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DC Circuits: Ch 19
• Voltage – Starts out at highest point at “+” end of
battery
• Voltage drops across lightbulbs and other
sources of resistance.
• Voltage increases again at battery.
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)
I
+
Voltage highest
Voltage zero
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 (V1,V2, V3)
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V = V1 + V2 + V3
V = IR1 + IR2 + IR3
V = I(R1 + R2 + R3)
Req = R1 + R2 + R3
V = IReq
Resistors in Parallel
• Current splits at the junction
• Same Voltage across all resistors
Calculate the equivalent resistance of two 200W
resistors placed in:
I = I1 + I2 + I3
I1 = V
R1
I=V
Req
1 = 1
Req R1
+
a. Series
b. Parallel
1
R2
+
1
R3
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Calculate the equivalent resistance for the circuit
below.
Which combination of auto headlights will produce
the brightest bulbs? Assume all bulbs have a
resistance of R.
For the Bulbs in Series:
Req = R + R = 2R
What current flows through each resistor in the
following circuit? (R = 100 W)
For the Bulbs in Parallel
1 = 1
+
1
Req R
R
1 = 2
Req R
Req = R/2
The bulbs in parallel have less resistance and
will be brighter
0.120 A
Calculate the current through this circuit, and the
voltage drop across each resistor.
Vab = (0.0174A)(400W)
Vab = 6.96 V
Vbc = (0.0174A)(290W)
Vbc = 5.04 V
Req = 690 W
I = 0.0174 A
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What current flows through each of the
resistors in this circuit? (Both are
100 W)
What current will flow through the circuit shown?
(I =0.48 A, I1=I2=0.24 A)
Req = 400 W + 290 W
Req = 690 W
V = IR
I = V/R
I = 12.0 V/690 W
I = 0.017 A or 17 mA
The voltage through the resistors in parallel will be:
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
Calculate the equivalent resistance in the following
circuit.
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
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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)
What resistance would be present between points A
and B?
(ANS: 41/15 R)
EMF and Terminal Voltage
• Batteries - source of emf (Electromotive Force),
E (battery rating)
A 12-V battery has an internal resistance of 0.1 W.
If 10 Amps flow from the battery, what is the
terminal voltage?
• All batteries have some internal resistance r
Vab = E – Ir
Vab = terminal(useful)voltage
E = battery rating
r = internal resistance
Vab = 11 V
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Calculate the current in the following circuit.
Req = 6 W + 2.7 W
Req = 8.7 W
1/Req = 1/10 W + 1/8.7 W
Req = 4.8 W
Everything is now in series
Req = 4.8 W + 5.0 W + 0.50 W
Req = 10.3 W
V = IR
I = V/R
I = 9.0 V/10.3 W
I = 0.87 A
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
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Kirchhoff’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
Kirchhoff Conventions
Kirchhoff Conventions
The “loop
current” is not a
current.
Just a direction
that you follow
around the loop.
Kirchhoff’s Rule Ex 1
Junction Rule
I1 = I2 + I3
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Loop Rule
Main Loop
6V – (I1)(4W) – (I3)(9W) = 0
Side Loop
(-I2)(5W) + (I3)(9W) = 0
I1 = I2 + I3
6V – (I1)(4W) – (I3)(9W) = 0
(-I2)(5W) + (I3)(9W) = 0
Solve Eqn 1
(-I2)(5W) + (I3)(9W) = 0
(I3)(9W) = (I2)(5W)
I3 = 5/9 I2
Kirchhoff’s Rule Ex 2
Eqn 3
Eqn 2
Eqn 3
Substitution into Eqn 2
6V – (I2 + I3)(4W) – (I3)(9W) = 0
6 – 4I2 -4I3 - 9I3 = 0
6 – 4I2 - 13I3 = 0
I3 = 5/9 I2 (from last slide)
6 – 4I2 - 13(5/9 I2) = 0
6 = 101/9 I2
I2 = 0.53 A
I3 = 5/9 I2 = 0.29 A
I1 = I2 + I3 = 0.53 A + 0.29 A = 0.82 A
I1 = I2 + I3
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Loop Rule
Main Loop
9V – (I3)(10W) – (I1)(5W) = 0
Side Loop
(-I2)(5W) + (I3)(10W) = 0
(-I2)(5W) + (I3)(10W) = 0
(I2)(5W) = (I3)(10W)
I2 = 2I3
I3 = 9/25 = 0.36 A
I2 = 2I3 = 2(0.36 A) = 0.72 A
I1 = I2 + I3 = 0.36A + 072 A = 1.08 A
9V – (I3)(10W) – (I1)(5W) = 0
9V – (I3)(10W) – (I2 + I3)(5W) = 0
9 –10I3 – 5I2 – 5I3 = 0
9 –15I3 – 5I2 = 0
9 –15I3 – 5(2I3) = 0
9 –25I3 = 0
I3 = 9/25 = 0.36 A
Calculate all the currents in the following circuit.
