How many currents on a DC circuit?

[Juan Pablo]

I understand pretty much everything related to this, except how many currents should I put when arbitrarly putting them. The number of resistors? The number of wires? The number of loops?

I'm seriously lost here, thanks!

 

[chroot]

Are you trying to analyze a circuit? You should start by showing us the circuit.

- Warren

 

[Juan Pablo]

No specific circuit here, just a general question. Is there a general answer?

 

[chroot]

If all you're doing is trying to find the dc current through a circuit composed of batteries and resistors, then all you need is Ohm's law. The current through each resistor is equal to the voltage across that resistor divided by its resistance. The current through one resistor is not necessarily related in any way to the current through any other.

- Warren

 

[Juan Pablo]

Yes, but I'm trying to analyze more complex circuits using Kirchhoff's laws. The first step on my book is assigning the direction of currents arbitrarly. However, I'm not sure how many different currents should I put. Thanks.

 

[collinsmark]

I like to use current "loops" so to speak, such that each loop starts and ends in the same place. If each component in the circuit has at least one current associated with it, you should be able to formulate your simultaneous equations. Note that some components may have multiple currents associated with it. So when calculating the voltage drops of components with multiple currents, use the impedance of the component, multiplied by the sum of the currents going through that component (always keeping track of the signs, because the currents can be in opposite directions).

So my advise is just to make sure that each component in the circuit has at least one current loop going through it. After that, you can stop defining currents and move on to your equations.

You'll find that there are often multiple ways/choices of defining current loops that meet this criterion. But it shouldn't affect the final results.

 

[willem2]

I understand pretty much everything related to this, except how many currents should I put when arbitrarly putting them. The number of resistors? The number of wires? The number of loops?

I'm seriously lost here, thanks!

There could be a different current through every battery or resistance, except if they are
in series, so all those currents will need a variable.

 

[collinsmark]

Allow me to elaborate on I wrote in post #6.

I've added some attachments that you can click on below for reference.

The first attachment shows an example circuit.

The next two attachments show different ways that we've going to solve this problem, using different current loops (same circuit, just different current loops). Assume that we are solving for the voltage across the 3 Ohm resistor.

Solution 1:
Please refer to second attachment, where there are two small loops. We sum together the voltages going around each loop. Since there are two current loops there are two equations:

(-1 V) + (1 \Omega)i_1 + (2 \Omega)(i_1 - i_2) = 0

(2 V) + (3 \Omega)i_2 + (2 \Omega)(i_2 - i_1) = 0

We can re-write each simultaneous equation, simplifying things somewhat,

(1 V) = (1 \Omega)i_1 + (2 \Omega)i_1 - (2 \Omega)i_2

(-2 V) = (3 \Omega)i_2 + (2 \Omega)i_2 - (2 \Omega)i_1

Factor the i_1s in the first equation and the i_2s in the second equation,

(1 V) = (3 \Omega)i_1 - (2 \Omega)i_2

(-2 V) = -(2 \Omega)i_1 + (5 \Omega)i_2

From here, there are a few ways we can go to solve this pair of simultaneous equations. There's linear algebra, substitution, and what I'm going to use here. We're really interested in i_2 since that will give us the voltage across R₃. But we need to eliminate i_1 somehow. So let's multiply everything in the top equation by 2 and everything in the bottom equation by 3, in anticipation of adding the two equations together later.

(2 V) = (6 \Omega)i_1 - (4 \Omega)i_2

(-6 V) = -(6 \Omega)i_1 + (15 \Omega)i_2

Now add them together.

(-4 V) = (11 \Omega)i_2

or simply i_2 = -\frac{4}{11} A

which gives us

V_{OUT} = (3 \Omega)(-\frac{4}{11} A) = -\frac{12}{11} V

Solution 2:
We can also solve the problem using totally different current loops. Refer to the third attachment that has one little loop to the left and another big loop that goes all the way around the circuit. The equations for these loops are (starting with the smaller loop):

(-1 V) + (1 \Omega)(i_1 + i_2) + (2 \Omega)i_1 = 0

(-1 V) + (1 \Omega)(i_1 + i_2) + (2 V) + (3 \Omega)i_2 = 0

Simplifying again like before gives us,

(1 V) = (1 \Omega)i_1 + (1 \Omega)i_2 + (2 \Omega)i_1

(-1 V) = (1 \Omega)i_1 + (1 \Omega)i_2 + (3 \Omega)i_2

Factor the i_1s in the first equation and the i_2s in the second equation,

(1 V) = (3 \Omega)i_1 + (1 \Omega)i_2

(-1 V) = (1 \Omega)i_1 + (4 \Omega)i_2

In an attempt to get rid of i_1, we multiply the bottom equation by -3, in anticipation of adding the equations together.

(1 V) = (3 \Omega)i_1 + (1 \Omega)i_2

(3 V) = (-3 \Omega)i_1 + (-12 \Omega)i_2

Now we add the equations together,

(4 V) = (-11 \Omega)i_2

making

i_2 = -\frac{4}{11} A

Thus

V_{OUT} = (3 \Omega)(-\frac{4}{11} A) = -\frac{12}{11} V

Summary:

Different current loops, same answer. Just make sure that each component in the circuit has at least one current loop going through it.

Oh, and you'll make things easier on yourself if you don't create any new current loops containing only components already associated with other loops. In other words, when defining current loops, its best to start with components that don't have any loops associated with them already. Once all the components in the circuit are associated with at least one loop, stop there (and then move on to writing down the simultaneous equations).

 

[Juan Pablo]

How kind! That was really useful collinsmark. Thanks a lot!