Section 21-5 presents Kirchhoff's Rules but also says these are simply applications of two conservation laws. It's important to understand where the conservation laws come in and how to apply them.

Series Circuits

Consider a circuit of 2 resistors of resistances R₁ and R₂ and a battery in a single loop as shown in Figure 1.  The current in the circuit is I (that's conventional, positive current) and is the same at all points.  We can say the latter, because electric charge is conserved.  All the charge that passes any point of the circuit in a given time interval must pass any other point in the same time interval.  If that didn't occur, charge would be lost or gained along the way.  For the single loop circuit, there's nowhere else for the charge to go.  If for some reason, the current changes--for example, this could be due to a charging or discharging capacitor--the change is the same in all parts of the circuit.

Figure 1	Figure 2

If we traverse the circuit counterclockwise in the direction of the current, then we encounter a potential rise across the battery and potential drops across the resistors.  We've denoted these changes in Figure 2 as DV_b, DV₁, and DV₂.  (The textbook uses the script upper-case symbol for the potential difference across the battery and calls this potential difference the emf.  That's certainly an acceptable practice.  However, any real battery has internal resistance, and the potential difference across its terminals is less than the emf and is equal to , where r is the internal resistance.)  The change in electrical potential energy of any particular amount of charge q passing through the battery is DE_b = q DV_b.  Note that this is a positive energy change or increase.  The change in electrical potential energy of charge q in the resistors is DE₁ + DE₂ = q(DV₁ + DV₂). This is a negative change or decrease. Making our usual assumption that the energy loss in the wires is negligible, conservation of energy applied to the circuit tells us that DE_b + DE₁ + DE₂ = 0.  Since q divides out from each term, we can also write:

DV_b + DV₁ + DV₂ = 0.

This is the loop rule.  We can state it more compactly as , where the summation is taken around the complete circuit. 

We can take this a step further to come up with the relationship for the equivalent resistance of resistors in series.  We'll use the relationship DV = -IR.  (You may at first find the negative sign confusing, since the textbook writes V = IR.  However, the textbook uses the conventional practice that V is measured in such a way that it is always positive. This is what you would get if you touched the positive probe of the multimeter to the higher potential side of the resistor. It's important to realize, however, that V = -DV for a resistor.  That's because the change in potential, DV, of a resistor is negative.)  Substituting,

V_b - IR₁ - IR₂ = 0.

V_b/I = R₁ + R₂.

The left-hand side, V_b/I, can be thought of as the resistance of a circuit with a single resistor whose resistance, R_eq, replaces that of R₁ and R₂ together.  The equivalent circuit is shown to the right.  Thus, R_eq = R₁ + R₂.

Parallel Circuits

Now let's apply conservation of energy and charge to a parallel circuit.  The circuit is shown to the left.  The current I splits at the junction point into two parts, I₁ and I₂.  Applying conservation of charge, q = q₁ + q₂.  All charge going into the junction in a particular time must leave the junction.  Dividing all terms by the same Dt allows us to apply the definition of current and write I = I₁ + I₂. This application of conservation of charge to a circuit is called the junction rule.

Junction rule:  The sum of currents going into a circuit junction is equal to the sum of currents going out of the junction.

Conservation of energy applies as well.  We can apply it to each loop of the circuit individually.  Note that there are actually 3 loops.  These are:

Loop 1:  including the battery and R₁
Loop 2:  including the battery and R₂
Loop 3:  including R₁ and R₂

Let's apply the loop rule to loops 1 and 2 and traverse the loops counterclockwise.

Loop 1:  DV_b + DV₁ = 0.
Loop 2:  DV_b + DV₂ = 0.

We see from this that DV_b = -DV₁ = -DV₂ .  Thus, the voltage rise across the battery is equal to the voltage drop across each resistor.  For completeness, let's look at loop 3.  If we traverse the loop counterclockwise, we have -DV₁ + DV₂ = 0.  Note the negative sign.  This is necessary, because we're traversing R₁ in a direction opposite to the current.  We then have  DV₁ = DV₂.  This, of course, is consistent with the result found from examining loops 1 and 2.

Let's apply these results to find the equivalent resistance of resistors in parallel.  We start with the junction rule:  I = I₁ + I₂.  Using DV = -IR or I = -DV/R,

DV_b/R_eq = -DV₁/R₁ - DV₂/R₂,

where the resistance we use on the left-hand side is that of the equivalent resistance that would replace the parallel combination of R₁ and R₂.  Now using the relationship found earlier, namely, DV_b = -DV₁ = -DV₂,

 1/R_eq = 1/R₁ + 1/R₂.

