DC Circuits: Ch 32 Resistors in Series

bahmotherElectronics - Devices

Oct 7, 2013 (4 years and 1 month ago)

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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
8
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