Chapter 6
Circuits

6.1.1  Kirchoff's Rules

We can start applying what we know about potential differences and currents to analyzing all kinds of circuits. First, let's review specific examples of how to handle particular electrical devices and how they are symbolized.

Device	Symbol	Direction of Positive Potential Change	Magnitude of Potential Change
emf source			Vb
capacitance			VC = q/C
resistance			VR = i*R

The blue arrow in the table above indicates the direction of a path across the device that would see a positive potential difference. Note that resistors require energy from an EMF source in order to get current to flow through them, so any path across a resistor that is in the direction of the current sees a negative change in potential.

Kirchoff's Rules apply to circuit elements and circuit junctions. We assume that a circuit is formed by having perfectly conducting lines connecting circuit elements like resistors, capacitors, or EMF sources. The first of these rules is reflected in the table above. This gives the amount of potential difference to apply for each circuit element when traced in the direction for positive potential difference.

The next rule defines what happens at circuit junctions. These are places where conducting lines meet. Since current is conserved, Kirchoff's junction rule states that the net current into a junction must equal the net current out of the junction. In the example below, we must have i₁ = i₂ + i₃

Figure 6.2: Current must be conserved at any junction.

6.1.2  Resistors in Series

With what has been defined, you can already do analysis of a number of circuits, some of them quite complex. To ease the burden of computation though, we can note some short cuts. The first is that if two resistors are arranged such that all the current through one resistor goes through the second, then we can replace this series combination with an equivalent resistance that changes nothing in terms of what the EMF source must do to provide current through the two original resistors. That is to say, the equivalent resistance will have the same potential difference across it as the two original resistors and will have the same current going through it.

To find what the value of this equivalent resistance must be, let's carry through the analysis using Kirchoff's rules. Consider the circuit below.

Figure 6.3: Resistors R₁ and R₂ connected in series.

The path we choose for evaluating potential differences is shown in magenta. We can start anywhere in the circuit, but let's start at the bottom right corner and go clockwise around the circuit until we reach the same point again. We will write the potential difference of the EMF source as V_b. The path shown yields, according to the rules above,

Vb - iR1 - iR2 = 0

(6.1.2.1)

Note that the current through the EMF source must be the same as through each of the resistors because there are no junctions and since all the current into each device must equal the current going out, there is no choice but to have all the currents for each circuit element be identical. To get to an equivalent circuit like the one below, we clearly want R_eq = R₁ + R₂.

Figure 6.4: The equivalent circuit to that shown in figure 6.3.

This result can be extended to any number of connected resistors. As long as all the current through each of N connected resistors goes through every resistor, the equivalent resistance can be written as

Req = R1 + R2 + R3 + ... + RN

(6.1.2.2)

6.1.3  Resistors in Parallel

If two resistors are connected together and arranged in a circuit so that the potential difference across them is the same, then those resistors are in parallel. Such resistor pairings can also be replaced by an equivalent resistance.

Figure 6.5: Two resistors in parallel in a circuit.

Following the clockwise loop around from the bottom left hand corner through the part of circuit containing R₁ and again through the part of the circuit containing R₂ (in both cases returning eventually to the bottom left hand corner) in this case gives

Vb - i1 = 0 Vb - i2R2 = 0 i = i1 + i2 (6.1.3.3)

Where the last equation comes from the junction rule applied to the point where i splits into i₁ and i₂. The equivalent resistance circuit appears below.

Figure 6.6: The equivalent circuit for the one in figure 6.5.

Note that the loop equation here is

Vb - iReq = 0

(6.1.3.4)

These equations can be related as follows

i = i1 + i2 Vb Req = Vb R1 + Vb R2 Þ 1 Req = 1 R1 + 1 R2 (6.1.3.5)

Again, this result can be extended to any number of resistors connected in parallel. For N such resistors, we would have

1 Req = 1 R1 + 1 R2 + 1 R3 + ... + 1 RN

(6.1.3.6)

With this being said, we can now go on to apply some of these ideas to typical problems of this genre.

Problem:

Find the equivalent resistance and the current through each resistor for the circuit in figure .

Figure 6.7: Four resistors in a circuit.

Solution:

First, let's note that conducting lines in circuits can be stretched or shrunk as long as no new connections between circuit elements are made and none of the existing connections are broken. So, we can perform the shifts as shown in figure 6.7. We see that resistors 2, 3, and 4 are in parallel, so

R243 = R2R4R3 R2R3 + R4R3 + R2R4

(6.1.3.7)

and R_eq = R₁ + R₂₄₃. In general, if no values are given, we assume that the potential provided by the EMF source and the values of the resistances are known. In this case, we are done as far as finding the equivalent resistance.

To find the current through each resistor, note that all the current through the battery goes also through R₁, so

i1 = Vb Req

(6.1.3.8)

For R₃ we have to use a loop equation to find the current. For a clockwise loop starting at the battery,

Vb - i1R1 - i2R2 = 0 Þ Vb - i1R1 = i2R2 Þ i2 = Vb - i1R1 R2 . (6.1.3.9)

Since all the values on the right hand side are assumed to be known or already calculated, we have solved for the current through resistor 2. A similar approach for resistor 4 gives

i4 = Vb - i1R1 R4

(6.1.3.10)

To get the current through resistor 3, use the junction rule

i3 = i1 - i2 - i4

(6.1.3.11)

You should verify that this last result is identical to what you get by going through the loop equation for resistor 3.

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Problem:

Find the current through each resistor in figure  and the potential difference between points a  and b. Assume that R₁ = 1 W, R₂ = 2 W, e₁ = 2 volts, and e₂ = e₃ = 4 volts.

Figure 6.8: Three batteries and five resistors in a circuit.

Solution:

We have the assumed directions for the currents as shown below. We note that these directions may not be correct (in fact they are not), but that simply does not matter.

Figure 6.9: Three batteries and five resistors in a circuit.

Let's start by using the loop equations tracing two clockwise paths from point b to point b.

loop1: -i1R1 + e1 - i1R1 -i2R2 - e2 = 0 loop2: e2 + i2R2 - i3R1 -e3 - i3R1 = 0 current: i2 + i3 = i1 (6.1.3.12)

Now we can take a number of approaches to solving for the currents. One is to go through the algebra from the above equations. Another is to take note that we can consider a loop over just the outside of the circuit. Starting from point b and going clockwise, this yields

-i1R1 + e1 - i1R1 - i3R1 -e3 - i3R1 = 0 Þ 2i3R1 = e1 - e3 - 2i1R1 Þ (6.1.3.13) i3 = e1 - e3 - 2i1R1 2R1 (6.1.3.14)

Similarly, we can solve the loop 1 equation from above to get

i2 = e1 - e2 - 2i1R1 R2

(6.1.3.15)

Now substitute into our current equation to get

i2 + i3 = i1 e1 - e2 - 2i1R1 R2 + e1 - e3 - 2i1R1 2R1 = i1 Þ e1 - e2 R2 + e1 - e3 2R1 = i1 + 2 i1R1 R2 + i1 (2 volts - 4 volts) 2 ohms + (2 volts - 4 volts) 2 ohms = 2i1 + 2i1 1 ohm 2 ohms -2 A = 3i1 Þ i1 = - 2 3  A (6.1.3.16)

Now we can find the other currents directly

i2 = (2 volts - 4 volts - 2(-2/3 A)(1 ohm)) 2 ohms = - 1 3  A i3 = i1 - i2 = - 2 3  A - - 1 3  A = - 1 3  A (6.1.3.17)

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