Two resistors connected to a battery

  • Thread starter Thread starter jay_buckets
  • Start date Start date
  • Tags Tags
    Battery Resistors
AI Thread Summary
In a circuit with a 100-ohm and a 200-ohm resistor connected in parallel to a 10-volt battery, the power dissipated by the 100-ohm resistor is calculated to be 1 watt. The current through the 200-ohm resistor can be determined using Ohm's Law, with the suggestion to first calculate the power dissipated by the 200-ohm resistor. The discussion emphasizes the importance of correctly applying relevant equations such as P=VI and P=V^2/R. Additionally, participants explore a bonus question regarding the same resistors connected in series, prompting further calculations. This thread highlights the collaborative effort to solve circuit problems using fundamental electrical principles.
jay_buckets
Messages
6
Reaction score
2
New user has been reminded to show their work on schoolwork questions
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.
 
Physics news on Phys.org
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?
 
  • Like
Likes berkeman
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!
 
  • Like
Likes berkeman
  • #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?
 
  • Haha
Likes berkeman
  • #13
Oops, thanks. Fixed it now. o0)
 
Back
Top