# 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 Problems Problem 1 Problem 1 – Parallel bulbs, ceiling lamps and you. Problem 2 Problem 2 – The Current Issue of Powerful Computing? Problem 3 Problem 3 – If a voltmeter were a car it would park like this? Problem 4 Problem 4 – It costs how much? Youve got to be kidding! SummarySummary - What did I learn? Here is what you should learn! Click on this icon to return to the this slide.

A. A.Clearly Sketch and label a circuit diagram modeling the ceiling lamp. C onsider 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 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 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. Answer 1AAnswer 1AAnswer 1AAnswer 1A

Problem 1 Problem 1 Problem 1 Problem 1 Problem 1 Problem 1 Problem 1 Problem 1 60 w R 1 75 w R 2 100 w R 3 120 V Power is the product of current and voltage therefore, current is the ratio of power to voltage. In parallel each bulb has the same voltage across it. I 1 = 0.5 amps I 2 = 0.625 ampsI 3 = 0.833 amps Answer 1B

Monitor 2.2 amps Speakers 1.5 amps H ow much power is used to operate this computer workstation? Each device operates at 120 volts. Computer 4.0 amps Printer.667 amps Scanner.52 amps Video Camera.375 amps Lamp 1.2 amps Problem 2

Answer 2 TOTAL CURRENT = 10.46 AMPS TOTAL POWER = 1255 WATTS (1.26 kW) Computer 480 watts Monitor 264 watts Lamp 144 watts Speakers 180 watts Scanner 62.4 watts Video Camera 45 watts Printer 80 watts

If Voltmeter Were a Car It Would Park In Parallel! Consider the adjacent Series Circuit: 1)Draw an equivalent circuit and calculate the equivalent resistance. 2)Calculate the current reading on the ammeter. 3)Calculate the voltage drop across each Resistor. R 1 =4.5 Ohms R 2 =7.5 Ohms R 3 =8 Ohms 15 V VVVV VV A R3R3R3R3 R2R2R2R2 R1R1R1R1

Answer 3 VV A R eq 15 V 1)The equivalent resistance is R eq = 20 Ohms. 2)The current in the circuit is I = 0.75 Ampere. 3)The Voltage Drops across: V R1 = 3.375 Volts V R1 = 3.375 Volts V R2 = 5.625 Volts V R2 = 5.625 Volts V R3 = 6.0 Volts V R3 = 6.0 Volts

Monitor 264 watts Speakers 180 watts Computer 480 watts Printer 80 watts Scanner 62.4 watts Video Camera 48 watts Lamp 144 watts C alculate the cost of operating this computer system for 1 month at \$ 0.10 per kwh. (24 hours x 30 days.) Problem 4Problem 4Problem 4Problem 4

Answer 4 \$ 90.36 per month

Summary of DC Circuits Parallel Circuits 1) 1)The voltage is constant across circuit elements in parallel. 2) 2)The current through circuit elements in parallel can change. 3) 3)The sum of individual currents add to equal the total system current. 4) 4)The sum of the individual elements power add to equal the total system power. 5) 5)Power = Energy / Time = I x V = V 2 /R = I 2 x R

Summary of DC Circuits Series Circuits 1) 1)The current circuit elements wired in series is constant. 2) 2)The voltage can change across a circuit element wired in series. 3) 3)The sum of voltage drops across individual circuit elements equals the voltage of the power supply. 4) 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 elements resistance.

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