Basic Circuit Theory: 60W Globe & Voltage Relationships

In summary, the conversation discusses the relationship between power, voltage, and resistance in a circuit and how it affects the output of a light bulb. It is noted that the power rating of a light bulb is specific to a certain voltage and changing the voltage can result in a different power output. It is also mentioned that applying a higher voltage than the bulb is designed for can cause it to blow. The conversation also touches on the concept of resistance and how it affects the power dissipated by the bulb. The conversation ends with a discussion on the non-linear nature of filament resistance and how it is typically ignored in introductory circuit problems.
  • #1
danago
Gold Member
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Hey all. I've never really done anything in the way of circuit theory up until now, where i have just started a unit on introduction to electrical/electronic engineering.

Lets say i have a 60W light globe. Does this mean that the maximum power it can dissipate is 60W, after which it will blow? Will it still be able to output less energy than stated?

For example, let's say i have a very basic circuit with a 60W globe and a 10V source voltage. According to KVL, the voltage across the globe will also be 10V, and then according to the relation P=VI, the current will be 6A? What if i now take a 0.01V source. Then by the same reasoning, a current 6000A would flow?

That definitely doesn't look right to me; it seems so very unrealisting for a 0.01V source to produce a current of 6000A, which would then tell me that the globe isn't actually outputting 60W anymore. If so, how can i find what power it is actually dissipating?

Am i on the right track, or have i completely missed something?

Thanks in advance,
Dan.
 
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  • #2
The power rating of a light bulb is for a specific voltage. Change to a different voltage, you'll get a different power.

For example, if the 60W rating is for a voltage of 120V across the lamp, then 10V would result in a lower power and less than 6A.
 
  • #3
Oh ok, makes sense. What if i apply a much higher voltage than it was designed for? Is this what will cause it to blow?
 
  • #4
danago said:
Oh ok, makes sense. What if i apply a much higher voltage than it was designed for? Is this what will cause it to blow?

Yes. I = V/R, so applying a higher RMS voltage will cause a higher current to flow through the resistance of the filimant, which will result in a higher power dissipation P = VI, which heats the filament beyond its operating temperature, and causes it to melt and fail open.
 
  • #5
Alright thanks for the replies :smile:

I was just doing some of the questions in my textbook and came across this one:

"A lightbulb rated at 60W will dissipate 60W as heat and light when connected with a 100V ideal voltage source. A lightbulb rated at 100W will dissipate 100W as heat and light when connected with the same voltage source.

If the bulbs are connected in series with the same 100V source, what is the power dissipated by each globe?
"

To produce 60W and 100W respectively with a 100V source, the currents through each individual circuit must be 0.6A and 1A respectively. Then, using ohm's law, i can calculate the resistance caused by each globe (166.67 Ohms and 100 Ohms respectively). When wired up in series, the effective resistance of the circuit will be 266.67 Ohms, and so the current flowing will be 0.375A. Using P=I^2 R, i can then calculate the power dissipated by each element as 23.4W and 14.06W.

Have i reasoned through that correctly?
 
  • #6
Yeah it should be ok, since I don't think the resistance of the bulbs are non-linear.
 
  • #7
Defennder said:
Yeah it should be ok, since I don't think the resistance of the bulbs are non-linear.

Alright cool, thanks :smile:
 
  • #8
The resistance of incandescent bulbs ARE nonlinear. Measure the resistance of a cold filament and you will see this is true.
 
  • #9
They can be considered ohmic within a given temperature range. Since the question doesn't give you any other information, how else would you do it?
 
  • #10
While filament resistance changes drastically with temperature*, that fact is typically ignored in solving this type of introductory circuit homework problems.

* R is about 15 times higher at operating temperature (2800-3000 K) than at room temperature (290-300 K)
 

1. What is the purpose of basic circuit theory?

Basic circuit theory is a fundamental concept in electrical engineering that describes the behavior and relationships of electrical components and their interactions within a circuit. It helps in understanding the flow of electricity and how different components work together to create a functioning circuit.

2. What is a 60W globe?

A 60W globe, also known as a light bulb, is a type of electrical component that converts electrical energy into light energy. The 60W refers to the amount of power it consumes, which is 60 watts, and determines the brightness of the light emitted by the globe.

3. How is voltage related to circuit theory?

Voltage is a key concept in circuit theory as it represents the electrical potential difference between two points in a circuit. It is measured in volts and plays a crucial role in determining the flow of current and the behavior of components within the circuit.

4. What are the relationships between voltage and current in a circuit?

According to Ohm's law, the relationship between voltage (V), current (I), and resistance (R) in a circuit is V=IR. This means that the voltage across a component is directly proportional to the current flowing through it, and inversely proportional to the resistance of the component.

5. How is basic circuit theory applied in real-world scenarios?

Basic circuit theory is applied in various real-world scenarios, such as designing and analyzing electrical circuits for different applications, troubleshooting circuit issues, and developing new technologies. It is also used in fields like telecommunications, power systems, and electronics to ensure efficient and safe operation of electrical devices and systems.

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