Calculating current with and without internal resistance

  • #1
ddobre
33
2

Homework Statement


KqTs2Gk.jpg

Calculate the magnitudes and directions of the currents in each of the 3 resistors in the circuit above, if...
(a) ...both batteries have zero internal resistance.
(b) ...each battery has an internal resistance of 1.0 Ω.

Homework Equations


I1 = I2 + I3 (Node Law)

The Attempt at a Solution


Y4q2Goy.jpg

I tried to use the node law to define the direction of the current, and for part a) I got I3 = .1452A,
I1 = .06398A, and I2 = .08122A. I'm not sure if this is correct but this is how I tried to solve part a). For part b), since the batteries now have internal resistance, I added two 1 ohm resistors to the circuit, on the left and right of the node at the top, (next to the negative terminals of the batteries). But, I had issues trying to solve the series of equations with 5 resistors in the circuit. I tried to solve for each anyway, but with poor results, (I couldn't narrow down one current). I would try to solve with an Req, but I need the current through each resistor. Any advice?
 

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  • #2
ddobre said:
I tried to use the node law to define the direction of the current
Don't use the node law to define directions. Just define. You are free to choose directions.

If your choise - as for an arrow direction - turns out to be opposite to the physical positive current direction, the resulting current will just be negative.

You can use the KVL law, as you have indicated, and you have two "windows" ( left and right ) that can be used as circular current paths.

If you choose the left and right current path to be counter clockwise, you will get the KVL equation for the left window:

12V - Ileft*35Ω - Ileft*48Ω + Iright*48Ω = 0

Now, write the equation as for the right window and calculate Ileft and Iright.

Ileft and Iright are virtual currents, not physical

The physical currents are calculated:

I1 = Ileft
I3 = Ileft - Iright
I2 = -Iright

You may alternatively use KCL to calculate node voltages, then use Ohms law to calculate currents.
 
  • #3
ddobre said:
But, I had issues trying to solve the series of equations with 5 resistors in the circuit. I tried to solve for each anyway, but with poor results, (I couldn't narrow down one current). I would try to solve with an Req, but I need the current through each resistor. Any advice?
So, will the current through the new resistors be different, in the new circuit, from the current through the exiting resistors that they are in series with. In other words, why are you worrying about 5 resistors when you only have 3?
 
  • #4
Hesch said:
Don't use the node law to define directions. Just define. You are free to choose directions.

If your choise - as for an arrow direction - turns out to be opposite to the physical positive current direction, the resulting current will just be negative.

You can use the KVL law, as you have indicated, and you have two "windows" ( left and right ) that can be used as circular current paths.

If you choose the left and right current path to be counter clockwise, you will get the KVL equation for the left window:

12V - Ileft*35Ω - Ileft*48Ω + Iright*48Ω = 0

Now, write the equation as for the right window and calculate Ileft and Iright.

Ileft and Iright are virtual currents, not physical

The physical currents are calculated:

I1 = Ileft
I3 = Ileft - Iright
I2 = -Iright

You may alternatively use KCL to calculate node voltages, then use Ohms law to calculate currents.

Would the right loop be: 9V - 25Iright-48Iright+48Ileft = 0
 
  • #5
phinds said:
So, will the current through the new resistors be different, in the new circuit, from the current through the exiting resistors that they are in series with. In other words, why are you worrying about 5 resistors when you only have 3?
No, current should remain the same since the resistors are in series with each other
 
  • #6
ddobre said:
Would the right loop be: 9V - 25Iright-48Iright+48Ileft = 0

Hesch said:
If you choose the left and right current path to be counter clockwise,

You have chosen the right circular current path to be clockwise, and that's ok except +48Ileft then becomes -48Ileft.

The reason why you have to choose directions ( arrows ) before setting up equations, is that you must respect your chosen directions! Otherwise you will get a wrong result.
 
  • #7
ddobre said:
No, current should remain the same since the resistors are in series with each other

Of course the currents through two resistors in series will be the same, but this common current will change, when an inner resistance in the battery is inserted.
 
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