The Magnetic Field of a Solenoid AP Physics C Montwood High School R. Casao.
RC Circuits AP Physics C Montwood High School R. Casao.
-
Upload
arleen-andra-johnston -
Category
Documents
-
view
236 -
download
5
Transcript of RC Circuits AP Physics C Montwood High School R. Casao.
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
Charging a Capacitor – Current Equation
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
Charging a Capacitor – Current Equation
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
Charging a Capacitor – Current Equation
Left side:
Right side:
max
max
I
I
I
I I
IlnIlnIlnIlndI
I
1maxmax
CRt
0tCR1
tCR1
dtCR1 t
0
t
0
Charging a Capacitor – Current Equation
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
Charging a Capacitor – Current Equation
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
Charging a Capacitor – Charge Equation
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
Charging a Capacitor – Charge Equation
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
Charging a Capacitor – Charge Equation
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
Charging a Capacitor – Charge Equation
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
Charging a Capacitor – Charge Equation
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
Charging a Capacitor – Charge Equation
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
Charging a Capacitor – Charge Equation
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
Discharging a Capacitor – Charge Equation
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
Discharging a Capacitor – Charge Equation
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
Discharging a Capacitor – Charge Equation
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
Discharging a Capacitor – Charge Equation
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
Discharging a Capacitor – Charge Equation
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
Discharging a Capacitor – Current Equation
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
Discharging a Capacitor – Current Equation
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
Discharging a Capacitor – Current Equation
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