Hard Kirchoff's Laws problem with batteries & resistors....

[Joe03]

Homework Statement

Homework Equations

Kirchhoff;s first and second law, V=IR

The Attempt at a Solution

Tried the question as follows:
V= 24/(2+6+4)
= 24/12 = 2V (pd from top cell)
then did 2=4 I3
therefore I3 = 0.5A
and then 2 = 6I2
I2 = 0.33A
and I1 = 0.5+0.33...
=0.83A?

 

[gneill]

Homework Equations

Kirchhoff;s first and second law, V=IR

The Attempt at a Solution

Tried the question as follows:
V= 24/(2+6+4)
= 24/12 = 2V (pd from top cell)

Which circuit law(s) did you employ for the above? How do you justify summing the resistances in that fashio? Are they really all in series?

then did 2=4 I3
therefore I3 = 0.5A
and then 2 = 6I2
I2 = 0.33A
and I1 = 0.5+0.33...
=0.83A?

Your solution doesn't take into account the effects that the second voltage source will have on the currents in the circuit.

Start again, writing KCL and KVL equations for the circuit: Identify the nodes where KCL applies, and loops where you will apply KVL, then write the equations for them.

 

[Joe03]

I actually don't know where to begin...
I tried to use V=E-Ir so did 24-2I1=4I3, 27-6I2=4I3 so 6I2-2I1=3?
Can you start me off on the correct path, bit derailed.

 

[gneill]

I actually don't know where to begin...
I tried to use V=E-Ir so did 24-2I1=4I3, 27-6I2=4I3 so 6I2-2I1=3?
Can you start me off on the correct path, bit derailed.

Those equations are correct, but you've got only two that originate from the circuit. The other is derived from those two, so doesn't provide any new information. Since you have three unknowns (I1, I2, and I3) you need three equations that originate in the circuit. These can come from KVL loop equations or KCL node equations or a combination of both.

Note that once the loops you've chosen have incorporated each component at least once, there's no new information that you can extract by writing more KVL loop equations. Then you must turn to KCL to supply any "new" relationships between the variables.

To work consistently you should set up a basic procedure to follow:
1. Identify the nodes and loops.
2. Label each component on the diagram with the potential drop expected from given current directions.
3. Write KVL for appropriate loops.
4. Write KCL at appropriate nodes.
5. Solve the simultaneous equations by whatever method you're comfortable with.

After that is done you can start to solve the simultaneous equations.

When writing KVL and KCL equations, it's often less prone to making sign errors if you write them as a sum of terms that sums to zero. That is, place all the terms on the left hand side and make the right hand size zero. When done this way you can write your KVL equations simply by taking a "KVL walk" around the loop from any starting point, ending back at the starting point. Just write down the potential rises and drops as you encounter them on the walk.

So your first equation, for example could be obtained by "walking" counterclockwise around the outer loop of the circuit starting from the negative terminal of the 24 V battery:

$+24 - 4 I_3 - 2 I_1 = 0$

 

[Joe03]

So far I have
24-4I3-2I1=0
27-4I3+6I2=0
and 24-27-6I2-2I1=0

cant seem to get any further my calculations keep saying things like -1.5+I2=-1.5+I2

 

[gneill]

So far I have
24-4I3-2I1=0
27-4I3+6I2=0
and 24-27-6I2-2I1=0

cant seem to get any further my calculations keep saying things like -1.5+I2=-1.5+I2

Then perhaps you missed this paragraph in my previous post:

Note that once the loops you've chosen have incorporated each component at least once, there's no new information that you can extract by writing more KVL loop equations. Then you must turn to KCL to supply any "new" relationships between the variables.

Your third equation above only goes over previously covered ground in the circuit.

 

[Joe03]

i solved them to get I1=3A,
I2=-1.5A
I3=4.5A

 

[gneill]

i solved them to get I1=3A,
I2=-1.5A
I3=4.5A

Looks good.

 

[Joe03]

Thanks for your help by the way have 7 more questions like this, will be more manageable now