6/3/2013
1
DC Circuits: Ch 32
•
Voltage
–
Starts out at highest point at “+” end of
battery
•
Voltage drops across lightbulbs and other
sources of resistance.
•
Voltage increases again at battery.
+

I
Voltage highest
Voltage zero
The following circuit uses a 1.5 V battery and
has a 15
W
lightbulb
.
a.
Calculate the current in the circuit
b.
Calculate the voltage drop across the
lightbulb
.
c.
Sketch a graph of voltage vs. path (battery, top
wire, resistor, bottom wire)
Resistors in Series
•
Same current (I) passes through all resistors
(bulbs)
•
All bulbs are equally bright (energy loss, not
current loss)
•
Voltage drop across each resistor (V
1
,V
2
, V
3
)
6/3/2013
2
V = V
1
+ V
2
+ V
3
V = IR
1
+ IR
2
+ IR
3
V = I(R
1
+ R
2
+ R
3
)
R
eq
= R
1
+ R
2
+ R
3
V = IR
eq
Resistors in Parallel
•
Current splits at the junction
•
Same Voltage across all resistors
I = I
1
+ I
2
+ I
3
I
1
=
V
R
1
I =
V
R
eq
1
=
1
+
1
+
1
R
eq
R
1
R
2
R
3
Which combination of auto headlights will produce
the brightest bulbs? Assume all bulbs have a
resistance of
R
.
For the Bulbs in Series:
R
eq
= R + R =
2R
For the Bulbs in Parallel
1
=
1
+
1
R
eq
R
R
1
=
2
R
eq
R
R
eq
= R/2
The bulbs in parallel have less resistance and
will be brighter
What current flows through each resistor in the
following circuit? (R = 100
W
)
R
eq
= R
1
+ R
2
R
eq
= 200
W
V = IR
eq
I = V/R
eq
I = 24.0 V/ 200
W
= 0.120 A
6/3/2013
3
Calculate the current through this circuit, and the
voltage drop across each resistor.
R
eq
= 400
W
+ 290
W
R
eq
= 690
W
V = IR
I = V/R
eq
I = 12.0 V/690
W
I = 0.0174 A
V
ab
= (0.0174A)(400
W
)
V
ab
= 6.96 V
V
bc
= (0.0174A)(290
W
)
V
bc
= 5.04 V
What current flows through each of the
resistors in this circuit? (Both are
100
W
)
(0.48 A, 0.24 A)
DC Circuits: Ex 4
What current will flow through the circuit shown?
1
=
1
+
1
R
p
= 500
W
700
W
R
p
= 290
W
R
eq
= 400
W
+ 290
W
R
eq
= 690
W
V = IR
I = V/R
I = 12.0 V/690
W
I = 0.017 A or 17 mA
Example 4
Calculate the equivalent resistance in the following
circuit.
6/3/2013
4
DC Circuits: Ex 5
What current is flowing through
just
the 500
W
resistor?
First we find the voltage drop across the first
resistor:
V = IR = (0.017 A)(400
W
)
V = 6.8 V
The voltage through the resistors in parallel will be:
12.0 V
–
6.8 V = 5.2 V
To find the current across the 500
W
resistor:
V = IR
I = V/R
I = 5.2 V/500
W
=
0.010 A = 10 mA
DC Circuits: Ex 6
Which bulb will be the brightest in this
arrangement (most current)?
Bulb C (current gets split running through A and
B)
What happens when the switch is opened?
–
C and B will have the same brightness (I is constant
in a series circuit)
DC Circuits: Ex 5
What resistance would be present between points A
and B?
(ANS: 41/15 R)
6/3/2013
5
EMF and Terminal Voltage
•
Batteries

source of
emf
(Electromotive Force),
E
(battery rating)
•
All batteries have some internal resistance
r
V
ab
=
E
–
Ir
V
ab
= terminal(useful)voltage
E
= battery rating
r = internal resistance
EMF: Example 1
A 12

