Electrical current & Kirchhoff’s laws

[roam]

Homework Statement

[PLAIN]http://img84.imageshack.us/img84/3711/imagode.gif

Consider the electrical circuit shown above. Using Kirchhoff’s Rules determine the currents I₁ and I₂.

(Correct answers: I₁=-0.727, I₂ = 2.33)

The Attempt at a Solution

We can see from the network diagram that there are 3 closed loops, a left inner loop containing a 40 v battery, a right inner loop with a 80 v battery, and an outer loop that contains both batteries. There is also a 60 v battery in the midle that belongs to left and right loops (but not the outer loop).

I'm confused, which (left/right) loop do I need to use for working out I₂? each of them gives different results:

Assuming the direction assigned to the loop is counter-clockwise, left loop gives:

I_2 = \frac{\epsilon_1 + \epsilon 2}{R_1 + R_2} = \frac{100}{115}= 0.86 which is wrong.

Going around the right loop gives:

I_2 = \frac{\epsilon_2 + \epsilon_3}{R_2}=\frac{140}{60}=2.33

which is the correct answer... but why?

Similarly for finding I₁ how do you decide whether you should use the left loop or the big loop? Any guidance is really appreciated.

 

[fuzzywolf]

I think you may have made a mistake in trying to skip a step on Kirchoff's rules. The expression you got for the left loop is only valid if R1 and R2 are in series. Are they?

Go back to the basic rules: \SigmaI = 0 at a junction and \SigmaV = 0 around a loop. Those should give you three equations to solve for the three unknowns in your problem.

 

[nivaOne]

I have changed the drawing a little bit.

Try to find out what the voltages are over every component in the left and middle arm !

Realise the following in this circuit:
- between points A and B, there can only be a "single" potential difference.
- the sum of the voltages in an arm equals the total voltage over that arm, this is valid for all three arms of this circuit.
- voltages can be either negative or positive. Have a good look at the battery symbol !

Once that you have determined the exact voltage over both resistors, use Ohm's law to calculate the I's.

 

[roam]

I can use \sum v= 0 to work out I₁ and when I find I₂, I can use \sum I=0 at a junction for finding I₃. But the real problem is finding I₂, do you just decide to use the right loop because in the left loop there is another resistor which is not in series with R₂? Why?

 

[nivaOne]

Hi Roan,

The answer lies in the fact that in this case

"I1 and I2 are determined by R1 and R2 respectively"

If you can determine the voltages across these resistors then you can calculate the currents. There is no resistor in the arm containing DV3. So the voltage across the left and middle arm is the full 80V. ( = DV3). ( see my second drawing)

According to your drawing then : I1 = I2 + I3. This is just for the reference !

(Which means that the outcome of your calculations will determine the "sign" of the calculated value. The correct answer provided before states that I1 = -0.727 A. The minus sign indicates that the current flows in the opposite direction of the arrow on the drawing. Turn the arrow around and you need to state that I1= 0.727A.)

Back to the math now. You need to find the voltages across R1 and R2 first !

Use my second drawing to determine the values, then use Ohm’s law to determine I1 and I2.
Have a good look at the direction of the arrows first ! ( add versus subtract)

The left drawing will allow you to calculate the voltage across R1 and the right one across R2.

In short:
1. You need to find I1 and I2 before you can fill in I1 = I2 + I3. (which means that the current law is not of much use at first )
2. You do not know I1 nor I2, so you need to use Kirchhoff’s voltage laws first to determine the voltage across R1 and R2.
3. Then use Ohm’s law to determine I1 and I2. ( U = I * R )
4. Once you know I1 and I2 you can determine I3 as well.