Chapter 18: Electric Current and Circuits •Electric current •EMF •Current & Drift Velocity •Resistance & Resistivity •Kirchhoff’s Rules •Series & Parallel Circuit Elements •Applications of Kichhoff’s Rules •Power & Energy •Ammeters & Voltmeters •RC Circuits 1
§18.1 Electric Current e
ee
e
ee
e
A metal wire.
e
Assume electrons flow to the right.
Δq Current : I = Δt
Current is a measure of the amount of charge that passes though an area perpendicular to the flow of charge.
Current units: 1C/sec = 1 amp = 1 A 2
A current will flow until there is no potential difference. The direction of current flow in a wire is opposite the flow of the electrons.
I e
ee
e
ee
ee
ve
3
Example: If a current of 80.0 mA exists in a metal wire, how many electrons flow past a given crosssection of the wire in 10.0 minutes?
Δq I= Δt Δq = IΔt = 80.0 × 10 −3 A (600 sec ) = 48.0 C
(
)
q # of electrons = charge per electron 48.0 C = 1.60 × 1019 C/electron 20 = 3.00 × 10 electrons 4
§18.2 EMF and Circuits
An ideal battery maintains a constant potential difference. This potential difference is called the battery’s EMF(ε).
The work done by an ideal battery in pumping a charge q is W=qε.
5
At high potential The circuit symbol for a battery (EMF source) is

+
At low potential
Batteries do work by converting chemical energy into electrical energy. A battery dies when it can no longer sustain its chemical reactions and so can do no more work to move charges.
6
§18.3 Microscopic View of Current in a Metal Electrons in a metal might have a speed of ~106 m/s, but since the direction of travel is random, an electron has vdrift = 0.
7
Only when the ends of a wire are at different potentials (E≠0) will there be a net flow of electrons along the wire (vdrift ≠ 0). Typically, vdrift < 1 mm/sec.
8
Calculate the number of charges (Ne) that pass through the shaded region in a time Δt: l
l = vD Δt
n: number of “free” charge/ unit volume l
N e = n( Al ) = nA(vd Δt ) Δq eN e = = neAvd The current in the wire is: I = Δt Δt
Crosssectional area of wire 9
Example (text problem 18.19): A copper wire of crosssectional area 1.00 mm2 has a constant current of 2.0 A flowing along its length. What is the drift speed of the conduction electrons? Copper has 1.10×1029 electrons/m3.
I = neAvd I 2.0 A vd = = neA 1.10 ×10 29 m 3 1.60 × 1019 C 1.00 × 10 −6 m 2
(
)(
)(
)
= 1.1× 10 − 4 m/sec = 0.11 mm/sec
10
§18.4 Resistance and Resistivity A material is considered ohmic if ΔV∝I, where
“Ohm’s Law”
ΔV = IR The proportionality constant R is called resistance and is measured in ohms (Ω; and 1 Ω = 1 V/A).
Nonohmic
11
The resistance of a conductor is:
L R=ρ A
where ρ is the resistivity of the material, L is the length of the conductor, and A is its cross sectional area. With R∝ρ a material is considered a conductor if ρ is “small” and an insulator if ρ is “large”. Note: Resistivity is the reciprocal (inverse) of electrical coductivity
12
The resistivity of a material depends on its temperature:
ρ = ρ 0 (1 + α (T − T0 )) where ρ0 is the resistivity at the temperature T0, and α is the temperature coefficient of resistivity. A material is called a superconductor if ρ=0.
13
T/TC
Example (text problem 18.28): The resistance of a conductor is 19.8 Ω at 15.0 °C and 25.0 Ω at 85.0 °C. What is the temperature coefficient of resistivity? Values of R are given at different temperatures, not values of ρ. But the two quantities are related.
L R = ρ (1) A
ρ = ρ 0 (1 + α (T − T0 )) (2)
Multiply both sides of equation (2) by L/A and use equation (1) to get:
R = R0 (1 + α (T − T0 )) (3) 14
Example continued:
Solve equation (3) for α and evaluate using the given quantities:
R 25.0 Ω −1 −1 R0 −3 1 19 . 8 Ω α= = 3.75 × 10 °C = 85.0 °C − 15.0 °C ΔT
15
The Electric Circuit
16
§18.5 Kirchhoff’s Rules Junction rule: The current that flows into a junction is the same as the current that flows out. (Charge is conserved) A junction is a place where two or more wires (or other components) meet.
I1=I2+I3
Loop rule: The sum of the voltage dropped around a closed loop is zero. (Energy is conserved.) 17
For a resistor: If you cross a resistor in the direction of the current flow, the voltage drops by an amount IR (write as –IR). There is a voltage rise if you cross the other way (write as +IR).
