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W12D1:
RC and LR Circuits
Reading Course Notes: Sections 7.7-7.8, 7.11.3, 11.4-11.6,
11.12.2, 11.13.4-11.13.5
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Announcements
Math Review Week 12 Tuesday 9pm-11 pm in 26-152
PS 9 due Week 13 Tuesday April 30 at 9 pm in boxes outside 32-082 or
26-152
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Outline
DC Circuits with Capacitors
First Order Linear Differential Equations
RC Circuits
LR Circuits
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DC Circuits with Capacitors
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Sign Conventions - Capacitor
Moving across a capacitor from the negatively to
positively charged plate increases the electric
potential
DV = Vb - Va
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Power - Capacitor
Moving across a capacitor from the positive to
negative plate decreases your potential. If current
flows in that direction the capacitor absorbs power
(stores charge)
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Pabsorbed
dQ Q d Q
dU
 I V 


dt C dt 2C dt
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RC Circuits
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(Dis)Charging a Capacitor
1. When the direction of current flow is toward
the positive plate of a capacitor, then
dQ
I=+
dt
2. When the direction of current flow is away from
the positive plate of a capacitor, then
dQ
I=dt
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Charging a Capacitor
What happens when we close switch S at t = 0?
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Charging a Capacitor
Circulate clockwise
Q
å DVi = e - C - IR = 0
i
dQ
I=+
dt
First order linear
inhomogeneous differential
equation
dQ
`1
=(Q - Ce )
dt
RC
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Energy Balance: Circuit Equation
dQ
I=+
dt
Q
e - - IR = 0
C
dQ
Multiplying by I = +
dt
2
æ
Q dQ
d 1Q ö
2
2
eI = I R +
= I R+ ç
÷
C dt
dt è 2 C ø
(power delivered by battery) = (power dissipated through resistor)
+ (power absorbed by the capacitor)
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RC Circuit Charging: Solution
(
dQ
1
=Q - Ce
dt
RC
)
Solution to this equation when switch is closed at t = 0:
dQ
I(t) = +
Þ I(t) = I0 e-t / t
Q(t) = Ce (1- e )
dt
t = RC : time constant
(units: seconds)
-t / t
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Demonstration
RC Time Constant
Displayed with a Lightbulb (E10)
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=E%2010&show=0
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Review Some Math:
Exponential Decay
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Math Review: Exponential Decay
Consider function A where:
A decays exponentially:
dA
1
=- A
dt
t
A(t) = A0 e
-t t
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Exponential Behavior
Slightly modify diff. eq.:
A “grows” to Af:
dA
1
= - ( A - Af )
dt
t
A(t) = Af (1- e
-t / t
)
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Homework: Solve Differential
Equation for Charging and
Discharging RC Circuits
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Concept Question:
Current in RC Circuit
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Concept Question: RC Circuit
An uncharged capacitor is
connected to a battery,
resistor and switch. The
switch is initially open but at t
= 0 it is closed. A very long
time after the switch is
closed, the current in the
circuit is
1.
2.
3.
Nearly zero
At a maximum and
decreasing
Nearly constant but non-zero
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Concept Q. Answer: RC Circuit
Answer: 1. After a long time the current is 0
Eventually the capacitor gets
“completely charged” – the
voltage increase provided by
the battery is equal to the
voltage drop across the
capacitor. The voltage drop
across the resistor at this
point is 0 – no current is
flowing.
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Discharging A Capacitor
At t = 0 charge on capacitor is Q0. What happens
when we close switch S at t = 0?
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Discharging a Capacitor
Circulate clockwise
Q
å DVi = C - IR = 0
i
dQ
I=Þ
dt
First order linear
differential equation
dQ
Q
=dt
RC
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RC Circuit: Discharging
dQ
1
-t / RC
=Q Þ Q(t) = Qoe
dt
RC
Solution to this equation when switch is closed at t = 0
with time constant t = RC
Qo -t / t Qo -t / t
dQ
I=Þ I(t) =
e =
e
dt
t
RC
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Concept Questions:
RC Circuit
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Concept Question: RC Circuit
Consider the circuit at right,
with an initially uncharged
capacitor and two identical
resistors. At the instant the
switch is closed:
1. I R = IC = 0
2. I R = e / 2R , I C = 0
3. I R = 0 , I C = e / R
4.
