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Thursday, 04/09/15 Put this into your C-otes! TEKS: By the end of today, IWBAT… P.5F: Design, construct, and calculate in terms of current through, potential difference across, resistance of, and power used by electric circuit elements connected in both series and parallel combinations. Calculate current levels resistance, voltage drops and power dissipated across series circuits. Essential Question: Topic: How do we determine current, resistance, voltage drop and power dissipated in a series circuit? Series Circuit Calculations 1.) A 20 Ohm lamp and a 5 Ohm lamp are connected in series & placed across a potential difference of 50 V. • Draw the diagram for the circuit above! • a.) What is the equivalent resistance of the circuit? • b.) What is the current in each circuit? • c.) What is the voltage drop across each lamp? • d.) What is the power dissipated in each lamp? 1.) A 20 Ohm lamp and a 5 Ohm lamp are connected in series & placed across a potential difference of 50 V. • a.) What is the equivalent resistance of the circuit? • Req = R1+R2+R3 • Req = 25 Ω 1.) A 20 Ohm lamp and a 5 Ohm lamp are connected in series & placed across a potential difference of 50 V. • b.) What is the current in the circuit? • R = V/I • 25 Ω = 50 V I • 50V = I 25 Ω • I=2A Rearrange: 25 Ω (I) = 50 V then 1.) A 20 Ohm lamp and a 5 Ohm lamp are connected in series & placed across a potential difference of 50 V. • c.) What is the voltage drop across each lamp? • R = V/I • 20 Ω = V_ 2A • Rearrange: V = (20 Ω)(2A) = 40 V • R = V/I • 5 Ω = V_ 2A • Rearrange: V = (5 Ω)(2A) = 10 V 1.) A 20 Ohm lamp and a 5 Ohm lamp are connected in series & placed across a potential difference of 50 V. • d.) What is the power dissipated in each lamp? • P = VI • P = 20 W + 80 W • P = 100 W 2.) Three identical lamps are connected in series to a 6 V battery. What is the voltage drop across each lamp? • 6 V_ 3 • V = 2V 3.) The load across a 12-V battery consists of a series combination of three resistors of 15 Ω, 21 Ω, and 24 Ω. a.) What is the total resistance of the load? b.) What is the current in the circuit? 3.) The load across a 12-V battery consists of a series combination of three resistors of 15 Ω, 21 Ω, and 24 Ω. a.) What is the total resistance of the load? • Rt = R1+R2+R3 • Rt = 15Ω + 21Ω + 24Ω. • Rt = 60 Ω 3.) The load across a 12-V battery consists of a series combination of three resistors of 15 Ω, 21 Ω, and 24 Ω. b.) What is the current in the circuit? • R = V/I • 60 Ω = 12 V I • I = .2 A 4.) The load across a battery consists of two resistors, with values of 15 Ω, and 45 Ω connected in series. a.) What is the total resistance of the load? b.) What is the voltage of the battery if the current in the circuit is 0.10 A? 4.) The load across a battery consists of two resistors, with values of 15 Ω, and 45 Ω connected in series. a.) What is the total resistance of the load? • Rt = R1+R2+R3 • Rt = 15Ω + 45Ω • Rt = 60 Ω 4.) The load across a battery consists of two resistors, with values of 15 Ω, and 45 Ω connected in series. b.) What is the voltage of the battery if the current in the circuit is 0.10 A? • R = V/I • 60 Ω = _V_ .1 A • V=6V 5.) A lamp having a resistance of 10 Ω is connected across a 15-V battery. a.) What is the current through the lamp? b.) What resistance must be connected in series with the lamp to reduce the current to 0.50 A? 5.) A lamp having a resistance of 10 Ω is connected across a 15-V battery. a.) What is the current through the lamp? • R = V/I • 10 Ω = 15 V I • I = 1.5 A 5.) A lamp having a resistance of 10 Ω is connected across a 15-V battery. b. What resistance must be connected in series with the lamp to reduce the current to 0.50 A? • R = V/I • R = 15 V .5 A • R = 30 Ω