DC Circuit – Practice DC Circuit – Practice Problems Problems Problem 1 – Parallel bulbs, ceiling lamps and you. Problem 2 – The Current Issue of Powerful Computing? Problem 3 – If a voltmeter were a car it would park like this? Problem 4 – It costs how much? You’ve got to be kidding! Summary - What did I learn? Here is what you should learn! Click on this icon to return to the this slide.
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DC Circuit – Practice Problems Problem 1 Problem 1 – Parallel bulbs, ceiling lamps and you. Problem 2 Problem 2 – The Current Issue of Powerful Computing?
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DC Circuit – Practice ProblemsDC Circuit – Practice Problems
Problem 1 – Parallel bulbs, ceiling lamps and you.
Problem 2 – The Current Issue of Powerful Computing?
Problem 3 – If a voltmeter were a car it would park like this?
Problem 4 – It costs how much? You’ve got to be kidding!
Summary - What did I learn? Here is what you should learn!
Click on this icon to return to the this slide.
A. Clearly Sketch and label a circuit diagram modeling the ceiling lamp.
Consider a ceiling lamp made from 3 bulbs wired in parallel. The bulbs are rated 100, 75 and 60 watts respectively and operate at 120 volts.
Problem 1 Problem 1
The lamps are the only elements using energy in the circuit . The circuit is protected by a 15 amp circuit breaker.
B. Calculate the current through each individual bulb.
C. Discuss relationships that exists between individual bulb currents, total lamp current & the circuit breaker.
Problem 1 Problem 1
Problem 1 Problem 1
The sum of individual bulb currents add to equal the total lamp current.
If the lamp current exceeds the limit set by the circuit breaker the circuit will open resulting in no current. I = 0 amps.
The current through each individual bulb depends on the power rating of the bulb. The power rating on each bulb measures the rate of which the bulb can transform electrical energy into heat and light energy: the greater the power, the greater the current.
The greatest current is through the 100 watt bulb and least through the 60 watt bulb.
AAnnsswweer r 11AA
Problem 1 Problem 1
Problem 1 Problem 1
60 w 60 w R R11
75 w 75 w
RR22
100 w 100 w R R33
120 V 120 V
Power is the product of current and voltage therefore, current Power is the product of current and voltage therefore, current is the ratio of power to voltage. In parallel each bulb has the is the ratio of power to voltage. In parallel each bulb has the same voltage across it.same voltage across it.
15 V15 V1)1) The equivalent resistance The equivalent resistance
is Ris Reqeq = 20 Ohms. = 20 Ohms.
2)2) The current in the circuit The current in the circuit is I = 0.75 Ampere.is I = 0.75 Ampere.
3)3) The Voltage Drops across:The Voltage Drops across:
VVR1R1 = 3.375 Volts = 3.375 Volts
VVR2R2 = = 5.625 Volts5.625 Volts
VVR3R3 = = 6.0 Volts6.0 Volts
Monitor 264 watts
Speakers 180 watts
Computer 480 watts
Printer 80 watts
Scanner 62.4 watts
Video Camera 48 watts
Lamp 144
watts
Calculate the cost of operating this computer system for 1 month at $ 0.10 per kwh. (24 hours x 30 days.)
PPrroobblleemm 44
Answer 4Answer 4
$ $ 90.36 90.36 per per monthmonth
Summary of DC CircuitsSummary of DC Circuits
Parallel Circuits
1) The voltage is constant across circuit elements in parallel.
2) The current through circuit elements in parallel can change.
3) The sum of individual currents add to equal the total system current.
4) The sum of the individual element’s power add to equal the total system power.
5) Power = Energy / Time = I x V = V2/R = I2 x R
Summary of DC Circuits
Series Circuits
1) The current circuit elements wired in series is constant.
2) The voltage can change across a circuit element wired in series.
3) The sum of voltage drops across individual circuit elements equals the voltage of the power supply.
4) The equivalent resistance of a circuit with more than one circuit element wired in series is equal to the sum of the individual circuit element’s resistance.