18
For batteries (or other sources of EMF): If you move from the positive to the negative terminal the potential drops by ε (write as –ε). The potential rises if you cross in the other direction (write as +ε).
19
A current will only flow around a closed loop.
A
Applying the loop rule:
B VAB is the terminal voltage.
VAB − IR = 0
ε − Ir − IR = 0 20
In a circuit, if the current always flows in the same direction it is called a direct current (DC) circuit.
21
§18.6 Series and Parallel Circuits Resistors: The current through the two resistors is the same. It is not “used up” as it flows around the circuit! These resistors are in series. Apply Kirchhoff’s loop rule: ε − IR1 − IR2 = 0
ε = IR1 + IR2 = I ( R1 + R2 ) = IReq 22
The pair of resistors R1 and R2 can be replaced with a single equivalent resistor provided that Req=R1 + R2.
In general, for resistors in series
Req = R1 + R2 + … + Rn n
= ∑ Ri . i =1
23
Current only flows around closed loops. When the current reaches point A it splits into two currents. R1 and R2 do not have the same current through them, they are in parallel.
Apply Kirchhoff’s loop rule:
ε − I1 R1 = 0 ε − I 2 R2 = 0 The potential drop across each resistor is the same.
24
Applying the junction rule at A: I =I1+I2. From the loop rules:
ε = I1 R1 = I 2 R2
Substituting for I1 and I2 in the junction rule:
I=
ε R1
+
ε R2
I
1 1 1 = + = ε R1 R2 Req 25
The pair of resistors R1 and R2 can be replaced with a single equivalent resistor provided that
1 1 1 = + . Req R1 R2
In general, for resistors in parallel
1 1 1 1 = + +…+ Req R1 R2 Rn n
1 =∑ . i =1 Ri 26
Example (text problem 18.40): In the given circuit, what is the total resistance between points A and B? R1= 15Ω A
R 2= 12 Ω
R 3= 24 Ω
B
R2 and R3 are in parallel. Replace with an equivalent resistor R23.
1 1 1 = + R23 R2 R3 R23 = 8 Ω 27
Example continued:
The circuit can now be redrawn: R1= 15Ω A
R23=8 Ω
The resistors R23 and R1 are in series:
R123 = R1 + R23 = 23 Ω = Req
B A
R123 =23 Ω
Is the equivalent circuit and the total resistance is 23 Ω.
B 28
Capacitors: C1
C2
For capacitors in series the charge on the plates is the same. ε
Apply Kirchhoff’s loop rule:
ε− ε Q
Q Q − =0 C1 C2
=
1 1 1 + = C1 C2 Ceq
29
The pair of capacitors C1 and C2 can be replaced with a single equivalent capacitor provided that
1 1 1 = + . Ceq C1 C2
In general, for capacitors in series 1 1 1 1 = + +…+ Ceq C1 C2 Cn n
=∑ i =1
1 . Ci 30
C2
C1
ε
Apply Kirchhoff’s loop rule: Q1 ε − =0 C1 Q2 ε− =0 C2
For capacitors in parallel the charge on the plates may be different. Here
Qeq = Q1 + Q2
ε Ceq = ε C1 + ε C2 31
The pair of capacitors C1 and C2 can be replaced with a single equivalent capacitor provided that Ccq= C1 + C2.
In general, for capacitors in parallel n
Ceq = C1 + C2 + … + Cn = ∑ Ci . i =1
32
Example (text problem 18.49): Find the value of a single capacitor that replaces the three in the circuit below if C1 = C2 = C3 = 12 μF. C1 A
C2 and C3 are in parallel C3
B
C2
C23 = C2 + C3 = 24 μF
33
Example continued:
The circuit can be redrawn: C1
The remaining two capacitors are in series.
A
1 1 1 = + C123 C1 C23
C23
B
C123
1 1 = + 12 μF 24 μF = 8 μF
A C123 B
Is the final, equivalent circuit. 34
§18.7 Circuit Analysis Using Kirchhoff’s Rules To solve multiloop circuit problems:
1. Assign polarity (+/) to all EMF sources. 2. Assign currents to each branch of the circuit. 3. Apply Kirchhoff’s rules. 4. Solve for the unknowns.
35
Example (text problem 18.53): Find the three unknown currents (the current in each resistor). I3 ε2
+ 
R3 I2 R2
ε1
+
I1 R1 36
Example continued:
Loop EDCFE:
ε 1 − I1 R1 − I 2 R2 = 0
Loop AFCBA:
ε 2 + I 2 R2 − I 3 R3 = 0 Note: Could also use Loop AFEDCBA
Junction C:
I1 = I 2 + I 3 Note: could also use junction F 37
Example continued:
The point is to write down three equations for the three unknown currents.
