# How many currents on a DC circuit?

1. Feb 17, 2010

### 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!

2. Feb 17, 2010

### chroot

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

- Warren

3. Feb 17, 2010

### Juan Pablo

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

4. Feb 17, 2010

### chroot

Staff Emeritus
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

5. Feb 17, 2010

### 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.

6. Feb 17, 2010

### 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.

7. Feb 18, 2010

### willem2

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.

8. Feb 18, 2010

### 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.

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

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

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

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).

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Last edited: Feb 18, 2010
9. Feb 18, 2010

### Juan Pablo

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