A = 0.417 A
B = 0.26 A
C = 0.156 A
Calculate the currents in
the following circuit.
2 Amps, 3 Amps, -1 Amps
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Bottom Loop (clockwise)
10V – (6W)I1 – (2W)I3 = 0
Top Loop (clockwise)
-14V +(6W)I1 – 10 V -(4W)I2 = 0
Work with Bottom Loop
10V – (6W)I1 – (2W)I3 = 0
I1 + I2 = I3
10 – 6I1 – 2(I1 + I2) = 0
10 – 6I1 – 2I1 - 2I2 = 0
10 - 8I1 - 2I2 = 0
10 = 8I1 + 2I2
5 = 4I1 + I2
I2 = 5 - 4I1
Working with Top Loop
-14V +(6W)I1 – 10 V -(4W)I2 = 0
24 = 6I1 - 4I2
12 = 3I1 - 2I2
12 = 3I1 - 2(5 - 4I1)
22 = 11I1
I1 = 22/11 = 2.0 Amps
Calculate all the currents in the following circuit.
I2 = 5 - 4I1
I2 = 5 – 4(2) = -3.0 Amps
A = 1.56A
B = 0.522 A
C = 1.04 A
I1 + I2 = I3
I3 = -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)
Extra Kirchhoff Problems
I1 = -0.864 A
I2 = 2.6 A
I3 = 1.73 A
Batteries in Parallel
• Provide large current when
needed (Same voltage)
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a. Calculate the equivalent resistance. (2.26 W)
b. Calculate the current in the upper and lower
wires. (3.98 A)
c. Calculate I1, I2, and I3 (0.60 A, 2.25 A, 1.13 A)
d. Sketch a graph showing the voltage through the
circuit starting at the battery.
Calculate I (0.5 Amps)
Capacitors in Parallel
• I1 = +2.00 A I2 = +1.50 A 13 = −0.50 A
Voltage is constant in parallel
Q = CV
Q = C1V + C2V
Q = CeqV
Ceq = C1 + C2 + C3 +…
C1
C2
(like using a larger plate)
I1 = +2.00 A I2 = +1.50 A 13 = −0.50 A
Capacitors in Series
Charge is constant in series
1= 1
Ceq C1
+
C1
1
C2
+
C2
1
C3
Eq. Capacitance Ex. 1
Determine the equivalent capacitance for the circuit
below if each capacitor is 5 mF.
