How to Find the Current in a 6 Ohm Resistor?

AI Thread Summary
To find the current in a 6 Ohm resistor, first determine the equivalent resistance of the circuit, which is calculated to be 3.2 Ohms. The voltage across this equivalent resistance is 32 volts, leading to a current of 8 A through the right branch. The voltage across the 16 Ohm resistor is also 32V, resulting in 2A flowing through it. The remaining current of 8A is directed into the other resistors, allowing for the application of Kirchhoff's Voltage Law to find the voltage across the 6 Ohm resistor. This process effectively leads to the calculation of power dissipated in the resistor.
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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
 

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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.
 
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.
 
Tom Mattson said:
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?
 
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.
 
Tom Mattson said:
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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