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W10D2DC Circuits
Today’s Reading Assignment W10D2 DC Circuits & Kirchhoff’s Loop Rules Course Notes: Sections 7.1-7.5
Announcements
PS 8 due Week 11 Tuesday at 9 pm in boxes outside 32-082 or 26-152
Next Reading Assignment W10D3 PS07: PhET: Building a Circuit 7.1-7.5, 7.10
Exam 3 Thursday April 18 7:30 pm –9:30 pm
2
3
Outline
DC Circuits
Kirchoff’s Laws
Electrical Power
Measuring Voltage and Current
4
DC Circuits
5
Examples of Circuits
6
Symbols for Circuit Elements
Battery
Resistor
Capacitor
Switch
Equipotential
Junction
Branch
7
Electromotive Force (EMF)
The work per unit charge around a closed path done is called electromotive force EMF. This is a bad name because it is not a force. Let denote the force per unit charge, then the EMF is
If a conducting closed path is present then
8
Ideal Battery
1) Inside Battery: chemical force , non-zero inside battery (zero outside), moves charges through a region in which static electric field opposes motion
2) Ideal battery: net force on charges is zero
3) Potential difference between its terminals
4) Extend path through external circuit where
9
Concept Question
The electric field inside the battery points to
1. The left
2. The Right
10
Concept Question Answer
The electric field inside the battery points to
1. The left
Goes from positive to negative charge
11
Sign Conventions - BatteryMoving from the negative to positive terminal of a battery increases your potential
Moving from the positive to negative terminal of a battery decreases your potential
12
Concept Question
The electric field inside the resistor points to
1. The left
2. The Right
13
Concept Question Answer
The electric field inside the resistor points to
2. The right
Electric field pushes charge in the same direction, so current in direction of E
This is cause and effect!
14
Sign Conventions - ResistorMoving across a resistor in the direction of current decreases your potential
Direction of current
Is the same as direction of electric field which points to lower electric potential
Moving across a resistor in the direction of current decreases your potential
Internal Resistance
Real batteries have an internal resistance, r, which is small but non-zero
Terminal voltage:
(Even if you short the leads you don’t get infinite current)
16
Series vs. Parallel
Series Parallel
17
Resistors In SeriesThe same current I must flow through both resistors
18
Current Conservation
Sum of currents entering any junction in a circuit must equal sum of currents leaving that junction.
19
Resistors In ParallelVoltage drop across the resistors must be the same
20
Concept Q: Resistors In Parallel
Suppose that . Then the equivalent resistance
1.
2.
3.
4.
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Concept Q: Resistors In Parallel
Answer 1. When
Then the equivalent resistance is the smaller resistance. This is an important circuit design principle.
22
Group Problem: Four Resistors
Four resistors are connected to a battery
as shown in the figure. The current in the battery is I,
the battery emf is ε, and the resistor values are R1 = R, R2 = 2R, R3 = 4R, R4 = 3R.
Determine the current in
each resistor in terms of I.
23
Measuring Voltage & Current
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Measuring Potential Difference
A voltmeter must be hooked in parallel across the element you want to measure the potential difference across
Voltmeters have a very large resistance, so that they don’t affect the circuit too much
25
Measuring Current
An ammeter must be hooked in series with the element you want to measure the current through
Ammeters have a very low resistance, so that they don’t affect the circuit too much
P18-26
Concept Question: Measuring Current
If R1 > R2, compare the currents measured by the three ammeters:
1. I1 > I2 > I3
2. I2 > I1 > I3
3. I3 > I1 > I2
4. I3 > I2 > I1
5. I3 > I1 = I2
6. None of the above7. Not enough information is given.
P18-27
Concept Question Answer: Measuring Current
Answer: 4. I3 > I2 > I1
The total current must add to the two individual currents, so I3 must be largest. Most current prefers to go through the smaller resistor so I2 > I1 .
28
Measuring Resistance
An ohmmeter must be hooked in parallel across the element you want to measure the resistance of
Here we are measuring R1
Ohmmeters apply a voltage and measure the current that flows. They typically won’t work if the resistor is powered (connected to a battery)
P18-29
Concept Question: Bulbs & Batteries
An ideal battery is hooked to a light bulb with wires. A second identical light bulb is connected in parallel to the first light bulb. After the second light bulb is connected, the current from the battery compared to when only one bulb was connected.