I R = e / 2R , I C = e / R
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Concept Question Answer: RC
Circuit
Answer: 3. I R = 0 IC = e R
Initially there is no charge on
the capacitor and hence no
voltage drop across it – it
looks like a short. Thus all
current will flow through it
rather than through the
bottom resistor. So the circuit
looks like:
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Concept Q.: Current Thru Capacitor
In the circuit at right the
switch is closed at t = 0.
At t = ∞ (long after) the
current through the
capacitor will be:
1.
2.
3.
IC = 0
.
IC = e R
IC = e 2R
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Con. Q. Ans.: Current Thru Capacitor
Answer 1.
IC = 0
After a long time the capacitor becomes “fully
charged.” No more current flows into it.
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Concept Q.: Current Thru Resistor
In the circuit at right the
switch is closed at t = 0.
At t = ∞ (long after) the
current through the lower
resistor will be:
1. I R = 0
2. I R = e R
3. I = e 2R
.
R
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Concept Q. Ans.: Current Thru Resistor
Answer 3.
Since the capacitor is “fullly charged” we can
remove it from the circuit, and all that is left is the
battery and two resistors. So the current
is I R = e 2R
.
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Group Problem: RC Circuit
For the circuit shown in the
figure the currents through the
two bottom branches as a
function of time (switch closes
at t = 0, opens at t = T>>RC).
State the values of current
(i) just after switch is closed at
t = 0+
(ii) Just before switch is
opened at t = T-,
(iii) Just after switch is opened
at t = T+
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Concept Q.: Open Switch in RC Circuit
Now, after the switch has
been closed for a very long
time, it is opened. What
happens to the current
through the lower resistor?
1.
2.
3.
4.
5.
6.
It stays the same
Same magnitude, flips direction
It is cut in half, same direction
It is cut in half, flips direction
It doubles, same direction
It doubles, flips direction
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Con. Q. Ans.: Open Switch in RC Circuit
Answer: 1. It stays the same
The capacitor has been
charged to a potential of
VC = e 2
so when it is responsible for
pushing current through the
lower resistor it pushes a
current of I R = e 2R, in the
same direction as before (its
positive terminal is also on the
left)
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LR Circuits
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Inductors in Circuits
Inductor: Circuit element with self-inductance
Ideally it has zero resistance
Symbol:
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Non-Static Fields
dF B
E
×
d
s
=
ò
dt
E is no longer a static field
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Kirchhoff’s Modified 2nd Rule
dF B
å D Vi = - ò E × d s = + d t
i
dF B
Þ å D Vi =0
dt
i
If all inductance is ‘localized’ in inductors then
our problems go away – we just have:
dI
å D Vi - L d t = 0
i
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Ideal Inductor
• BUT, EMF generated
by an inductor is not a
voltage drop across
the inductor!
ò
dI
e = -L
dt
E × ds = 0
ideal
inductor
Because resistance is 0, E must be 0!
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Non-Ideal Inductors
Non-Ideal (Real) Inductor: Not only L but also some R
=
In direction of current:
e
dI
= - L - IR
dt
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Circuits:
Applying Modified Kirchhoff’s
(Really Just Faraday’s Law)
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Sign Conventions - Inductor
Moving across an inductor in
the direction of current
contributes

dI
 L
dt
Moving across an inductor
opposite the direction of
current contributes
e
dI
= +L
dt
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LR Circuit
Circulate clockwise
dI
e - IR - L = 0 Þ
dt
First order linear
inhomogeneous
differential equation
ö
dI
Ræ
e
=- çI - ÷
dt
Lè
Rø
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RL Circuit
ö
dI
Ræ
e
e
-t / ( L / R)
= - ç I - ÷ Þ I(t) = (1- e
)
dt
Lè
Rø
R
Solution to this equation when switch is closed at t = 0:
I(t) =
e
R
(1- e
-t / t
)
L
t = : time constant
R
(units: seconds)
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RL Circuit
t=0+: Current is trying to change. Inductor works as
hard as it needs to to stop it
t=∞: Current is steady. Inductor does nothing.
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Group Problem: LR Circuit
For the circuit shown in the
figure the currents through the
two bottom branches as a
function of time (switch closes
at t = 0, opens at t = T>>L/R).
State the values of current
(i) just after switch is closed at
t = 0+
(ii) Just before switch is
opened at t = T-,
(iii) Just after switch is opened
at t = T+
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