Summary

Let's summarize now what we've found.

1. For all circuits, the loop rule and junction rule apply, since these are generally valid conservation laws.
2. For series circuits, the current is the same in all parts.  (Note that this is not the same as saying the current is constant.  Constant implies no change in time.  The current can be changing in all parts of the circuit at the same time.  This is what happens, for example, when capacitors charge and discharge.)
3. For series circuits, the potential rise across the battery is equal to the sum of the potential drops across the resistors.
4. For parallel circuits, the potential rise across the battery is equal to the potential drop across each branch of the circuit.
5. For parallel circuits, the current from the battery is the algebraic sum of the currents in the branches.  (We specify algebraic sum in case we select current directions in such a way that one or more are negative.)
6. Finally, we have the formulas for calculating equivalent resistance of series and parallel circuits.

Energy and Power in Series and Parallel Circuits

Let's take this discussion a bit further and look at energy produced in the battery and then converted to thermal and radiant energy (for bulbs) in the resistors.  Considering once again the series circuit of 2 resistors (shown to the right for convenience),  DE_b + DE₁ + DE₂ = 0 or DE_b = -(DE₁ + DE₂).  The minus sign is needed, because energy is gained by charge in the battery and lost in the resistors.  In a given time interval, Dt,

DE_b/Dt = -(DE₁ + DE₂)/Dt.

DE/Dt is just power.  We're going to write a power equation next, but we have to warn you first that we're going to write it as a magnitude equation.  Once again, that's the conventional way to do it.  We realize that the subject of circuits has the potential to trip you up with signs (currents that we say go one way when they really go the other way, V's used to represent potential differences, etc).  It's unfortunate that the subject of circuits developed that way, but we have to live with it now. Here's the power equation.

P_b = P₁ + P₂. 

We can now apply the textbook formulas, P = VI, P = I²R, and P = V²/R.  However, we must be very careful to identify the V's clearly by indicating with subscripts across what part of the circuit the V is measured.  One possibility is to write:

V_bI = I²R₁ + I²R₂.

We use the same I in all terms, because the current is the same in all parts of a series circuit.  Equivalently, we could write:

V_bI = V²/R₁ + V²/R₂.

Now consider energy in the parallel circuit (again shown to the left for convenience).  Let's approach this in a different way.  Starting with the junction rule:

I = I₁ + I₂.

Let's rewrite this as:  V_b/R = V₁/R₁ + V₂/R₂.   Now P = V²/R, so P/V = V/R.  Making that substitution for all terms:

P_b/V_b = P₁/V₁ + P₂/V₂.

Since this is a parallel circuit, we know that V_b = V₁ = V₂.  Therefore, the denominators divide out, leaving:  P_b = P₁ + P₂.  Perhaps you're surprised that this is the same result as for the series circuit.  However, it makes sense from the standpoint of energy conservation.  All the energy that's produced in the battery is converted to other forms in the resistors.  That's independent of how the resistors are wired into the circuit.  However, the way that the energy is distributed between the resistors does depend on whether they're in series or parallel.  This will be investigated later in a video exercise on series and parallel light bulbs.

A Misconception to Be Aware Of

A common misconception is that an electrical energy source always provides the same amount of current and energy no matter what the circuit is.  In fact, the current from the energy source increases or decreases depending on the equivalent resistance of the circuit.  Consider the following three cases for two resistors of identical resistance, R.  The potential difference across the battery is V_b in every case.  However, the total currents, I, will be different, because the equivalent resistance of the circuit is different for each case.

Case 1	Case 2	Case 3

Case 1.  The equivalent resistance is simply R_eq = R.  Thus, I = V_b/R_eq = V_b/R.  The power dissipated by the resistor is P = I²R_eq = V_b²/R.

Case 2.  The equivalent resistance is R_eq = 2R.  Thus, I = V_b/R_eq = V_b/2R = ½(V_b/R).  The total current is half as much as for Case 1.  The total power dissipated by both resistors together is P = I²R_eq = [½(V_b/R)]²(2R) = ½V_b²/R. So that's half as much as for Case 1.  (Note that each resistor alone dissipates a quarter as much power as the resistor in Case 1.)

Case 3.  The equivalent resistance is R_eq = R/2.  Thus, I = V_b/R_eq = V_b/(R/2) = 2(V_b/R).  The total current is twice as much as for Case 1.  The total power dissipated by both resistors together is P = I²R_eq = [2(V_b/R)]²(R/2) = 2V_b²/R. So that's twice as much as for Case 1.  (Note that each resistor alone dissipates the same power as the resistor in Case 1.)