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Ohm’s Law V = I R + _ V I R

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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

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Power and Energy in Electric Circuits Rate at which energy is supplied to a circuit is Power P = VI measured in Watts Can also be stated as: P = VI = (IR) I = I 2 R or P = VI = V (V/R) = V 2 /R

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EXAMPLES 1)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:P heat = I 2 R = (2.0 A) 2 (150 Ω) = 600 W

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Energy (W) Recall: Power = Work / time Therefore Work (energy) = Power x time W = Pt = VIt = I 2 RT = V 2 t / R Unit is joules

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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

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Series Circuits Or why old Christmas lights used to all go out when only one bulb was broken

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Series Circuit Has only one current path and if that path is interrupted, the entire circuit ceases to operate.

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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

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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

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For series circuits I = I 1 = I 2 = I 3 = … V = V 1 + V 2 + V 3 + … R eq = R 1 + R 2 + R 3 + … - this is the equivalent resistance of the circuit

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Example – calculate the meter readings 24 V 3Ω3Ω6Ω6Ω9Ω9Ω VtVt

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Solution First find equivalent resistance R eq = R 1 + R 2 + R 3 + … = 3Ω + 6Ω + 9Ω = 18Ω The total potential difference, V T = 24 V since the source supplies the entire circuit The current through circuit (I) is V=I R eq 24 V = I (18Ω) I = 1.33 A

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Solution cont’d Potential difference across each resistance can be found using Ohm’s Law V=IR V 1 = (1.33 A) (3Ω) = 4 V V 2 = (1.33 A) (6Ω) = 8 V V 3 = (1.33 A) (9Ω) = 12 V

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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.

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Example Suppose a fourth resistance of 18Ω is added to the series circuit. Calculate (a) equivalent resistance of circuit R eq = R 1 + R 2 + R 3 + … = 3Ω + 6Ω + 9Ω + 18Ω = 36Ω (b) the current through the circuit V=I R eq 24 V = I (36Ω) I = 0.67 A

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Parallel Circuits

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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 //.

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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)

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example In the diagram below, what are the magnitude and the direction of the current in wire X? 1A 2A 4A X

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For any // circuit V = V 1 = V 2 = V 3 = … = V n voltage is constant I = I 1 + I 2 + I 3 + … + I n current through entire circuit is equal to the sum of the currents through all resistances (Kirchoff’s 2 nd rule) V n = I n R n

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example Calculate (a) the equivalent resistance (b) Currents I 1 and I 2 (c) Total current of the following circuit. 3 Ω 6 Ω 24 V I1I1 ITIT I2I2

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Solution a) To find equivalent resistance:

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Solution (cont’d) b) To find I 1 and I 2 and c) I T 3 Ω 6 Ω 24 V I1I1 ITIT I2I2

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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.

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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 I1I1 ITIT I2I2 2 Ω I (2Ω)

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