1. Is Higher2. Is Lower3. Is The Same4. Don’t know
P18-30
Concept Question Answer: Bulbs & Batteries
There are several ways to see this:
(A) The equivalent resistance of the two light bulbs in parallel is half that of one of the bulbs, and since the resistance is lower the current is higher, for a given voltage.
(B) The battery must keep two resistances at the same potential I doubles.
Answer: 1. More current flows from the battery
P18-31
Concept Question: Bulbs & Batteries
An ideal battery is hooked to a light bulb with wires. A second identical light bulb is connected in series with the first light bulb. After the second light bulb is connected, the current from the battery compared to when only one bulb was connected. 1. Is Higher
2. Is Lower
3. Is The Same
P18-32
Concept Question Answer: Bulbs & Batteries
The equivalent resistance of the two light bulbs in series is twice that of one of the bulbs, and since the resistance is higher the current is lower, for the given voltage.
Answer: 2. Less current flows from the battery
33
Electrical Power
Power is change in energy per unit time
So power to move current through circuit elements:
VIP
34
Power - Battery
I
IVIP supplied
Moving from the negative to positive terminal of a battery increases your potential. If current flows in that direction the battery supplies power
35
Power – Resistor (Joule Heating)
Moving across a resistor in the direction of current decreases your potential. Resistors always dissipate power
P18-36
Concept Question: PowerAn ideal battery is hooked to a light bulb with wires. A second identical light bulb is connected in parallel to the first light bulb. After the second light bulb is connected, the power output from the battery (compared to when only one bulb was connected)
1. Is four times higher
2. Is twice as high
3. Is the same
4. Is half as much
5. Is ¼ as much
P18-37
Concept Question Answer: Power
Answer: 2. Is twice as high
The current from the battery must double (it must raise two light bulbs to the same voltage difference) and
P18-38
Concept Question: Power
An ideal battery is hooked to a light bulb with wires. A second identical light bulb is connected in series with the first light bulb. After the second light bulb is connected, the light (power) from the first bulb (compared to when only one bulb was connected)
1. Is four times higher
2. Is twice as high
3. Is the same
4. Is half as much
5. Is ¼ as much
P18-39
Concept Question Answer: Power
Answer: 5. Is 1/4 as bright
R doubles current is cut in half. So power delivered by the battery is half what it was. But that power is further divided between two bulbs now.
40
Kirchhoff’s Loop Rules
41
Recall: Current Conservation
Sum of currents entering any junction in a circuit must equal sum of currents leaving that junction.
42
Sum of Potential Differences Around a Closed Path
Sum of potential differences across all elements around any closed circuit loop must be zero.
0i statici Closed
Path
V d E s
NO TIME DEPENDENCE! 0Bd
dt
43
Sum of Potential Differences Around a Closed Path
Sum of potential differences across all elements around any closed circuit loop must be zero.
0i statici Closed
Path
V d E s
44
Demonstration:Five Different Types of Resistance
F13
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=F 13&show=0
45
Steps of Solving Circuit Problem
1. Straighten out circuit (make squares)
2. Simplify resistors in series/parallel
3. Assign current loops (arbitrary)
4. Write loop equations (1 per loop)
5. Solve
46
Worked Example: Circuit
What is current through each branch of the circuit shown in the figure below?
47
Worked Example: Simple Circuit
Upper Loop: Start at a and circulate counterclockwise. Then
Current conservation at a:
Lower Loop: Start at a and circulate clockwise. Then
Solve for I2: and
48
Demonstration:Wheatstone Bridge F14
http://tsgphysics.mit.edu/front/?page=demo.php&letnum=F 14&show=0
49
Group Problem: Wheatstone Bridge
A circuit consisting of two resistors with R1=6.0 ohms and R2=1.5 ohms, a variable resistor, with resistance Rvar, a resistor of unknown value Ru, and 9.0 volt battery, are connected as shown in the figure. When Rvar is adjusted to 12 ohms, there is zero current through the ammeter. What is the unknown resistance Ru?