to the

circuit after the switch is closed:

I ·R is the potential drop across the resistor.

q/C is the potential drop across the capacitor.

I and q are instantaneous values of the current and charge as the capacitor charges.

0C

qRIE0VRIE

capacitor

Charging a Capacitor At t = 0 s, when the switch is closed, the

charge on the capacitor is 0 C and the initial current is:

At t = 0 s, the potential drop is entirely across the resistor.

As the capacitor is charged to its maximum value Q, the charges quit flowing and the current in the circuit is 0 A and the potential drop is entirely across the capacitor.

R

EI

max

Charging a Capacitor

Maximum charge: From t = 0 s until the capacitor is fully

charged and the current stops, the amount of current in the circuit decreases over time and the amount of charge on the capacitor increases over time.

To determine values for the current in the circuit and for the charge on the capacitor as functions of time, we have to use a differential equation.

ECQmax

Charging a Capacitor – Current Equation

Beginning with Kirchhoff’s loop equation, differentiate the equation with respect to time:

dt

dIR

dt

dq

C

10

dt

dq

C

1

dt

dIR

dt

dq

C

1

dtCq

d

dt

dIR

dt

RId0

dt

dE

0dt

Cq

RIEd


Replace dq/dt with I:

Get the current terms on one side of the equation and the other terms on the other side of the equation:

Integrate both sides of the equation.

dt

dIR

C

I

dt

dIRI

C

1

dtCR

1-IdI


The limits of integration for the current side of the equation is from Imax (at t = 0 s) to the current value at time t.

The limits of integration for the time side of the equation is from t = 0 s to time t.

t

0

I

Idt

CR1-

IdI

max


Left side:

Right side:

max

max

I

I

I

I I

IlnIlnIlnIlndI

I

1maxmax

CRt

0tCR1

tCR1

dtCR1 t

0

t

0


Combining both sides of the integration:

To eliminate the natural log term (ln), we can use the terms as exponents for the base e. From the properties of logarithms:

CRt

II

lnmax

xxln e

CR

t

max

CR

tI

Iln

tosimplifiesmax

eI

Iee


Current in an RC circuit as a function of time:

Graph of Current vs. time for a charging capacitor:

CR

t

CR

t

max R

EII(t)

ee

Charging a Capacitor – Charge Equation

To find the charge on the capacitor as a function of time, begin with Kirchhoff’s loop equation:

Replace I with dq/dt:

0C

qRIE0VRIE

capacitor

0C

qR

dt

dqE


Get the dq/dt term on one side of the equation:

Divide both sides by R to solve for dq/dt:

Common demominator, R·C:

C

qER

dt

dq

CR

q

R

E

dt

dq

CR

qCE

dt

dq

CR

q

CR

CE

dt

dq


Multiply both sides by dt:

Divide both sides by E·C-q to get the charge terms together on the same side of the equation:

In problems involving capacitance, q is positive, so multiply both sides by –1 to make the charge positive:

dtCR

qCEdq

CR

dt

qCE

dq

dtCR

1-

qCE

dq


Integrate both sides of the resulting differential equation. – For the charge side of the equation, the

limits of integration are from q = 0 at t = 0 s to q at time t. I rearranged the equation to put the positive q first followed by the negative E·C.

– For the time side of the equation, the limits of integration are from 0 s to t.

t

0

q

0dt

CR

1-

CEq

dq


Left side:

Right side:

CE

CEqlnCElnCEqln

CE0lnCEqlnCEqlnCEq

dq q

0

q

0

CR

t0t

CR

1

tCR

1dt

CR

1dt

CR

1 t

0

t

0

t

0


Combining both sides of the integrals:

To eliminate the natural log term (ln), we can use the terms as exponents for the base e.

CR

t

CE

CEqln

CRt

CECEq

ln

ee


Simplify:

Solve for q:– Multiply both sides by –E·C.– Add E·C to both sides.

Factor out the E·C:

CR

t

CE

CEq

e

CR

t

CECEq

e

)(1CEq CR

t

e


Substitute: Qmax = E·C

Graph of Charge vs. time for a charging capacitor:

)e(1Qq(t) CR

t

max

Charging A Capacitor

Current has its maximum value of I max = E/R at t = 0 s and decays exponentially to 0 A as t infinity.

Charging A Capacitor

The charge on a capacitor is 0 C at t = 0 s and approaches a maximum value of Qmax = C·E as t infinity.

Charging Capacitor Graphs

current voltage charge

Time Constant RTime Constant R·C·C The quantity R·C, which appears in the

exponential component of the charge and current equations is called the time constant of the circuit.

The time constant is a measure of how quickly the capacitor becomes charged.

The time constant represents the time it takes the: current to decrease to 1/e of its initial value. charge to increase from 0 C to C·E·(1-e-1) =

0.63·C·E. The unit for the time constant is seconds.