C1
C3
C3
C2
+Q -Q
+Q -Q
+Q -Q
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Given the following setup:
Parallel first:
Ceq = C1 + C2
Ceq = 5mF + 5mF = 10 mF
Series:
1 = 1
+
Ceq
10
a. Calculate the charge on each capacitor (52.8
mC, 28.8 mC)
1
5
=
3/10
Ceq = 3.33 mF
Given the circuit below:
a. Calculate the equivalent capacitance
b. Calculate the charge across each capacitor
c. Calculate the voltage on each capacitor
(HINT: Combine them for part a, and then split
them back up for b and c)
Parallel capacitors
Ceq = 5 mF +1 mF = 6 mF
Now both are series
1/ Ceq = 1/6 + 1/3
Ceq = 2 mF
Q = CV = (2 mF)(12 V) = 24 mC
For 3 mF capacitor
Q = CV
V = Q/C = 24 mC/3 mF = 8 V
For Combined Parallel Capacitors (same
voltage)
V = Q/C = 24 mC/6 mF = 4 V
For 5 mF Capacitor
Q = CV = (5 mF)(4 V) = 20 mC
For 1 mF Capacitor
Q = CV = (1 mF)(4 V) = 4 mC
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Given the following setup:
a. Calculate the equivalent
capacitance. (2.0 mF)
b. Calculate the total
charge that leaves the
battery. (8.0 mC)
c. Calculate the charge and
voltage on each
capacitor. (Q1 = 8.0 mC,
2.7 V: Q2 = Q3 = 4.0
mC, 1.3 V)
Given the following setup:
a. Calculate the equivalent capacitance. (0.609 mF)
b. Calculate the total charge that leaves the battery.
(14.6 mC)
c. Calculate the charge and voltage on each capacitor.
(Q1 = 14.6 mC, 14.6 V, Q2 = 14.6 mC, 7.30 V,
Q3 = 6.26 mC, 2.09 V, Q4 = 8.36 mC, 2.09 V )
a. Calculate the equivalent capacitance. (18 mF)
b. Calculate the total charge that leaves the battery. (162 mC)
c. Calculate the charge and voltage on each capacitor.
(Q1 = 45 mC, 9 V, Q2 = 90 mC, 9 V, Q3 = 27 mC, 9 V)
a. Equivalent capacitance. (1.58 mF)
b. Calculate the total charge that leaves the battery. (14.2 mC)
c. Calculate the charge and voltage on each capacitor. (Q1 =
14.2 mC, 2.84 V, Q2 = 14.2 mC, 1.42 V, Q3 = 14.2 mC,
4.73 V)
a. Equivalent capacitance. (7.31 mF)
b. Total charge that leaves the battery. (43.8 mC)
c. Calculate the charge and voltage on each capacitor. (Q1 =
30 mC, 6 V, Q2 = 13.8 mC, 1.38 V, Q3 = 13.8 mC, 4.6 V)
a. Equivalent capacitance. (6.41 mF)
b. Calculate the total charge that leaves the battery. (22.4
mC)
c. Calculate the charge and voltage on each capacitor. (Q1 =
22.4 mC, 2 V, Q2 = 15 mC, 1.5 V, Q3 = 7.5 mC, 1.5 V)
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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 off
Camera flashes
Difibralator
DVc + DVr = 0
Q - IR = 0
C
(Divide by R)
Q - I =0
RC
(I = -dQ/dt)
Q + dQ = 0
RC dt
 = time constant (time to reach 63% of full
voltage)
 = RC
A versatile relationship
Charging the Capacitor (for capacitor)
V = Vo (1-e-t/RC)
Q = Qo (1-e-t/RC)
I = Io e-t/RC
Discharging the Capacitor
V = Vo e-t/RC
Q = Qo e-t/RC
I = Ioe-t/RC
RC Circuits: Ex 1
What is the time constant for an RC circuit of
resistance 200 kW and capacitance of 3.0 mF?
 = (200,000 W)(3.0 X 10-6 F)
 = 0.60 s
(lower resistance will cause the capacitor to charge
more quickly)
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RC Circuits: Ex 2
What will happen to the bulb (resistor) in the
circuit below when the switch is closed (like a
car door)?
RC Circuits: Ex 3
An uncharged RC circuit has a 12 V battery, a 5.0
mF capacitor and a 800 kW resistor. Calculate
the time constant.
