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Phy 213: General Physics III
Chapter 27: Electric Circuits
Lecture Notes
Electromotive Force & Internal Resistance
1.
The intrinsic potential difference associated with a power
source is referred to as the “electromotive force” or emf
a.
b.
2.
Real power sources are limited in their ability to deliver
power output, due to factors such as the maximal rate of
the internal chemical reaction, the input power (in an AC
plug-in DC power source), etc…
a.
3.
For a battery, the rate of reaction is dependent on the
conditioning and corrosion of electrodes and depletion of
internal reactants. This results establishes an effective
internal resistance, within the voltage source.
As an electrochemical power source is utilized and is run
down, the decline in performance output is reflected in the
increased in internal resistance
a.
+
In an ideal power source, the voltage across its terminals is
its emf
For a real power source, such as a battery, the emf is
determined by the net electrochemical potential due to its
internal redox reaction BUT the actual voltage across its
terminals is slightly lower due to its internal resistance (Rint).
The output voltage will wane as more of the potential drops
across Rint even though the emf remains constant
Rint
V
+
emf
-
-
Kirchoff’s Voltage & Current Laws
Kirchoff’s Voltage Law (aka the Loop Law):
• For any closed loop in a circuit, the total
voltage around the loop is equal to zero:
N
V
i=1
i
+
10V
=0
V
i=1
i
R3
=V-VR1 -VR2 -VR3 =0  VR1 +VR2 +VR3 =3VR =10V
VR
= 0.333A
R
Kirchoff’s Current Law (aka the Node Law):
• The total current through any node is equal to zero:
VR = 3.33V
and
iloop=
i2
i1
i3
R2
-
Example: A single loop circuit where R1=R2=R3=10W
N
R1
N
i
i=1
i
=0
Series Circuits
1. In series wiring, circuit
elements (loads) are
connected end to end
2. The combined load or
resistance (Req) in the series
is
R eq=R1+R 2+...+R N=
N
R
i=1
i
3. Across each resistance, the
potential difference (V) drops
V
4. The current i that flows through R1 also flows through R2
V = V1 + V2 = iR1 + iR2
or
V = i(R1 + R2)= iReq
Parallel Circuits
1. Circuit elements (loads) are
connected with ends attached
2. The combined load or resistance
(Rp) in the parallel is
1
1
1
1
= + +...+ =
R eq R1 R 2
RN
1

i=1 R i
N
3. Across each resistance, the
potential difference (V) is the
same
i1
i2
V=
=
R1
R2
4. The total current drawn through
the circuit is:
i = i1 + i2
or
1
V V
1 
V
i=
R1
+
R2
= V  + =
 R1 R2  R eq
Analyzing Circuits 1 (using Kirchoff’s Laws)
Consider the following 2 loop circuit, with 6 equal value
resistors (R=R1=R2=R3=R4=R5=R6=10W):
1. What is the current & voltage for each R?
a. Kirchoff’s Laws:
Loop 1: VR +VR +VR =10V
1
2
Loop 2: VR +VR +VR =VR
5
6
10V +
-
1
i1= i2+ i3
solving for i’s:
i3
R2
i2
2
R5
3
i3 R 4+R 5+R 6 )=i2R 2
Node:
R4
i1
3
i1R1+i2R2+i1R3=10V
4
R1
i1
R3
R6
1
i3= 11
A=0.091 A  VR 4 =VR5 =VR6 =10W )0.091A )=0.91V
i2=3i3=0.273 A  VR2 =10W )0.273A )=2.73V
i1=i2+i3=0.364 A  VR1 =VR3 =10W )0.364A )=3.64V
Analyzing Circuits 2 (using equivalent resistances)
Consider the same circuit:
2. What is the current & voltage for each R?
a. Solve for Req1:
R eq1= R 4+R5+R 6= 30W
b. Solve for Req2:
R eq2=
R 2R eq1
R 2+R eq1
10V
= 7.5W
R1
R1
+
-
R2
R3
Req1
+
-
Req2
R3
c. Solve for Req: R eq=R1+R eq2+R 3=27.5W
d. Use Req to get i1 & Req2 to get VR1& VR3
i1=
10V
=0.364A  VR1=VR3=0.364A )10V )=3.64V
27.5W
e. Solve for the rest:
VR2=Veq2=(10-7.27)V=2.73V
 i2=
2.73V
2.73V
=0.273A and i3=
=0.091A
10W
30W
+
-
Req
RC Circuits
1.
2.
3.
A circuit containing a capacitor and resistor(s)
is called an RC circuit
A resistor in series with a capacitor will limit
the rate (not quantity) at which charge
accumulates in the capacitor
dV
When V is constant across a capacitor dt
(
=
0) no current will flow through this branch of
the circuit since:
dq
dV
i=
4.
dt
=C
dt
R
C
=0
When a fully charged capacitor is discharged, the rate of charge loss
is limited by the voltage across it and is limited by on the resistance:
dq dq
q = CV = CiR = RC

= RCdt
dt
q
q
t
 q 
dq
1
1
integrating both sides  
=
dt

ln
=
t



q
RC 0
RC
 qmax 
qmax
solving for q
-
 q(t) = qmaxe
t
RC
Charging a Capacitor
The voltage equation around a loop with resistor and capacitor
in series with a constant voltage source is given by:
R
V - VR - Vcap = 0
V
C
q
dq
q
V
=0 
+
=
C
dt
RC
R
q
dq 

where
V
=
and
i=


cap
C
dt


V - iR -
Re-arranging the equation leads to a 1st order linear nonhomogeneous differential equation. This can be solved by
applying the separation of variables technique:
qmax q
dq
V q
dq
1
= =

=
dt
dt
R RC
RC RC
qmax -q
RC
t

- t 
 qmax -q  t
dq
1
0 qmax -q = RC 0 dt  ln qmax = RC  q(t) = qmax 1 - e RC 


q
Capacitive Charging & Discharging
(C=0.1F, Vmax=10V & R=10W)
-t


RC
q(t) = qmax 1 - e 


1.2
q (C)
1
0.8
q (charge)
0.6
q (discharge)
0.4
0.2
q(t) = qmaxe
0
0
5
10
time (s)
15
20
-t
RC
Capacitive Charging in RC circuit
(Effects of increasing R on Vcap)
Charging
Discharging
12
12
10
8
6
10 ohms
30 ohms
60 ohms
4
2
0
V (volts)
V (volts)
10
10 ohms
30 ohms
60 ohms
8
6
4
2
0
0
10
time (s)
20
0
10
time (s)
20
Capacitive Charging
(i vs t)
Charging
Discharging
What should these graphs should look like?