DC Circuits: Ch 32 Voltage – Starts out at highest point at “+” end of battery Voltage drops...
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Transcript of DC Circuits: Ch 32 Voltage – Starts out at highest point at “+” end of battery Voltage drops...
![Page 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.](https://reader030.fdocuments.us/reader030/viewer/2022032706/56649dff5503460f94ae7f5c/html5/thumbnails/1.jpg)
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.
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+ -
I
Voltage highest Voltage zero
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The following circuit uses a 1.5 V battery and has a 15 W lightbulb.
a. Calculate the current in the circuit (P = IV)
b.Calculate the voltage drop across the lightbulb.
c. Sketch a graph of voltage vs. path (battery, top wire, resistor, bottom wire)
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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
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Resistors in Parallel
• Current splits at the junction
• Same Voltage across all resistors
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I = I1 + I2 + I3
I1 = V
R1
I = V
Req
1 = 1 + 1 + 1
Req R1 R2 R3
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Which combination of auto headlights will produce the brightest bulbs? Assume all bulbs have a resistance of R.
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For the Bulbs in Series:
Req = R + R = 2R
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
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What current flows through each resistor in the following circuit? (R = 100 )
Req = R1 + R2
Req = 200
V = IReq
I = V/Req
I = 24.0 V/ 200 = 0.120 A
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Calculate the current through this circuit, and the voltage drop across each resistor.
Req = 400 + 290
Req = 690
V = IR
I = V/Req
I = 12.0 V/690 I = 0.0174 A
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Vab = (0.0174A)(400)
Vab = 6.96 V
Vbc = (0.0174A)(290)
Vbc = 5.04 V
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What current flows through each of the resistors in this circuit? (R = 100 )
1 = 1 + 1
Req 100 100
1/Req = 2/100
Req = 50
I = V/Req = 24.0 V/50 = 0.48 A
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DC Circuits: Ex 4
What current will flow through the circuit shown?
1 = 1 + 1
Rp = 500 700
Rp = 290
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Req = 400 + 290
Req = 690
V = IR
I = V/R
I = 12.0 V/690 I = 0.017 A or 17 mA
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Example 4
Calculate the equivalent resistance in the following circuit.
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DC Circuits: Ex 5
What current is flowing through just the 500 resistor?
First we find the voltage drop across the first resistor:
V = IR = (0.017 A)(400 )
V = 6.8 V
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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 resistor:
V = IR
I = V/R
I = 5.2 V/500 = 0.010 A = 10 mA
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DC Circuits: Ex 6
Which bulb will be the brightest in this arrangement (most current)?
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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)
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DC Circuits: Ex 5
What resistance would be present between points A and B?
(ANS: 41/15 R)
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EMF and Terminal Voltage
• Batteries - source of emf (Electromotive Force), E (battery rating)
• All batteries have some internal resistance r
Vab = E – Ir
Vab = terminal(useful)voltage
E = battery rating
r = internal resistance
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EMF: Example 1
A 12-V battery has an internal resistance of 0.1 . If 10 Amps flow from the battery, what is the terminal voltage?
Vab = E – Ir
Vab = 12 V – (10 A)(0.10 )
Vab = 11 V
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EMF: Example 2
Calculate the current in the following circuit.
1/Req = 1/8 + ¼
Req = 2.7
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Req = 6
Req = 8.7
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1/Req = 1/10
Req = 4.8
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Everything is now in series
Req = 4.8
Req = 10.3
V = IR
I = V/R
I = 9.0 V/10.3 I = 0.87 A
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EMF: Example 2a
Now calculate the terminal(useful)voltage.
V = E – Ir
V = 9.0 V – (0.87 A)(0.50 )
V = 8.6 V
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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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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
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Kirchoff Conventions
The “loop current” is not a current.Just a direction that you follow around the loop.
