Chapter 27 Subjects
Î Emf
devices (review)
Î Resistors
Î Power
in series and parallel
(examples)
Î Multiloop
circuits
Î Ammeters
Î RC
and voltmeters
circuits and time constant
PHY2049: Chapter 27
1
More Complicated Circuits
(More than one battery, not bundled)
ÎParallel
ÎUse
and series rules are not enough!
Kirchhoff’s rules
9V
22 Ω
15 Ω
6V
8Ω
PHY2049: Chapter 27
2
Kirchhoff’s Rules
ÎJunction
ÎLoop
rule
rule
‹Example
a
a
V1= iR1
ℇ
a
V2= iR2
iR1
ℇ
iR2
ℇ – iR1 – iR2=0
Analogy: battery=pump, resistors=narrow tubes,
potential=water pressure.
PHY2049: Chapter 27
3
Kirchhoff’s rules
ÎDetermine
the magnitudes and directions of the
currents through the three resistors in the figure
below.
‹Take
two loops, 1 and 2, as shown
Junction rule: I1 + I2 through 8 Ω
b 22 Ω
Loop rule:
9V
I2 a ->a 6 – 8(I + I ) – 15I =0
1
2
1
2
b ->b 9 + 15I1– 22I2 =0
I1 15 Ω
a
6V
I1+I2
1
8Ω
I1= 30/313 = 0.096 A
I2= 297/626 = 0.474 A
I1+ I2= 357/626 = 0.570 A
PHY2049: Chapter 27
4
Multiloop Circuit
ÎWhich
shown?
of the equations is valid for the circuit
1Ω
‹(a)
2 – I1 – 2I2 = 0
‹(b) 2 – 2I1 – 2I2 – 4I3 = 0
‹(c) 2 – I1 – 4 – 2I2 = 0
‹(d) I3 – 2I2 – 4I3 = 0
‹(e) 2 – 2I1 – 2I2 – 4I3 = 02 V
I2
4V
I1
1Ω
PHY2049: Chapter 27
2Ω
6V
I3
3Ω
5
Deceptively Simple Problem
ÎFind Req of the resistance network shown
‹This network cannot be broken down into parallel and
series resistors. (Try and convince yourself)
‹Must use Kirchhoff’s rules.
‹That requires adding a battery,
10 kΩ
which drives current through 3 kΩ
the network.
10 kΩ
‹Req then means
I
ℇ
network
10 kΩ
10 kΩ
ℇ = I Req
I So find I, for given ℇ.
PHY2049: Chapter 27
6
(continued)
ÎKirchhoff’s
rules
junction rule,
reduce the number of
unknown currents to 3:
for instance i1, i2, i.
‹Choose 3 loops and
apply loop rule to them. ℇ
-i2R2-iR+i1R=0
R=10 kΩ R2=3 kΩ
‹Using
R2
i2
i1
R
i
R
-(i2-i)R+(i1+i)R+iR=0
i2–i
ℇ-i2R2-(i2-i)R=0
‹Solve for i1, i2, i
i=ℇ(R-R2)/(3R+5R2)R
i1=ℇ(R+3R2)/(3R+5R2)R
R
1
2
R
i1+i
i2=ℇ 4/(3R+5R2)
PHY2049: Chapter 27
7
(continued)
the total current I
into, and out of, the
network is i1+ i2. So
ℇ = (i1+i2)Req
‹Substitute:
ℇ
ℇ = ℇ (5R+3R2)/(3R+5R2)R * Req
i1+i2
‹Note:
Req= (3R+5R2)R/(5R+3R2)
=450/59=7.63 kΩ
R
R2
i2
i1
R
i
R
R
i2–i
i1+i
i1+i2
PHY2049: Chapter 27
8
Problem Solving Using Kirchhoff’s Rules
ÎLabel
the current in each branch of the circuit
‹Choice of direction is arbitrary
‹Signs will work out in the end (if you are careful!!)