Ω · F = (V/A)(C/V) = C/(C/s) = s

Charge and Current during the Charge and Current during the Charging of a CapacitorCharging of a Capacitor

Time, t

Qmaxq

Rise in Rise in ChargeCharge

Capacitor

0.63 I

Time, t

I i

Current Current DecayDecay

Capacitor

0.37 I

In a time In a time of one time constant, the of one time constant, the charge q rises to 63% of its maximum, charge q rises to 63% of its maximum, while the current i decays to 37% of its while the current i decays to 37% of its maximum value.maximum value.

Discharging a Capacitor Removing the battery from the circuit

while keeping the switch open leaves us with a circuit containing only a charged capacitor and a resistor.

Discharging a Capacitor When the switch is open, there is a

potential difference of Q/C across the capacitor and 0 V across the resistor since I = 0 A.

If the switch is closed at time t = 0 s, the capacitor begins to discharge through the resistor and a current flows through the circuit.

At some time during the discharge, current in the circuit is I and the charge on the capacitor is q.

Discharging a Capacitor – Charge Equation

To find the charge on the capacitor as a function of time, begin with Kirchhoff’s loop equation. There is no E term in the equation because the battery has been removed. The I·R term is negative because the energy carried by the charges is dissipated in the resistor.

C

qRI

0C

qRI0VRI

capacitor


Replace I with -dq/dt because the Replace I with -dq/dt because the current in the circuit is decreasing as current in the circuit is decreasing as the capacitor discharges over time:the capacitor discharges over time:

Get the charge terms on one side of Get the charge terms on one side of the equation and the remaining the equation and the remaining variables on the other side of the variables on the other side of the equation.equation.

C

q

dt

dqR

dtCR

1

q

dq


Integrate both sides of the resulting differential equation. – For the charge side of the equation, the limits of

integration are from q = Qmax at t = 0 s to q at time t.

– For the time side of the equation, the limits of integration are from 0 s to t.

t

0

q

Q

t

0

q

Q

dtCR

1dq

q

1

dtCR

1

q

dq

max

max


Left side:

Right side:

max

max

q

Q

q

Q Q

qlnQlnqlnqlndq

q

1maxmax

CR

t0t

CR

1t

CR

1dt

CR

1 t

0

t

0


Combining both sides of the integrals:

To eliminate the natural log term (ln), we can use the terms as exponents for the base e.

CR

t

Q

qln

max

CR

tQ

qln

max

ee


Simplify:

Solve for q:

CR

t

maxQ

q

e

CR

t

maxQtq

e

Discharging a Capacitor – Current Equation

To find the current on the capacitor as a function of time, begin with the charge equation.

Current I = dq/dt, so take the derivative of the charge equation with respect to time:

CR

t

maxQtq

e

dt

eQd

dt

dqCR

t

max


Left side:

Right side:

Idt

dq

Idt

dq

CR

t-max

CR

t-

maxCR

t-

max

CR

-t

max

CR

-t

max

CR

Q-dt

dt

CR

1Q

dt

dQ

dt

dQ

dt

Qd

CR

t

e

ee

ee


Combining the two sides of the integral:

Qmax = C·V, substituting:

CR

tmax

CR

QI

e

CR

t

max

CR

t

CR

t

ItI

R

V

CR

VCtI

e

ee


The negative sign indicates that the direction of the discharging current is opposite to the direction of the charging current.

The voltage V across the capacitor is equal to the EMF of the battery since the capacitor is fully charged at the time of the switch is closed to discharge the capacitor through the resistor.

Discharging Capacitor Graphs

voltage charge current

Bonus Equations! I have never seen these equations in

any textbook and had never been asked to find the voltage across the capacitor as a function of time. I got these equations from the E & M course I took Fall 06.

Discharging Capacitor:

CR

t

CR

tCR

t

max

EV(t)

C

CE

C

Q

C

tqtV

e

ee

Bonus Equations! Charging Capacitor:

CR

t

CR

t

CR

t

max

1EV(t)

1C

CE

C

1Q

C

tqtV

e

ee

Energy Conservation in Charging a Capacitor

During the charging process, a total charge Q = EC flows thru the battery.

The battery does work W = QmaxE or W = CE2.

Half of this work is accounted for by the energy stored in the capacitor: U = 0.5QV = 0.5QmaxE = 0.5CE2

The other half of the work done by the battery goes into Joule heat in the resistance of the circuit.

– The rate at which energy is put into the resistance R is:

– Using the equation for current in a charging capacitor:

– Determine the total Joule heat by integrating from t = 0 s to t = :

RIdt

dW 2R

CRt22

2

CRt

R eRE

ReRE

dtdW

dteRE

W CRt2

0

2

R

Substitute: let

dx2CR

dt

dtCR

2dxthen,

CRt2

x

o

x2

0

x2

R

0

x2

R

x

0

2CRt2

0

2

R

e2

CEdxe

2CE

W

dxe2CR

RE

W

dx2CR

eRE

dteRE

W

The result is independent of the resistance R; when a capacitor is charged by a battery with a constant EMF, half the energy provided by the battery is stored in the capacitor and half goes into thermal energy.