 = RC
 = (5.0 X 10-6 F)(800,000 W)
 = 4.0 s
Calculate:
a. Time constant (6 ms)
b. Maximum charge on the capacitor (3.6 mC)
c. Maximum current (0.60 mA)
d. Time to reach 99% of maximum charge (28 ms)
e. Current when charge = ½ Qmax (0.300 mA)
f. The charge when the current is 20% of the maximum
value. (2.9 mC)
Answer: Bulb will glow brightly initially, then dim
as capacitor nears full charge.
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 mC
What is the voltage and charge on the capacitor
after 1 time constant (when discharging)?
Discharging the RC Circuit
V = Vo e-t/RC
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An RC circuit has a charged capacitor C = 35
mF and a resistance of 120W. How much time
will elapse until the voltage falls to 10 percent
of its original (maximum) value?
V = Vo e-t/RC
0.10Vo =Voe-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)(120W)(35 X 10-6 F)
t = 0.0097 s or 9.7 ms
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?
A fully charged 1.02 mF 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 ms.
a. Calculate the charge on the capacitor at t=0
(20.4 mC)
b. Calculate the resistance R (57 W)
c. Calculate the charge at t = 60 ms (7.3 mC)
(t = 1.39RC)
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 mC)
b. Calculate the charge on the
capacitor after 5.0 ms while
discharging. (5.5 mC)
c. Calculate the voltage after 5.0
ms while discharging. (5.5 V)
d. Calculate the current through
the resistor after 5.0 ms while
discharging. (0.55 A)
Meters
• Galvanometer
– Can only handle a small current
• Full-scale Current Sensitivity (Im)
– Maximum deflection
• Ex:
– Multimeter
– Car speedometer
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Measuring I and V
Measuring Current
– Anmeter is placed in series
– Current is constant in series
Anmeter (Series)
Voltmeter (parallel)
Measuring Voltage
– Voltmeter placed in parallel
– Voltage constant in parallel circuits
– Measuring voltage drop across a resistor
DC Anmeter
• Uses “Shunt” (parallel) resistor
• Shunt resistor has low resistance
• Most of current flows through shunt, only a little
through Galvanometer
• IRR = IGr
IRR = Igr
(1 A)(R) = (50 X 10-6 A)(30 W)
R = 1.5 X 10-3 W or 1.5 m W
Meters: Ex 1
What size shunt resistor should be used if a
galvanometer has a full-scale sensitivity of 50 mA
and a resistance of r= 30 W? You want the meter
to read a 1.0 A current.
Voltage same through both (V=IR)
IRR = Igr
Since most of the current goes through the shunt
IR ~ 1 A
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.
IRR = Igr
(12 A)(R) = (1 X 10-3 A)(50 W)
R = 4.2 X 10-3 W or 4.2 m W
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DC Voltmeter
• Resistor in series
• Large R for resistor (keeps current low in
Galvanometer)
• V = I(R + r)
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)
26.
28.
30.
32.
34.
36.
38.
40.
42.
44.
183 W
13 V, 5 V, 0 V, -2 V
2 ms
6.9 ms
0.87 K W
A> D = E > B = C
1.99 m
4.0 W
7W
10 W and 60 V
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 30W and a full scale
deflection of 50 mA.
V = I(R + r)
15 V = (50 X 10-6 A)(R + 30W)
R + 30W = 15 V
50 X 10-6 A
R + 30W = 300,000 W
R ~ 300,000 W
2. Clockwise (12 V. 50 W, (10 mF 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 106 J
14. 25 W
18. 3.2 %
20. R/4
22. 12 W
24. 24 W
46. 9.0 V
50.1.0 A, 2.0 A, 15 V
56.a) 8 Amps (8 Vab)
58. a) R =0.505 W
62.Resistor(W)
4
6
8
24(bottom)
24 (right)
b) 9 Amps (0 Vab)
b) Req = 0.500 W
(V)
Current (A)
8
2
8
1.3
8
1
8
0.33
16
0.66
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I2 = -0.33 A
I1 = 0.54 A
I3 = 0.875 A
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