V battery has an internal resistance of 0.1
W
.
If 10 Amps flow from the battery, what is the
terminal voltage?
V
ab
=
E
–
Ir
V
ab
=
12 V
–
(10 A)(0.10
W
)
V
ab
=
11 V
EMF: Example 2
Calculate the current in the following circuit.
1/R
eq
= 1/8
W
+ ¼
W
R
eq
= 2.7
W
R
eq
= 6
W + 2.7 W
R
eq
= 8.7
W
1/R
eq
= 1/10
W + 1/8.7 W
R
eq
= 4.8
W
Everything is now in series
R
eq
= 4.8
W + 5.0 W + 0.50 W
R
eq
= 10.3
W
V = IR
I = V/R
I = 9.0 V/10.3
W
I = 0.87 A
6/3/2013
6
EMF: Example 2a
Now calculate the terminal(useful)voltage.
V
=
E
–
Ir
V = 9.0 V
–
(0.87 A)(0.50
W
)
V = 8.6 V
Grounded
•
Wire is run to the ground
•
Houses have a ground wire at main circuit box
•
Does not affect circuit behavior normally
•
Provides path for electricity to flow in
emergency
Kirchoff’s Rules
1.
Junction Rule

The sum of the
currents entering a junction must
equal the sum of currents leaving
2.
Loop Rule

The sum of the changes
in potential around any closed path =
0
Kirchoff Conventions
The “loop
current” is
not
a
current.
Just a direction
that you follow
around the loop.
Kirchoff Conventions
Kirchoff’s Rule Ex 1
6/3/2013
7
Junction Rule
I
1
= I
2
+ I
3
Loop Rule
Main Loop
6V
–
(I
1
)(4
W
)
–
(I
3
)(9
W
) = 0
Side Loop
(

I
2
)(5
W
) + (I
3
)(9
W
) = 0
I
1
= I
2
+ I
3
Eqn 3
6V
–
(I
1
)(4
W
)
–
(I
3
)(9
W
) = 0
Eqn 2
(

I
2
)(5
W
) + (I
3
)(9
W
) = 0
Eqn 3
Solve Eqn 1
(

I
2
)(5
W
) + (I
3
)(9
W
) = 0
(I
3
)(9
W
) = (I
2
)(5
W
)
I
3
= 5
/9
I
2
Substitution into Eqn 2
6V
–
(I
2
+ I
3
)(4
W
)
–
(I
3
)(9
W
) = 0
6
–
4I
2

4I
3

9I
3
= 0
6
–
4I
2

13I
3
= 0
I
3
= 5
/9
I
2
(from last slide)
6
–
4I
2

13(5
/9
I
2
) = 0
6 = 101/9 I
2
I
2
= 0.53 A
I
3
=
5/9 I
2
=
0.29 A
I
1
=
I
2
+ I
3
= 0.53 A + 0.29 A =
0.82 A
6/3/2013
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Kirchoff’s Rule Ex 2
I
1
= I
2
+ I
3
Loop Rule
Main Loop
9V
–
(I
3
)(10
W
)
–
(I
1
)(5
W
) = 0
Side Loop
(

I
2
)(5
W
) + (I
3
)(10
W
) = 0
(

I
2
)(5
W
) + (I
3
)(10
W
) = 0
(I
2
)(5
W
) = (I
3
)(10
W
)
I
2
= 2I
3
9V
–
(I
3
)(10
W
)
–
(I
1
)(5
W
) = 0
9V
–
(I
3
)(10
W
)
–
(I
2
+ I
3
)(5
W
) = 0
9
–
10I
3
–
5I
2
–
5I
3
= 0
9
–
15I
3
–
5I
2
= 0
9
–
15I
3
–
5(2I
3
) = 0
9
–
25I
3
= 0
I
3
= 9/25 = 0.36 A
I
3
=
9/25 =
0.36 A
I
2
=
2I
3
= 2(0.36 A) =
0.72 A
I
1
= I
2
+ I
3
= 0.36A + 072 A =
1.08 A
6/3/2013
9
Kirchoff’s Rules: Ex 3
Calculate the currents in
the following circuit.
Bottom Loop (clockwise)
10V
–
(6
W
)I
1
–
(2
W
)I
3
= 0
Top Loop (clockwise)

