time level of concentration 5.00pm syllabus reactive components: inductors and capacitors. ohms law,...
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Time
Level of concentration
5.00pm
Syllabus
Reactive components: Inductors and Capacitors.
Ohms law, resistors in series and in parallel.
Power.
Ideal and realistic Voltage and Current sources.
Kirchoff’s Laws.
Resistive networks.
AC circuits.
Moving charges and electric currents
Battery
+ –
Electric current I
€
I =dq
dt1 ampere = 1 A= 1 Coulomb per second = 1 C/s
Current is scaler
I1
I2
I3
€
I1 = I2 + I3
The arrows that we use in diagrams are indicative of the direction of the current flow.The arrows point in the direction in which positively charged particles would be forced to move by the electric field.
1 A2 A
2 A2 A
3 A4 A
i
Find the direction and magnitude of current i in thecircuit above.
Current Density
€
J =I
A(Am−2)
Resistance
Pump
Water current proportional to potential differenceWater current inversely proportional to pipe resistance
€
Flow current =Potential difference
resis tance
Electrical resistance R
Battery = Charge pump
High potential
Low potential
Poential difference = VI
€
I =V
R
If V= 1 V, I= 1 A, then R=1
Electrical resistance R
Electrical Power
IV
In time dt, charge dq moves through the resistor. (potential drop V). Thus energydrop is (dq V) .
To maintain the current, the battery must lift charge dq to the potential V, performing work
€
dU = dq ×V = idt ×VTo work done in unit time is the power of the battery:
€
P =dU
dt= iV = i2R =
V 2
R
emf devices and internal resistanceA device, such as a battery, which can maintain a potential difference by pumping charges is calledan emf device.
emf devices: electric generator uses mechanical energy to pump charges.Solar cells
An ideal emf device : the potential difference between
the terminals of an ideal emf device is always
A real emf device, has internal resistance, so that the potential difference between its two terminals dependson what load is connected.
a
IRr
+
–
b
Rr
a ab
i
iriR
Potential
The voltage available for R: - ir.The internal consumption:ir, proportional to i.
The loop rule:The sum of the changes in potential encountered in acomplete traversal of any loop of a circuit must be zero.
This is also known as Kirchhoff’s loop rule.
For a move through a resistance in the directionof the current, the change in potential is –iR.
For a move through an ideal emf device in the directionof the emf arrow, the potential change is .
a
IRr
+
–
b
Apply Kirchhoff’s loop rule to the above circuit:
Solving for the current:
€
i =ε
R + rThe power delivered to the external load:
€
P = I2R = (ε
R + r)2 R P reaches maximum for R=r
€
−ir − iR = 0
Resistances in series
i
+
–
Req
€
−iR1 − iR2 − iR3 = 0
€
i =ε
R1 + R2 + R3
R1
+
–
R2
R3
€
Req = R1 + R2 + R3
Kirchhoff’s Junction rule
The sum of the current entering any junction must be equal to the sum of the current leaving the junction.
I1
I3I2
I4
€
I1 + I2 = I3 + I4
Resistances in parallel
R1
+
–
R2 R3
i
i1 i2i3
i2+i3 The potential differenceacross all three resistorsis the same
From Ohm’s law:
€
i1 =ε
R1
,i2 =ε
R2
,i3 =ε
R3
€
i = i1 + i2 + i3 = ε(1
R1
+1
R2
+1
R3
)
+
–
Req
i
R1
+
–
R2 R3
i
i1 i2i3
i2+i3
€
i = i1 + i2 + i3 = ε(1
R1
+1
R2
+1
R3
)
€
i =ε
Req
Therefore:
€
1
Req
= (1
R1
+1
R2
+1
R3
)
For two resistances in parallel:
€
1
Req
= (1
R1
+1
R2
) =R1 + R2
R1R2
€
Req =R1R2
R1 + R2
Sample problem 1
iR1
+
–
R3
R2
1=4.4 V, 2=2.1 V, R1=2.3 , R2=5.5 , R3 1.8 Find i.
Solution: use Kirchhoff’s loop rule
€
1 − iR1 − iR2 − iR3 −ε2 = 0
€
1 − iR1 − iR2 − iR3 −ε2 = 0
€
i =ε1 −ε2
R1 + R2 + R3
=4.4 − 2.1
2.3+ 5.5 +1.8= 0.24A
R1
i
i
+
–
R3
R2
€
1 + iR1 + iR2 + iR3 −ε2 = 0
€
i = −ε1 −ε2
R1 + R2 + R3
= −4.4 − 2.1
2.3 + 5.5 +1.8= −0.24 A
Sample problem 2Each resistance is 1, find the equivalent resistance
+
–
+
–
Req=4/3
+
–
Req=4/3
Some practical circuits
2. The Ammeter and the voltmeter
1.The Wheastone bridge
The Wheastone bridgeR1 and R2 are standardResistors with resistance values known.
Rs is a variable resistance,its resistance value can be adjusted by sliding the contact
+ –
R1
R2
RsRx
R0
a
b
A
When point a and b are atthe same potential:
€
Rx = Rs
R2
R1
⎛
⎝ ⎜
⎞
⎠ ⎟
+ –
R1
R2
RsRx
R0
a
b
A
I1
I2
Potential drop across R1:I1R1
Potential drop across RsI2Rs
I1R1=I2Rs
For the same reason:I1R2=I2Rx
€
I1R1
I1R2
=I2Rs
I2Rx
€
Rx = Rs
R2
R1
⎛
⎝ ⎜
⎞
⎠ ⎟
The Ammeter and the voltmeter
The instrument to measure electrical currents is calledan ammeter.
The instrument to measure electrical potential differences is called a voltmeter.
+
–
R
I
+
–
R
I
+
–
R
IA
+
–
R
IA
V
Resistance of an Ammeter must be very small
Resistance of a voltmeter must be very large
Current source
i
Current source is a device that deliversa specified current I regardless of the load
R
i
VR
€
VR = iRCurrent source is a device that delivers a specified current I regardless of VR
+- i
R1
R2
Find voltage V
V
i1 i2
€
i = i1 + i2
Kirchhoff’s loop rule around the red loop
€
i2R2 − i1R1 −ε = 0
€
i2R2 − (i − i2)R1 −ε = 0
€
i2(R2 + R1) = ε + iR1
€
i2 =ε + iR1
(R2 + R1)
€
V = i2R2 =ε + iR1
(R2 + R1)R2
The organic battery: electric eel
R
500 electroplaques per row
140 rows
Physicists’ electric eel
Total emf per row: 5000 x 0.15 V= 750 VTotal resistance per row: 5000 x 0.25 = 1250
The equivalent circuit per row:
row= 750V
Rrow=1250
The equivalent circuit of many rows:
row= 750V Rrow=1250
Req=?
eq= ?V
the electric eel stuns
Why doesn’t the eel get itself stunned?
Because the electric eel is FAT.
When it comes to stunning, the current density is important, not just the current.
The electric eel is not an eel!!!
Neotropical knifefishes
Weight: 25 kg. 500 W electric power does not come from nothing.
Voltage divider
R1
R2
Vin
Vout
€
Vout =Vin
R1 + R2
R2
Network analysis
Superposition theorem
Eliminate all but one source at a time.
Take away B2, and calculate currents from B1 only
R2//R3:I1= 6A, I2=2A, I3=4A.
Take away B1
R1//R2:I3= 3A, I2=2A, I1=1A.
Now add algebraically:Current through R1:5A left to right;Current through R2: 4A down. Current through R3:3A
R1
+
–R2
R1
R1
R1
R1
+
–R2
R1
R1
R1
i1 i2
i3
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