The thermal energy includes the energy that goes into the internal resistance of the battery.

2CE

W

102

CE2

CEW

2

R

20

2

R

ee

Capacitors and Resistors in Parallel

The capacitor in the figure is initially uncharged when the switch S is closed.

Immediately after the switch is closed, the potential is the same at points c and d.

An uncharged capacitor does not resist the flow of current and acts like a wire.

No current flows thru the 8 Ω resistor; the capacitor acts as a short circuit between

points c and d. Apply Kirchhoff’s loop rule to the outer loop

abcdefa: 12 V – 4 Ω·I0 = 0; I0 = 3 A

After the capacitor is fully charged, no more charge flows onto or off of the plates; the capacitor acts like a broken wire or open in the circuit.

Apply Kirchhoff’s loop rule to loop abefa:12 V – 4 Ω·If – 8 Ω·If = 0; 12 V – 12 Ω·If; If = 1 A

RC Circit Problem ExampleRC Circit Problem Example You will see this problem on your You will see this problem on your

homework.homework.

Charging: When the switch S is first Charging: When the switch S is first closed and the current begins to move closed and the current begins to move through the circuit, it will move through the circuit, it will move through the 2 through the 2 ΩΩ resistor onto the resistor onto the capacitor. capacitor.

Charge will move off of the capacitor Charge will move off of the capacitor through the 22 through the 22 ΩΩ to the battery. to the battery.

No charge moves through the 18 No charge moves through the 18 ΩΩ or or 38 38 ΩΩ resistor while the capacitor resistor while the capacitor charges.charges.

Once the voltage across the 22 Once the voltage across the 22 μμF F capacitor reaches its maximum capacitor reaches its maximum voltage, no additional charge will voltage, no additional charge will move onto the capacitor and the move onto the capacitor and the current in the circuit will now begin to current in the circuit will now begin to move through all four resistors.move through all four resistors.

The 18 The 18 ΩΩ and 22 and 22 ΩΩ resistors are in resistors are in series; total resistance = 40 series; total resistance = 40 Ω Ω ..

The 2 The 2 ΩΩ and 38 and 38 ΩΩ resistors are in resistors are in series; total resistance = 40 series; total resistance = 40 ΩΩ. .

The 40 The 40 ΩΩ resistors are in parallel with resistors are in parallel with each other; total resistance of the each other; total resistance of the circuit:circuit:

Total current:Total current:

The current in each branch is:The current in each branch is:

11 1

40 40 20 20

120

TT

T

V VI A

R

200.5

40

V VI A

R

The voltage drop across the 18 The voltage drop across the 18 ΩΩ resistor resistor isis

V = 0.5 A·18 V = 0.5 A·18 ΩΩ = 9 V. = 9 V. The voltage drop across the 2 The voltage drop across the 2 ΩΩ resistor resistor

is V = 0.5 A·2 is V = 0.5 A·2 ΩΩ = 1 V. = 1 V. The difference in The difference in

the voltage the voltage

between the two between the two

points is equal to points is equal to

the voltage acrossthe voltage across

the capacitor: 9V – 1 V = 8 V. the capacitor: 9V – 1 V = 8 V.

Check: determine the voltage Check: determine the voltage difference between the 22 difference between the 22 ΩΩ and the and the 38 38 ΩΩ resistor: resistor:

ΔV = 19 V – 11 V = 8 VΔV = 19 V – 11 V = 8 V The difference in the voltage between The difference in the voltage between

these two resistors is equal to the these two resistors is equal to the voltage across the capacitor.voltage across the capacitor.

38

22

0.5 38 19

0.5 22 11

V A V

V A V

Discharging: when the switch S is Discharging: when the switch S is opened and the battery is removed opened and the battery is removed from the circuit, the capacitor will from the circuit, the capacitor will discharge current Idischarge current I11 through the 18 through the 18 ΩΩ and 2 and 2 ΩΩ resistors and current I resistors and current I22 through the 22 through the 22 ΩΩ and 38 and 38 ΩΩ resistors. resistors.

The 18 The 18 ΩΩ and 2 and 2 ΩΩ resistors are in resistors are in series; total resistance = 20 series; total resistance = 20 ΩΩ..

The 22 The 22 ΩΩ and 38 and 38 ΩΩ resistors are in resistors are in series; total resistance = 60 series; total resistance = 60 ΩΩ..

The 20 The 20 ΩΩ and and 60 and and 60 ΩΩ resistances are resistances are in parallel with each other.in parallel with each other.

Total resistance:Total resistance:

Time constant Time constant ττ = R·C = 15 = R·C = 15 Ω·Ω·22 x 1022 x 10--

66 F = 3.3 x 10 F = 3.3 x 10-4-4 s s From here, you can use the From here, you can use the

discharging voltage equation to solve discharging voltage equation to solve for the time t. for the time t.

11 1

20 60 15TR

t

R CV t Vc e

rc circuits

Documents