Electric Circuit Homework: Series vs. Parallel Light Bulbs

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SUMMARY

The discussion focuses on calculating the power dissipated by two light bulbs rated at 60 W and 75 W when connected in series to a 220 V source. Key equations include P=IV, V=IR, and the relationship between power and current. The participants clarify that power is the time derivative of work, and current can be derived from power and voltage. The primary recommendation is to first calculate the resistance of each bulb before analyzing the differences between series and parallel configurations.

PREREQUISITES
  • Understanding of electrical power calculations (P=IV)
  • Knowledge of Ohm's Law (V=IR)
  • Familiarity with the concepts of series and parallel circuits
  • Basic grasp of units of measurement: Watts (W), Volts (V), Amperes (A), and Ohms (Ω)
NEXT STEPS
  • Calculate the resistance of light bulbs using the formula R = V^2 / P
  • Explore the differences in voltage and current distribution in series vs. parallel circuits
  • Learn about the implications of connecting resistors in series and parallel on total resistance
  • Study the impact of bulb ratings on overall circuit performance and safety
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone seeking to understand the principles of circuit design and power distribution in electrical systems.

poskhare
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Homework Statement


Two light bulbs are rated as 60 W and 75 W at 220 V. If these are connected in series to a source of 220 V, what will the power in each be? Assume a constant resistance for the light bulbs.


Homework Equations


W=VQ P=IV V=IR


The Attempt at a Solution


I was solving a similar question, which was dealing with bulbs connected in parallel. By dividing the work (W) by the Volt (V) I got the current (A), which seems a little strange to me since I thought I'd get the charge. Could anyone explain this please? (I then used V=IR and then P=IV to figure out the power dissipated) When it comes to light bulbs connected in series I don't know how to approach the problem or what differs between the bulbs that are connected in series and those that are connected in parallel.
 
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Hello poskhare,

Welcome to physics Forums!
poskhare said:
I was solving a similar question, which was dealing with bulbs connected in parallel. By dividing the work (W) by the Volt (V) I got the current (A), which seems a little strange to me since I thought I'd get the charge. Could anyone explain this please?
But there isn't any 'work' value given in the problem. There are power values however: the 60 [W] and 75 [W].

Power is the time derivative of work. In other words, power is the change in energy per unit time.

So when you divide the power by the emf (a.k.a. voltage), you get the change in charge per unit time. Charge per unit time is current.

In summary, the important variables to this problem involve some or all of the following:

o Power, P (and two different values were given, one for each bulb). Power has units of Watts [W].
o EMF (sometimes called 'voltage'), V. Voltage has units of Volts [V].
o Current, I. Current has units of Amperes (sometimes abbreviated as 'Amps') [A].
o Resistance R. Resistance has units of Ohms [Ω].
(I then used V=IR and then P=IV to figure out the power dissipated) When it comes to light bulbs connected in series I don't know how to approach the problem or what differs between the bulbs that are connected in series and those that are connected in parallel.
My first recommendation is to find the resistance of each bulb, based on each bulbs rated power and rated voltage.

Then study the difference between putting resistors in parallel and putting them in series. It makes a difference. :wink:
 
Okay, now I get it. Thanks a lot!
 

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