14V +(6
W
)I
1
–
10 V

(4
W
)I
2
= 0
Work with Bottom Loop
10V
–
(6
W
)I
1
–
(2
W
)I
3
= 0
I
1
+ I
2
= I
3
10
–
6I
1
–
2(I
1
+ I
2
) = 0
10
–
6I
1
–
2I
1

2I
2
= 0
10

8I
1

2I
2
= 0
10 = 8I
1
+ 2I
2
5 = 4I
1
+ I
2
I
2
= 5

4I
1
Working with Top Loop

14V +(6
W
)I
1
–
10 V

(4
W
)I
2
= 0
24 = 6I
1

4I
2
12 = 3I
1

2I
2
12 = 3I
1

2(5

4I
1
)
22 = 11I
1
I
1
= 22/11 =
2.0 Amps
I
2
= 5

4I
1
I
2
= 5
–
4(2) =

3.0 Amps
I
1
+ I
2
= I
3
I
3
=

1.0 A
Batteries in Series
•
If + to

, voltages add (top
drawing)
•
If + to +, voltages subtract
(middle drawing = 8V, used
to charge the 12V battery as
in a car engine)
Batteries in Parallel
•
Provide large current when
needed (Same voltage)
Extra Kirchoff Problems
I
1
=

0.864 A
I
2
= 2.6 A
I
3
= 1.73 A
6/3/2013
10
A Strange Example
Calculate I (0.5 Amps)
a.
Calculate the equivalent resistance. (2.26
W
)
b.
Calculate the current in the upper and lower
wires. (3.98 A)
c.
Calculate I
1
, I
2
, and I
3
(0.60 A, 2.25 A, 1.13 A)
d.
Sketch a graph showing the voltage through the
circuit starting at the battery.
RC Circuits
•
Capacitors store energy (flash in a camera)
•
Resistors control how fast that energy is released
•
Car lights that dim after you shut them
•
Camera flashes
D
V
c
+
D
V
r
= 0
Q

IR = 0
(Divide by R)
C
Q

I = 0
(I =

dQ/dt)
RC
Q
+
dQ
= 0
RC dt
= time constant (time to reach 63% of full
voltage)
= RC
6/3/2013
11
A versatile relationship
Charging the Capacitor (for capacitor)
V = V
o
(1

e

t/RC
)
Q = Q
o
(1

e

t/RC
)
I = I
o
e

t/RC
Discharging the Capacitor
V = V
o
e

t/RC
I = I
o
(e

t/RC
)
Q = Q
o
e

t/RC
RC Circuits: Ex 1
What is the time constant for an RC circuit of
resistance 200 k
W
and capacitance of 3.0
m
F?
= (200,000
W
)(3.0 X 10

6
F)
= 0.60 s
(lower resistance will cause the capacitor to charge
more quickly)
RC Circuits: Ex 2
What will happen to the bulb (resistor) in the
circuit below when the switch is closed (like a
car door)?
Answer: Bulb will glow brightly initially, then dim
as capacitor nears full charge.
RC Circuits: Ex 3
An uncharged RC circuit has a 12 V battery, a 5.0
m
F capacitor and a 800 k
W
resistor. Calculate
the time constant.
= RC
= (5.0 X 10