(1)
I1 R1 + I 2 R2 = ε 1
(2)
I 3 R3 − I 2 R2 = ε 2
(3)
I1 = I 2 + I 3 38
Example continued:
Substitute (3) into (1):
(R1 + R2 )I 2 + R1I 3 = ε1
(4)
R3 I 3 − R2 I 2 = ε 2
(2)
Multiply the top equation by –R3 and the bottom equation by +R1, add the equations together, then solve for I2.
− R3 (R1 + R2 )I 2 − R1 R3 I 3 + R1 R3 I 3 − R1 R2 I 2 = R1ε 2− R3ε 1 R1ε 2− R3ε 1 I2 = − R3 (R1 + R2 ) − R1 R2 I 2 = −0.123 amps 39
Example continued:
Substitute I2 = 0.123 amps in to (2):
I3 =
ε 2 + I 2 R2 R3
I 3 = +0.199 amps Now substitute the known values of I2 and I3 into (3):
I1 = I 2 + I 3 = −0.123 amps + 0.199 amps = +0.076 amps 40
Example continued:
The negative sign on I2 means that instead of the current going from right to left (from point C to point F) in the branch with resistor 2, it really goes from left to right. It is essential to keep the negative sign when evaluating your equations numerically. Make the correction only when the problem is finished.
41
§18.8 Power and Energy in Circuits
The energy dissipation rate is:
For an EMF source:
ΔU q P= = ΔV = IΔV Δt Δt
P = Iε
2 V Δ For a resistor: P = IΔV = I 2 R = R
42
§18.9 Measuring Currents and Voltages Current is measured with an ammeter. An ammeter is placed in series with a circuit component. A1
An ammeter has a low internal resistance.
R1
A1 measures the current through R1.
R2
A2 measures the current through R2.
A2
A3
ε
A3 measures the current drawn from the EMF.
43
A voltmeter is used to measure the potential drop across a circuit element. It is placed in parallel with the component. A voltmeter has a large internal resistance. V
The voltmeter measures the voltage drop across R1. R1
R2
ε 44
§18.10 RC Circuits Switch
R ++
C

Close the switch at t=0 to start the flow of current. The capacitor is being charged.
ε
Apply Kirchhoff’s loop rule:
Q ε − IR − = 0 C ΔQ Note : I = Δt 45
The current I(t) that satisfies Kirchhoff’s loop rule is:
I (t ) = I 0 e
where
I0 =
ε R
−t
τ
and τ = RC.
τ is the RC time constant and is a measure of the charge (and discharge) rate of a capacitor.
46
The voltage drop across the capacitor is:
−t ⎞ ⎛ VC (t ) = ε ⎜1 − e τ ⎟ ⎝ ⎠
The voltage drop across the resistor is:
VR (t ) = I (t )R
The charge on the capacitor is:
Q C (t ) = CVC (t )
Note: Kirchhoff’s loop rule must be satisfied for all times. 47
Plots of the voltage drop across the (charging) capacitor and current in the circuit.
48
While the capacitor is charging S2 is open. After the capacitor is fully charged S1 is opened at the same time S2 is closed: this removes the battery from the circuit. Current will now flow in the right hand loop only, discharging the capacitor. I
Apply Kirchhoff’s loop rule:
C
Q − IR + = 0 C
R
S1
S2
+
ε
The current in the circuit is
I (t ) = I 0 e
−t
τ
.
But the voltage drop across the capacitor is now VC (t ) = ε e
−t
49
τ
.
The voltage drop across the discharging capacitor:
50
Example (text problem 18.83): A capacitor is charged to an initial voltage of V0=9.0 volts. The capacitor is then discharged through a resistor. The current is measured and is shown in the figure.
51
Example continued:
(a) Find C, R, and the total energy dissipated in the resistor. Use the graph to determine τ. I0=100 mA; the current is I0/e = 36.8 mA at t= 13 msec.
I 0 = 100 mA =
ε
R τ = 13 msec = RC Since ε = V0 = 9.0 volts, R = 90 Ω and C = 144 μF. All of the energy stored in the capacitor is eventually dissipated by the resistor.
1 2 U = CV0 = 5.8 ×10 −3 J 2
52
Example continued:
(b) At what time is the energy in the capacitor half of the initial value?
1 1 2 U (t ) = CV (t ) and U (t = 0) = CV02 2 2
1 1 Want: U (t ) = U (t = 0) = CV02 2 4 1 1 2 2 CV (t ) = CV0 2 4 1 V (t ) = V0 2 53
Example continued:
Solve for t:
V (t ) = V0 e
−t
τ
1 = V0 2
t = τ ln 2 = (13 msec) ln 2 = 4.5 msec
54
Summary •Current & Drift Velocity •Resistance & Resistivity •Ohm’s Law •Kirchhoff’s Rules •Series/Parallel Resistors/Capacitors •Power •Voltmeters & Ammeters • RC Circuits 55