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Kirchoff Conventions
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Kirchoff’s Rule Ex 1
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Junction Rule
I1 = I2 + I3
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Loop Rule
Main Loop
6V – (I1)(4) – (I3)(9) = 0
Side Loop
(-I2)(5) + (I3)(9) = 0
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I1 = I2 + I3 Eqn 3
6V – (I1)(4) – (I3)(9) = 0 Eqn 2
(-I2)(5) + (I3)(9) = 0 Eqn 3
Solve Eqn 1
(-I2)(5) + (I3)(9) = 0
(I3)(9) = (I2)(5)
I3 = 5I2
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Substitution into Eqn 2
6V – (I2 + I3)(4) – (I3)(9) = 0
6 – 4I2 -4I3 - 9I3 = 0
6 – 4I2 - 13I3 = 0
I3 = 5I2 (from last slide)
6 – 4I2 - 13(5I2) = 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
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Kirchoff’s Rule Ex 2
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I1 = I2 + I3
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Loop Rule
Main Loop
9V – (I3)(10) – (I1)(5) = 0
Side Loop
(-I2)(5) + (I3)(10) = 0
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(-I2)(5) + (I3)(10) = 0
(I2)(5) = (I3)(10)
I2 = 2I3
9V – (I3)(10) – (I1)(5) = 0
9V – (I3)(10) – (I2 + I3)(5) = 0
9 –10I3 – 5I2 – 5I3 = 0
9 –15I3 – 5I2 = 0
9 –15I3 – 5(2I3) = 0
9 –25I3 = 0
I3 = 9/25 = 0.36 A
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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
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Kirchoff’s Rules: Ex 3
Calculate the currents in the following circuit.
I1 + I2 = I3
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Bottom Loop (clockwise)
10V – (6)I1 – (2)I3 = 0
Top Loop (clockwise)
-14V +(6)I1 – 10 V -(4)I2 = 0
Work with Bottom Loop
10V – (6)I1 – (2)I3 = 0
I1 + I2 = I3
10 – 6I1 – 2(I1 + I2) = 0
10 – 6I1 – 2I1 - 2I2 = 0
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10 - 8I1 - 2I2 = 0
10 = 8I1 + 2I2
5 = 4I1 + I2
I2 = 5 - 4I1
Working with Top Loop
-14V +(6)I1 – 10 V -(4)I2 = 0
24 = 6I1 - 4I2
12 = 3I1 - 2I2
12 = 3I1 - 2(5 - 4I1)
22 = 11I1
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I1 = 22/11 = 2.0 Amps
I2 = 5 - 4I1
I2 = 5 – 4(2) = -3.0 Amps
I1 + I2 = I3
I3 = -1.0 A
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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
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Extra Kirchoff Problems
I1 = -0.864 A
I2 = 2.6 A
I3 = 1.73 A
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A Strange Example
Calculate I (-0.5 Amps (we picked wrong direction))
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a. Calculate the equivalent resistance. (2.26 )
b. Calculate the current in the upper and lower wires. (3.98 A)
c. Calculate I1, I2, and I3 (0.60 A, 0.225 A, 1.13 A)
d. Sketch a graph showing the voltage through the circuit starting at the battery.
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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
• Camera flashes
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Vc + Vr = 0
Q - IR = 0 (Divide by R)
C
Q - I = 0 (I = -dQ/dt)
RC
Q + dQ = 0
RC dt
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= time constant (time to reach 63% of full voltage)
= RC
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A versatile relationship
V = Vo e-t/RC
I = Io e-t/RC
Q = Qo e-t/RC
I generally find voltage, then use V=IR and Q=VC
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RC Circuits: Ex 1
What is the time constant for an RC circuit of resistance 200 k and capacitance of 3.0 F?
= (200,000 )(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)?
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Answer: Bulb will glow brightly initially, then dim as capacitor nears full charge.
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RC Circuits: Ex 3
An uncharged RC circuit has a 12 V battery, a 5.0 F capacitor and a 800 k resistor. Calculate the time constant.
= RC
= (5.0 X 10-6 F)(800,000 W)
= 4.0 s
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What is the maximum charge on the capacitor?
Q = CV
Q = (5.0 X 10-6 F)(12 V)
Q = 6 X 10-5 F or 60 F
What is the voltage and charge on the capacitor after 1 time constant?
V = (0.632)(12 V) = 7.584 V
Q = (0.632)(60 F) = 38 F
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Consider the circuit below. Calculate:
a. The time constant (6 ms)
b. Maximum charge on the capacitor (3.6 C)
c. Time to reach 99% of maximum charge (28 ms)
d. Current when charge = ½ Qmax (300 A)
e. Maximum current (600 A)
f. The charge when the current is 20% of the maximum value. (2.9 C)
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Discharging the RC Circuit
V = Vo e-t/RC
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RC Circuits: Ex 4
An RC circuit has a charged capacitor C = 35 F and a resistance of 120. 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
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2.3 = t/RC
t = 2.3RC
t = (2.3)(120)(35 X 10-6 F)
t = 0.0097 s or 9.7 ms
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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?