6
F)(800,000 W)
= 4.0 s
What is the maximum charge on the capacitor?
Q = CV
Q = (5.0 X 10

6
F)(12 V)
Q = 6 X 10

5
C or 60
m
C
What is the voltage and charge on the capacitor
after 1 time constant (when discharging)?
6/3/2013
12
Consider the circuit below, which is being charged.
Calculate:
a.
The time constant (6 ms)
b.
Maximum charge on the capacitor (3.6
m
C
)
c.
Maximum current (600
m
A
)
d.
Time to reach 99% of maximum charge (28 ms)
e.
Current when charge = ½
Q
max
(300
m
A
)
f.
The charge when the current is 20% of the maximum
value. (2.9
m
C
)
Discharging the RC Circuit
V = V
o
e

t/RC
RC Circuits: Ex 4
An RC circuit has a charged capacitor C = 35
m
F
and a resistance of 120
W
. How much time will
elapse until the voltage falls to 10 percent of its
original (maximum) value?
V = V
o
e

t/RC
0.10V
o
=V
o
e

t/RC
0.10 =e

t/RC
ln(0.10) = ln(e

t/RC
)

2.3 =

t/RC
2.3 = t/RC
t = 2.3RC
t = (2.3)(120
W
)(35 X 10

6
F)
t = 0.0097 s or 9.7 ms
RC Circuits: Ex 5
If a capacitor is discharged in an RC circuit, how
many time constants will it take the voltage to
drop to ¼ its maximum value?
(t = 1.39RC)
A fully charged 1.02
m
F capacitor is in a circuit
with a 20.0 V battery and a resistor. When
discharged, the current is observed to
decrease to 50% of it’s initial value in 40
m
s.
a.
Calculate the charge on the capacitor at t=0
(20.4
m
C
)
b.
Calculate the resistance R (57
W
)
c.
Calculate the charge at t = 60
m
s (7.3
m
C
)
6/3/2013
13
The capacitor in the drawing has been fully
charged. The switch is quickly moved to
position b (camera flash).
a.
Calculate the initial charge on the capacitor. (9
m
C
)
b.
Calculate the charge on the capacitor after 5.0
m
s.
(5.5
m
C
)
c.
Calculate the voltage after 5.0
m
s (5.5 V
d.
Calculate the current through the resistor after
5.0
m
s (0.55 A)
Meters
•
Galvanometer
–
Can only handle a small current
•
Full

scale Current Sensitivity (
I
m
)
–
Maximum deflection
•
Ex:
–
Multimeter
–
Car speedometer
Measuring I and V
Measuring Current
–
Anmeter is placed in series
–
Current is constant in series
Measuring Voltage
–
Voltmeter placed in parallel
–
Voltage constant in parallel circuits
–
Measuring voltage drop across a resistor
Anmeter (Series)
Voltmeter (parallel)
DC Anmeter
•
Uses “Shunt” (parallel) resistor
•
Shunt resistor has low resistance
•
Most of current flows through shunt, only a little
through Galvanometer
•
I
R
R = I
G
r
Meters: Ex 1
What size shunt resistor should be used if a
galvanometer has a full

scale sensitivity of 50
m
A
and a resistance of r= 30
W
? You want the meter
to read a 1.0 A current.
Voltage same through both (V=IR)
I
R
R = I
g
r
Since most of the current goes through the shunt
I
R
~ 1 A
6/3/2013
14
I
R
R = I
g
r
(1 A)(R) = (50 X 10

6
A)(30
W
)
R = 1.5 X 10

3
W
or 1.5 m
W
Meters: Ex 2
Design an anmeter that can test a 12 A vacuum
cleaner if the galvanometer has an internal
resistance of 50
W
and a full scale deflection of 1
mA.
I
R
R = I
g
r
(12 A)(R) = (1 X 10

3
A)(50
W
)
R = 4.2 X 10

3
W
or 4.2 m
W
DC Voltmeter
•
Resistor in series
•
Large R for resistor (keeps current low in
Galvanometer)
•
V = I(R + r)
Meters: Ex 3
What resistor should be used in a voltmeter that
can read a maximum of 15 V? The galvanometer
has an internal resistance of 30
W
and a full scale
deflection of 50
m
A.
V = I(R + r)
12 V = (50 X 10

