# DC Circuits - Part B

## Resistors in Combination Circuits

Here, we will combine series circuits and parallel circuits. These are known as combination circuits. No new equations will be learned here.

We can imagine a branch in a parallel circuit, but which contains two resistors in series. For example, between points \(A\) and \(B\) in Figure 1.

In this situation, we could calculate the equivalent resistance of branch \(AB\) using our rules for series circuits. So,

\(R_{AB} = R_1 +R_2\)

Now, we can replace the two resistors with a single, equivalent resistor with no effective change to the circuit.

As can be seen in Figure 2, the circuit is now a parallel circuit, with resistors \(R_{AB}\) and \(R_3\) in parallel. This circuit can be solved using the same rules as any other parallel circuit. (See Resistors in Parallel)

Another combination circuit can occur with parallel circuits connected in series. Figure 3 shows a typical example of two parallel circuits (\(AB\) and \(CD\)) connected in series with another resistor, \(R_3\).

Here, the resistors in the parallel circuit \(AB\) can be replaced by an equivalent resistance. Again, we will use the equivalence rule for resistors connected in parallel:

\(\frac{1}{R_ \mathrm {equivalent}} = \sum \frac {1}{R_i}\)

This gives:

\(\frac{1}{R_{AB}} = \frac {1}{R_1} + \frac {1}{R_2}\)

So, the equivalent resistance between points \(A\) and \(B\) is \(R_{AB}\). Replacing the parallel circuit between these two points with \(R_{AB}\) gives the following circuit.

Similarly, we can replace the parallel circuit containing \(R_4\) and \(R_5\) (between points \(C\) and \(D\)) with its equivalent resistance, \(R_{CD}\), where

\(\frac {1}{R_{CD}} = \frac{1}{R_4} + \frac {1}{R_5}\)

Replacing the parallel circuit between \(CD\) with its equivalent resistance yields the circuit in Figure 5 (above).

Now, you can see that we have simplified Circuit 2 to one which contains resistors connected in series only. That is, this circuit now contains \(R_{AB}\), \(R_3\), and \(R_{CD}\) in series. The equivalent resistance for this circuit would be found using:

\(R_\mathrm{equivalent} = \sum R_i\)

or

\(R_ \mathrm {total} = R_{AB} + R_3 + R_{CD}\)

There are more complicated circuits which cannot be reduced to simply a parallel or series circuit using equivalent resistances. Instead, these need to be solved using to concepts: Kirchhoff's Current Law, and Kirchhoff's Voltage Law. Such complicated circuits will not be dealt with in this course, but are available in this tutorial.

## Kirchhoff's Current Law

This fundamental law results from the conservation of charge. It applies to a junction or node in a circuit -- a point in the circuit where charge has several possible paths to travel.

In Figure 6, we see that \(I_A\) is the only current flowing into the node. However, there are three paths for current to leave the node, and these current are represented by \(I_B\), \(I_C\), and \(I_D\).

Once charge has entered into the node, it has no place to go except to leave (this is known as conservation of charge). The total charge flowing into a node must be the same as the the total charge flowing out of the node. So,

\(I_B + I_C + I_D = I_A\)

Bringing everything to the left side of the above equation, we get

\((I_B + I_C + I_D) - I_A = 0\)

Then, the sum of all the currents is zero. This can be generalized as follows

Note the convention we have chosen here: current flowing into the node are taken to be negative, and currents flowing *out of* the node are positive. It should not really matter which you choose to be the positive or negative current, as long as you stay consistent. However, it may be a good idea to find out the convention used in your class.

## Kirchhoff's Voltage Law

Kirchhoff's Voltage Law (or Kirchhoff's Loop Rule) is a result of the electrostatic field being conservative. It states that the total voltage around a closed loop must be zero. If this were not the case, then when we travel around a closed loop, the voltages would be indefinite. So

In Figure 7 the total voltage around loop 1 should sum to zero, as does the total voltage in loop 2. Furthermore, the loop which consists of the outer part of the circuit (the path ABCD) should also sum to zero.

We can adopt the convention that potential gains (i.e. going from lower to higher potential, such as with an emf source) is taken to be positive. Potential losses (such as across a resistor) will then be negative. However, as long as you are consistent in doing your problems, you should be able to choose whichever convention you like. It is a good idea to adopt the convention used in your class.