V = Vo e-t/RC 1.39 = t/RC
0.25Vo =Voe-t/RC t = 1.39RC
0.25 =e-t/RC
ln(0.25) = ln(e-t/RC)
-1.39 = -t/RC
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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 s.
a. Calculate the charge on the capacitor at t=0 (20.4 C)
b.Calculate the resistance R (57 )
c. Calculate the charge at t = 60 s (7.3 C)
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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 C)
b.Calculate the charge on the capacitor after 5.0 s. (5.5 C)
c. Calculate the voltage after 5.0 s (5.5 V
d.Calculate the current through the resistor after 5.0 s (0.55 A)
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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
Measuring Voltage– Voltmeter placed in parallel– Voltage constant in parallel circuits– Measuring voltage drop across a resistor
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Anmeter (Series) Voltmeter (parallel)
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DC Anmeter
• Uses “Shunt” (parallel) resistor
• Shunt resistor has low resistance
• Most of current flows through shunt, only a little through Galvanometer
• IRR = IGr
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Meters: Ex 1What size shunt resistor should be used if a galvanometer has a full-scale sensitivity of 50 A and a resistance of r= 30 ? You want the meter to read a 1.0 A current.
Voltage same through both (V=IR)IRR = IgrSince most of the current goes through the shuntIR ~ 1 A
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IRR = Igr
(1 A)(R) = (50 X 10-6 A)(30 )
R = 1.5 X 10-3 or 1.5 m
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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 )
R = 4.2 X 10-3 or 4.2 m
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DC Voltmeter
• Resistor in series
• Large R for resistor (keeps current low in Galvanometer)
• V = I(R + r)
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Meters: Ex 3What resistor should be used in a voltmeter that
can read a maximum of 15 V? The galvanometer has an internal resistance of 30 and a full scale deflection of 50 A.
V = I(R + r)
12 V = (50 X 10-6 A)(R + 30)
R + 30= 12 V
50 X 10-6 A
R + 30= 300,000 R ~ 300,000
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Meters: Ex 4
Design a voltmeter for a 120 V appliance with and internal galvanometer resistance of 50 and a current sensitivity of 1 mA.
(ANS: R = 120,000 )
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Electric Power
• Watt
• 1 Watt = 1 Joule
1 second
P = I2R
“Twinkle, twinkle, little star. Power equals I2R”
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Power: Ex 1
Calculate the resistance of a 40-W auto headlight that operates at 12 V.
P = I2R
V = IR (so I =V/R)
P = V2R
R2
P = V2
R
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P = V2
R
R = V2 = (12 V)2 = 3.6 P 40 W
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Household Electricity
• Kilowatt-hour
• You do not pay for power, you pay for energy
1 kWh = (1000 J)(3600 s) = 3.60 X 106 J
1s
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Power: Ex 2
An electric heater draws 15.0 A on a 120-V line. How much power does it use?
P = I2R
V = IR so R = V/I
P = I2V
I
P = IV = (15 A)(120V) = 1800 W or 1.8 kW
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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
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Power: Ex 4
A lightening bolt can transfer 109 J of energy at a potential difference of 5 X 107V over 0.20 s. What is the charge transferred?
V = PE/Q
Q = E/V = 109 J/ 5 X 107V = 20 C
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What is the current?
I = Q/t
I = 20 C/0.2 s = 100 A
What is the power?
P = I2R
V = IR so R = V/I
P = I2V
I
P = IV = (100 A)(5 X 107V ) = 5 X 109 W
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Household Electricity
• Circuit breakers – prevent “overloading” (too much current per wire)
• Metal melts or bimetallic strip expands
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Household Electricity: Ex 1
Determine the total current drawn by all of the appliances shown.
P = IV
I = P/V
Ilight = 100W/120 V = 0.8 A
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Ilight = 100W/120 V = 0.8 A
Iheater = 1800W/120 V = 15 A
Istereo = 350W/120 V = 2.9 A
Ihair = 1200W/120 V = 10.0 A
Itotal = 0.8A + 15.0A + 2.9A + 10.0A = 28.7 A
This would blow the 20 A fuse
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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
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DC vs AC
http://www.ibiblio.org/obp/electricCircuits/AC/AC_1.html
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Jump Starting a Car
POSITIVE TO POSITIVE
(or your battery could explode)