Ohm’s Law V = I R + _ V I R. Ohm’s Law: V = IR A fundamental relationship in electric circuits....
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Transcript of Ohm’s Law V = I R + _ V I R. Ohm’s Law: V = IR A fundamental relationship in electric circuits....
Ohm’s Law
V = I R
+ _
V
IR
Ohm’s Law: V = IR
• A fundamental relationship in electric circuits. • Describes how much potential difference is
required to move charges through a resistance at a given current.
• Materials that have constant resistances are said to obey Ohm’s Law
Power and Energy in Electric Circuits
• Rate at which energy is supplied to a circuit is Power
P = VI measured in WattsCan also be stated as:
P = VI = (IR) I = I2Ror
P = VI = V (V/R) = V2/R
EXAMPLES1) Calculate the rate at which energy is supplied
by a 120 V source to a circuit if the current in the circuit is 5.5 A
Solution: P=VI = (120V)(5.5 A) = 660 W
2) A 150 Ω resistor carries a current of 2.0 A. Calculate the rate at which heat energy is produced in resistor
Solution: Pheat = I2R = (2.0 A)2(150 Ω) = 600 W
Energy (W)
• Recall: Power = Work / timeTherefore
Work (energy) = Power x time
W = Pt = VIt = I2RT = V2t / R
Unit is joules
Example
• How much energy is produced by a 50 V source that generates a current of 5.0 A for 2 minutes?
Solution: Don’t forget time must be in seconds!
W = VIt = (50 V)(5.0 A)(120 s) = 30,000 J = 30 kJ
Series Circuits
Or why old Christmas lights used to all go out when only one bulb was
broken
Series Circuit
• Has only one current path and if that path is interrupted, the entire circuit ceases to operate.
• The diagram represents a circuit containing three resistors in series with meters placed to measure various characteristics of the circuit
• - A - represents an ammeter, a very low resistance device that measures current in a circuit.
• - V - represents a voltmeter, a very high resistance device that measures potential difference across a circuit
In Series circuits…
• Current throughout the circuit is constant; therefore ammeter can be placed at any position.
• Potential Difference is equal to the SUM of the potential differences across all resistances– Known as Kirchhoff’s first rule (or simply the loop
rule)
• Ohm’s law holds for each resistance
For series circuits
• I = I1 = I2 = I3 = …
• V = V1 + V2 + V3 + …
• Req = R1 + R2 + R3 + … - this is the equivalent resistance of the circuit
Example – calculate the meter readings
24 V
3Ω 6Ω 9Ω
Vt
Solution
• First find equivalent resistanceReq = R1 + R2 + R3 + … = 3Ω + 6Ω + 9Ω = 18Ω
• The total potential difference, VT = 24 V since the source supplies the entire circuit
• The current through circuit (I) isV=I Req 24 V = I (18Ω) I = 1.33 A
Solution cont’d
• Potential difference across each resistance can be found using Ohm’s Law V=IR
V1 = (1.33 A) (3Ω) = 4 V
V2 = (1.33 A) (6Ω) = 8 V
V3 = (1.33 A) (9Ω) = 12 V
Important Fact!
• As the number of resistances in a series circuit increases, the equivalent resistance, of the circuit increases and the current through the circuit decreases.
Example
• Suppose a fourth resistance of 18Ω is added to the series circuit.
Calculate (a) equivalent resistance of circuitReq = R1 + R2 + R3 + …
= 3Ω + 6Ω + 9Ω + 18Ω = 36Ω(b) the current through the circuit
V=I Req 24 V = I (36Ω) I = 0.67 A
Parallel Circuits
Parallel Circuits
Parallel Circuits
• Have more than one current path.• If a segment of a // circuit is interrupted, the
result will not necessarily be that the entire circuit ceases to operate.
• House wiring is in //.
Parallel circuits
• Current separates into more than one path.• The point where separation occurs is known
as a junction • The sum of the currents entering a junction
must equal the sum of the currents leaving the junction
• This is Kirchoff’s second rule (or simply the junction rule)
example
• In the diagram below, what are the magnitude and the direction of the current in wire X?
1A
2A
4A
X
For any // circuit
• V = V1 = V2 = V3 = … = Vn voltage is constant
• I = I1 + I2 + I3 + … + In current through entire circuit is equal to the sum of the currents through all resistances (Kirchoff’s 2nd rule)
• Vn = InRn
• neq RRRRR
1...
1111
321
example
• Calculate (a) the equivalent resistance (b) Currents I1 and I2
(c) Total currentof the following circuit.
3 Ω
6 Ω
24 V
I1
IT
I2
Solution
a) To find equivalent resistance:
2
2
1
6
1
3
11
1...
1111
321
eq
eq
neq
R
R
RRRRR
Solution (cont’d)
b) To find I1 and I2 and c) IT
3 Ω
6 Ω
24 V
I1
IT
I2AAAIII
AV
R
VI
AV
R
VI
T 1248
46
24
83
24
21
22
11
Important Note• The equivalent resistance is less than any single
resistance in the circuit.• If more resistance is added in //, the equivalent
resistance decreases and the total current increases.– The result is roughly equivalent to increasing the cross-
sectional area of a conductor.• This is why overloading a household circuit by
connecting too many electrical appliances is dangerous. – As current increases, the amount of heat energy also
increases. This can lead to fires– Fuses and circuit breakers are designed to prevent such
fires from occurring.
example• A 2 – ohm resistor is added in parallel to the
previous circuit. Calculate (a) the equivalent resistance and (b) the total current of the altered circuit.
3 Ω
6 Ω
24 V
I1
IT
I2
2 ΩI(2Ω)
AV
R
VIb
R
R
RRRRRa
eq
eq
eq
neq
241
24)(
1
1
1
2
1
6
1
3
11
1...
1111)(
321