Two resistors connected to a battery

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SUMMARY

The discussion focuses on calculating the power dissipated by a 100-ohm resistor and the current in a 200-ohm resistor when both are connected in parallel to a 10.0-volt battery. The power dissipated by the 100-ohm resistor is determined to be 1 watt using the formula P = V^2/R. For the current in the 200-ohm resistor, Ohm's Law (V = IR) is recommended for calculation. Additionally, the discussion includes a bonus question regarding the same calculations for resistors in series.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Familiarity with power formulas: P = VI, P = I^2R, P = V^2/R
  • Basic knowledge of electrical circuits and resistor configurations
  • Ability to perform calculations involving resistors in parallel and series
NEXT STEPS
  • Learn how to calculate power in series circuits
  • Study the effects of resistor configurations on total resistance
  • Explore advanced circuit analysis techniques using Kirchhoff's laws
  • Investigate the implications of resistor tolerance on circuit performance
USEFUL FOR

This discussion is beneficial for students studying electrical engineering, hobbyists working on circuit projects, and educators looking for examples of resistor calculations in parallel and series configurations.

jay_buckets
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Homework Statement
In a circuit, a 100.-ohm resistor and a 200.-ohm resistor are connected in parallel to a 10.0-volt battery.

Determine the power dissipated by the 100.-ohm resistor.

Calculate the current in the 200.-ohm resistor.
Relevant Equations
P=VI
P=I^2R
P= V^2/R
I honestly have no idea how to do this problem whatsoever. My teacher didn't show an example like this before.
 
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jay_buckets said:
Homework Statement:: In a circuit, a 100.-ohm resistor and a 200.-ohm resistor are connected in parallel to a 10.0-volt battery.

Determine the power dissipated by the 100.-ohm resistor.

Calculate the current in the 200.-ohm resistor.
Relevant Equations:: P=VI
P=I^2R
P= V^2/R

I honestly have no idea how to do this problem whatsoever. My teacher didn't show an example like this before.
Welcome to the PF.

When the two resistors are connected in parallel to the battery, you can treat them as independent. Does that help? You have written the correct equations -- try applying them to the first question and show your work please. Thank you.
 
P= (10v^2)/(200.ohm) = 0.5 W. is that the correct answer for part 1?
 
jay_buckets said:
P= (10v^2)/(200.ohm) = 0.5 W. is that the correct answer for part 1?
I think part 1 asks about the 100 Ohm resistor, no? And good start, BTW. :smile:
 
ohhhh yeah it does oops. So would it be 1 W then?
 
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Yep. How about part 2 now? :smile:
 
would you use 1W=(I^2)100 ohms and then 1W/100ohms= I^2?
 
wait no would you calculate the power dissipated by the 200 ohm resistor and then use that formula?
 
For part 2, I would just use the simple form of Ohm's Law, V-IR.
 
  • #10
oh yeahhh thank you for your help!
 
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  • #11
You're welcome. And for bonus points, can you answer the same two questions for the case where the two resistors are connected in series across the 10 Ohm Volt battery? Hint -- start by calculating the series current by using the total series resistance and the battery voltage... :smile:
 
Last edited:
  • #12
berkeman said:
across the 10 Ohm battery?
Ohms, or Volts, or both?
 
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  • #13
Oops, thanks. Fixed it now. o0)
 

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