‹Apply the junction rule at each junction
‹Keep track of sign of currents entering and leaving
ÎApply
loop rule to each loop (follow in one direction only)
‹Resistors: if loop direction matches current direction,
voltage drop
‹Batteries: if loop direction goes through battery in
“normal” direction, voltage gain
ÎSolve
equations simultaneously
‹You need as many equations as you have unknowns
PHY2049: Chapter 27
9
Res-Monster Maze (p. 725)
All batteries are 4 V
All resistors are 4 Ω
Find current in R
PHY2049: Chapter 27
10
Ammeters and Voltmeters
Î Ideal ammeter
A
‹ Zero
internal resistance
Î Ideal voltmeter
‹ Infinite internal resistance
Î Internal
V
Actual ammeter
resistances
‹ In
ammeter: series with
“the ideal ammeter” inside
Actual voltmeter
the real ammeter
‹ In voltmeter: parallel with
“the ideal voltmeter inside
the real voltmeter
PHY2049: Chapter 27
A
Ideal ammeter
Ideal voltmeter
V
11
RC Circuit (Charging a Capacitor)
ÎCharging
a capacitor takes time in a real circuit
‹Resistance allows only a certain amount of current to
flow
‹Current takes time to charge a capacitor
ÎAssume
uncharged capacitor initially
‹Close switch at t = 0
‹Initial current: i = E /R
(since no charge on capacitor)
‹Initial
charge: q = 0
E
ÎWhen
fully charged
‹Final current: i = 0
‹Final charge: q = ℇ C
PHY2049: Chapter 27
12
(continued)
ÎWhat
happens in between? Use loop rule:
E–iR–q/C=0
ÎCurrent
ÎSo
and charge are related:
i = dq / dt
can recast first equation as “differential equation”
dq
q
E
E
+
=E
dt RC R
ÎGeneral
solution is
q = EEC + Ke −t / RC
‹ (Check
and see!)
‹ K = −E C (necessary to make q = 0 at t = 0)
ÎSolve
for charge q and current i
dq E −t / RC
−t / RC
= e
q = EEC 1 − e
i=
dt R
(
)
PHY2049: Chapter 27
13
Charge and Current vs Time
(
q ( t ) = q0 1 − e
−t / RC
)
q0: q at t →∞; q0=E C
i ( t ) = i0e
−t / RC
i0: i at t = 0; i0=E /R
PHY2049: Chapter 27
14
Exponential Behavior
ÎRC
is the “characteristic time” of any RC circuit
‹ Only
Ît
t / RC is meaningful
= RC
‹ Current
falls to 37% of maximum value
‹ Charge rises to 63% of maximum value
Ît
=2RC
‹ Current
falls to 13.5% of maximum value
‹ Charge rises to 86.5% of maximum value
Ît
=3RC
‹ Current
falls to 5% of maximum value
‹ Charge rises to 95% of maximum value
Ît
=5RC
‹ Current
falls to 0.7% of maximum value
‹ Charge rises to 99.3% of maximum value
PHY2049: Chapter 27
15
Discharging a Capacitor
ÎConnect
=0
V
+q
–q
ÎSolve
fully charged capacitor to a resistor at t
S
C
R
i
dq
q
+
=0
dt RC
for charge q
ÎDifferentiate
–iR–q/C=0
q = q0 e − t / RC
to find current i
q0 −t / RC
V0 −t / RC
dq
i=
=−
e
=− e
dt
RC
R
PHY2049: Chapter 27
16
Charge and Current vs Time
q ( t ) = q0e
−t / RC
q0: q at t = 0; q0=V0C
i ( t ) = i0e
−t / RC
i0: i at t = 0; i0=−V0/R
PHY2049: Chapter 27
17
Energy Consideration
ÎEnergy
initially stored in the capacitor (Sect. 25-5)
U = q02/ 2C
q0: q at t = 0; q0=V0C
ÎEnergy
dissipated (converted to thermal energy)
at the resistor
−t / RC i : i at t = 0; i =−V /R
i t =i e
0
0
0
()
‹Integrate
0
P=i2R from t=0 to ∞ to find
U = (i02R) RC/2 = [(q0/RC)2R] RC/2 = q02/2RC
‹Agrees
with energy initially stored in capacitor.
Confirms conservation of energy.
PHY2049: Chapter 27
18