Finding I1, I2, and power dissipated

[Josh225]

Homework Statement

See attatched image

Homework Equations

I am assuming to use the current divider rule... But when I use it, I do not get the correct answer

1/Rt = 1/R1 + 1/R2 + 1/R3 .. When I used this I got Rt= 2.18 ohms

I also tried Is= E/Rt... 24/2.18 =11 AI am not sure what to try next..

Thanks!

 

[The Electrician]

How many volts are across each resistor?

 

[Josh225]

How many volts are across each resistor?

To do that, I would have to apply ohms law: V= I x R , right?

So, i would need to find "I" so I= V/R

12 ohms = 24 v / 12 ohms = 2A
8 ohms = 24 v/ 8 ohms = 3 A
4 ohms = 24v / 4 ohms = 6 A

So now.. V= I x R

12 ohms : 2 A x 12 ohms = 24 Volts
8 ohms : 3 A x 8 ohms = 24 volts
4 ohms: 6 A x 4 ohms = 24 volts

Yeah... That can't be right. My brain must be dead right now or something

 

[cnh1995]

Yeah... That can't be right.

Why? It is wrong but why do you think that can't be right? Is there something wrong with the method or the numbers? See the diagram carefully.

 

[billy_joule]

To do that, I would have to apply ohms law: V= I x R , right?

Nope, just look more closely at the diagram, as others are hinting.
You've been using 24V, that would be correct if the other side of the resistors were at 0V but that's not what the diagram shows...

 

[Nidum]

 

[Nidum]

 

[Josh225]

I think I see where to go now. I would have to add the 24 V and -8 V to get 32 Volts. Right?

 

[Nidum]

That's it . Good !

 

[Josh225]

Strange.. So I after I determined that 32 volts went across, I applied the formula Is= E (32)/ Rt (2.18) and got 14.67 A.. Which is the correct answer for both I1 and I2. I don't understand why though since I was calculating for the source current and got the correct answer. I am kind of at that stage where I am seeing which formulas work, and then seeing why afterwards.

I think I get the correct answer because the 12ohm resistor goes with I2 and the 8 ohm and 4 ohm resistor (8+4=12) goes with I1.. So both equal 12 ohms. But that isn't always going to be the case. I didn't really arrive at the answer intentionally either.

 

[Josh225]

Then for the power dissipated by the 4 ohm resistor, I just do Px=V^2/Rx ... Px= 32^2 / 4 ohms = 256 Watts

 

[Nidum]

I just do Px=V^2/Rx ... Px= 32^2 / 4 ohms = 256 Watts

That's ok .

The diagram supplied with the problem is not very good . Try drawing a new diagram using modern conventions for the layout and symbols .
Two horizontal lines for the power rails with the three resistances set vertically between them . Imagine a ladder on it's side .

Your original method of finding the equivalent resistance and hence the total current is perfectly good .
You could also have found the total current by adding up the individual resistance currents .

 

[Josh225]

That's ok .

The diagram supplied with the problem is not very good . Try drawing a new diagram using modern conventions for the layout and symbols .
Two horizontal lines for the power rails with the three resistances set vertically between them . Imagine a ladder on it's side .

Your original method of finding the equivalent resistance and hence the total current is perfectly good .
You could also have found the total current by adding up the individual resistance currents .

Oh nice! Yeah, that involves a little less mess. But how do I know that BOTH I1 and I2 equal the same amount?

 

[Josh225]

Oh... Current going in = the current going out...

 

[Nidum]

That's it .

 

[Josh225]

Thank you!

 

[Aristotle]

Then for the power dissipated by the 4 ohm resistor, I just do Px=V^2/Rx ... Px= 32^2 / 4 ohms = 256 Watts

There is many ways in finding power dissipation--you can also do P_x=(I_across-4Ω)² x R_4Ω where
I_across-4Ω= 36 volts / 4Ω = 8 Amps (since they are in parallel connection, voltage is same across each resistor!)
Then square this and multiply it by the 4Ω..
and still get the correct answer--just pointing out the many different techniques in circuit analysis.

 

[Josh225]

There is many ways in finding power dissipation--you can also do P_x=(I_across-4Ω)² x R_4Ω where
I_across-4Ω= 36 volts / 4Ω = 8 Amps (since they are in parallel connection, voltage is same across each resistor!)
Then square this and multiply it by the 4Ω..
and still get the correct answer--just pointing out the many different techniques in circuit analysis.

Thanks! :) I've added that to my bag of tricks.