6
A)(R + 30
W
)
R + 30
W
=
12 V
50 X 10

6
A
R + 30
W
= 300,000
W
R ~ 300,000
W
Meters: Ex 4
Design a voltmeter for a 120 V appliance with and
internal galvanometer resistance of 50
W
and a
current sensitivity of 1 mA.
(ANS: R = 120,000
W
)
Electric Power
•
Watt
•
1 Watt =
1 Joule
1 second
P = I
2
R
6/3/2013
15
Power: Ex 1
Calculate the resistance of a 40

W auto headlight
that operates at 12 V.
P = I
2
R
(3.6
W
)
Household Electricity
•
Kilowatt

hour
•
You do not pay for power, you pay for energy
1 kWh =
(1000 J)(3600 s)
= 3.60 X 10
6
J
1s
Power: Ex 2
An electric heater draws 15.0 A on a 120

V line.
How much power does it use?
P = I
2
R
V = IR so R = V/I
P =
I
2
V
I
P = IV = (15 A)(120V) =
1800 W or 1.8 kW
Power: Ex 3
How much does it cost to run it for 30 days if it
operates 3.0 h per day and the electric company
charges 10.5 cents per kWh?
Hours = 30 days X 3.0 h/day = 90 h
Cost = (1.80 kW)(90 h)($0.105/kWh) =
$17
Power: Ex 4
A lightening bolt can transfer 10
9
J of energy at a
potential difference of 5 X 10
7
V over 0.20 s.
What is the charge transferred?
V = PE/Q
Q = E/V = 10
9
J/ 5 X 10
7
V = 20 C
What is the current?
I =
D
Q/
D
t
I = 20 C/0.2 s = 100 A
What is the power?
P = I
2
R
V = IR so R = V/I
P =
I
2
V
I
P = IV = (100 A)(5 X 10
7
V ) =
5 X 10
9
W
6/3/2013
16
Household Electricity
•
Circuit breakers
–
prevent “overloading” (too
much current per wire)
•
Metal melts or bimetallic strip expands
Household Electricity:
Ex 1
Determine the total current
drawn by all of the
appliances shown.
P = IV
I = P/V
I
light
= 100W/120 V = 0.8 A
I
light
= 100W/120 V = 0.8 A
I
heater
= 1800W/120 V = 15 A
I
stereo
= 350W/120 V = 2.9 A
I
hair
= 1200W/120 V = 10.0 A
I
total
= 0.8A + 15.0A + 2.9A + 10.0A =
28.7 A
This would blow the 20 A fuse
DC vs AC
DC
•
Electrons flow constantly
•
Electrons flow in only one
direction
•
Batteries
AC
•
Electrons flow in short burst
•
Electrons switch directions
(60 times a second)
•
House current
DC vs AC
http://www.ibiblio.org/obp/electricCircuits/AC/AC_1.html
Jump Starting a Car
POSITIVE TO POSITIVE
(or your battery could explode)
6/3/2013
17
2. Clockwise (12 V. 50
W
, (10
m
F and 100
W
parallel))
4. a) 0.10 A
b) Graph page 32

3
6. a) 5 V, 10 V
b) Graph page 32

4
8. 1.92 W, 2.9 W
10. 3.6 X 10
6
J
14. 25
W
18. 3.2 %
20. R/4
22. 12
W
24. 24
W
26. 183
W
28. 13 V, 5 V, 0 V,

2 V
30. 2 ms
32. 6.9 ms
34. 0.87 K
W
36. A> D = E > B = C
38. 1.99 m
40. 4.0
W
42. 7
W
44. 10
W
and 60 V
46. 9.0 V
50.
1.0 A, 2.0 A, 15 V
56.
a) 8 Amps (8
V
ab
)
b) 9 Amps (0
V
ab
)
58. a) R =0.505
W
b)
R
eq
= 0.500
W
62.
Resistor(
W
)
(V)
Current (A)
4
8
2
6
8
1.3
8
8
1
24(bottom)
8
0.33
24 (right)
16
0.66
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