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!

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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.

V

PI

IVP V

PI 11

V

PI 22

V

PI 33

II11 = 0.5 amps = 0.5 amps I I22 = 0.625 amps = 0.625 amps II33 = 0.833 = 0.833 ampsamps

Answer Answer 1B1B

Monitor 2.2 amps

Speakers 1.5 amps

How 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 2Problem 2

Answer 2Answer 2

TOTAL CURRENT = 10.46 AMPS TOTAL CURRENT = 10.46 AMPS

TOTAL POWER = 1255 WATTS (1.26 kW)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 Consider the adjacent SeriesSeries Circuit:Circuit:

1)1) Draw an equivalent circuit Draw an equivalent circuit and calculate the equivalent and calculate the equivalent resistance.resistance.

2)2) Calculate the current reading Calculate the current reading on the ammeter.on the ammeter.

3)3) Calculate the voltage drop Calculate the voltage drop across each Resistor.across each Resistor.

RR11=4.5 Ohms =4.5 Ohms RR22=7.5 Ohms =7.5 Ohms RR33=8 Ohms=8 Ohms

15 15 VV

VVVV

VVVV

VVVV

AARR33

RR22

RR11

Answer 3Answer 3

VVVV

AA

RReqeq

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.