Download 1.) A 20 Ohm lamp and a 5 Ohm lamp are connected in series

Thursday, 04/09/15
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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