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Power dissipated in a resistor

  • Thread starter RedDead
  • Start date
  • #1
8
0
hello


Find the power dissipated in the (6 ohm) resistor.

the circuit is in the attachments


i know that P in the resistor = (i^2)*R
but how can i find the current in that 6 ohm resistor?

thanks
 

Attachments

Answers and Replies

  • #2
Think about first combining resistors. Reply back to show your work for that. From there, it should be simple to get the I in the 6ohm resistor.
 
  • #3
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
7
Answer these questions in order and you'll get there.

1.) What is the equivalent resistance of this circuit?
2.) What is the voltage across that equivalent resistance?
3.) What is the current through the right branch of the circuit? (Mr. Ohm will help you here)

Now you're ready to compute the power dissipated in that resistor.
 
  • #4
8
0
Answer these questions in order and you'll get there.

1.) What is the equivalent resistance of this circuit?
2.) What is the voltage across that equivalent resistance?
3.) What is the current through the right branch of the circuit? (Mr. Ohm will help you here)

Now you're ready to compute the power dissipated in that resistor.
i got R equivalent = 3.2 ohm
the voltage across R equivalent = 32 volts
I in the right branch = 8 A
am i right?
 
  • #5
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
7
Ack! You know what, I made the problem a little too simple in my head. You're right about the equivalent resistance and the voltage across it. That means that the voltage across the 16 Ohm resistor is 32V, so there's 2A going through it. So the other 8A goes into the other resistors. You can apply Kirchhoff's Voltage Law at this point to get the voltage across the 6 Ohm resistor.
 
  • #6
8
0
Ack! You know what, I made the problem a little too simple in my head. You're right about the equivalent resistance and the voltage across it. That means that the voltage across the 16 Ohm resistor is 32V, so there's 2A going through it. So the other 8A goes into the other resistors. You can apply Kirchhoff's Voltage Law at this point to get the voltage across the 6 Ohm resistor.
thats exactly what i did ;)
thanks you!
 

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