What is the Power Loss in a 20 Ohm Resistor in a Circuit?

AI Thread Summary
The discussion focuses on calculating the power loss in a 20-ohm resistor within a circuit that includes a 6-ohm and a 4-ohm resistor. Participants emphasize the importance of reducing the circuit to an equivalent resistance and determining the current flowing through each resistor. The correct approach involves using Ohm's Law (V = IR) to find the voltage across the resistors and applying power equations (P = IV) to calculate power loss. There is confusion about how to treat the resistors in series and parallel, particularly regarding the current distribution. Ultimately, understanding the voltage drop across the 6-ohm resistor is crucial for accurately determining the power loss in the 20-ohm resistor.
poopoo16
Messages
5
Reaction score
0
ok consider the circuit. what is the power loss in the 20 ohm resistor in J/s?



ok so what i did was that... and i don't know where to go after... please help
 

Attachments

  • Untitled-1 copy.jpg
    Untitled-1 copy.jpg
    6.4 KB · Views: 519
Physics news on Phys.org
poopoo16 said:
ok consider the circuit. what is the power loss in the 20 ohm resistor in J/s?



ok so what i did was that... and i don't know where to go after... please help

That's a good start. Now what about the 6 ohm resistor?
 
Reduce the circuit to an equivalent resistor, find the flowing current, and from there use your power equations to find the power.

You are on the right track.
 
OlderDan said:
That's a good start. Now what about the 6 ohm resistor?


that s my question... don't know how to treat the 6 ohms... do i add it to the 4?

and then use P=IV where I=V/R?...if i do that, its give me 1...and that is not the right answer...
 
poopoo16 said:
that s my question... don't know how to treat the 6 ohms... do i add it to the 4?

Yes, because it's in series with the other two. Be careful, however, in the power calculation. Is the current flowing through the 20 ohm resistor the same as through the whole circuit?
 
well, after u combine the 2 parallel ones, its no longer going thru the 20 per se right, its going thru a 6 and a 4?...soo... it then a series right?
 
so...

R=10
V=10
I=V/R
hence I=10/10?
meanning
P=IV
P=(1)10...?
 
poopoo16 said:
that s my question... don't know how to treat the 6 ohms... do i add it to the 4?

and then use P=IV where I=V/R?...if i do that, its give me 1...and that is not the right answer...

What you have found IS the right answer for the current flowing through the power source and through the 6 ohm resistor. The next step is to figure out the voltage across the 20 ohm and 5 ohm resistors (both the same). You can do that by using the total current and the equivalent resistance of that parallel combination that you found in the first step, or you can calculate the voltage drop across the 6 ohm resistor when that 1 amp of current flows through it and subtract that from the 10 volt source voltage. Once you have that voltage, you can use a power equation in terms of V and R, or you can use it to calculate the current through each of those resistors and use the power equation in terms of I and V. You should note that the 1 amp of current gets divided into two parts that flow through the two paths (5 ohm and 20 ohm) and that the sum of those two parts must add up to the current flowing through the 6 ohm and the power source, the 1 amp you have already found.
 
how do i calcuate voltyage drop arcross the 6ohm? do i use P=I^2R?
 
  • #10
You used V = IR solved for I to find the current of 1amp. Now you know the current through the 6 ohm resistor. Use V = IR to find the voltage across that one resistor.
 

Similar threads

Replies
5
Views
4K
Replies
3
Views
1K
Replies
4
Views
2K
Replies
1
Views
2K
Replies
3
Views
667
Replies
14
Views
1K
Replies
4
Views
3K
Replies
35
Views
3K
Back
Top