30 Resistors in Series and Parallel

When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then R₁ in Figure 1(a) could be the resistance of the screwdriver’s shaft, R₂ the resistance of its handle, R₃ the person’s body resistance, and R₄ the resistance of her shoes. Figure 2 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubber-soled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.)

To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 2. According to Ohm’s law, the voltage drop, V, across a resistor when a current flows through it is calculated using the equation V = IR, where I equals the current in amps (A) and R is the resistance in ohms (Ω). Another way to think of this is that V is the voltage necessary to make a current I flow through a resistance R. So the voltage drop across R₁ is V₁ = IR₁, that across R₂ is V₂ = IR₂, and that across R₃ is V₃ = IR₃. The sum of these voltages equals the voltage output of the source; that is,

V = V₁ + V₂ + V₃.

This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation PE = qV, where q is the electric charge and V is the voltage. Thus the energy supplied by the source is qV, while that dissipated by the resistors is

qV₁ + qV₂ + qV₃.

Figure 3 shows resistors in parallel, wired to a voltage source. Resistors are in parallel when each resistor is connected directly to the voltage source by connecting wires having negligible resistance. Each resistor thus has the full voltage of the source applied to it. Each resistor draws the same current it would if it alone were connected to the voltage source (provided the voltage source is not overloaded). For example, an automobile’s headlights, radio, and so on, are wired in parallel, so that they utilize the full voltage of the source and can operate completely independently. The same is true in your house, or any building. (See Figure 3(b).)

To find an expression for the equivalent parallel resistance R_p, let us consider the currents that flow and how they are related to resistance. Since each resistor in the circuit has the full voltage, the currents flowing through the individual resistors are [latex]{I}_{1}=\frac{V}{{R}_{1}}\\[/latex], [latex]{I}_{2}=\frac{V}{{R}_{2}}\\[/latex], and [latex]{I}_{3}=\frac{V}{{R}_{3}}\\[/latex]. Conservation of charge implies that the total current I produced by the source is the sum of these currents:

Substituting the expressions for the individual currents gives

[latex]I=\frac{V}{{R}_{1}}+\frac{V}{{R}_{2}}+\frac{V}{{R}_{3}}=V\left(\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}\right)\\[/latex].

Note that Ohm’s law for the equivalent single resistance gives

[latex]I=\frac{V}{{R}_{p}}=V\left(\frac{1}{{R}_{p}}\right)\\[/latex].

The terms inside the parentheses in the last two equations must be equal. Generalizing to any number of resistors, the total resistance R_p of a parallel connection is related to the individual resistances by

[latex]\frac{1}{{R}_{p}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{\text{.}3}}+\text{.}\text{...}\\[/latex]

This relationship results in a total resistance R_p that is less than the smallest of the individual resistances. (This is seen in the next example.) When resistors are connected in parallel, more current flows from the source than would flow for any of them individually, and so the total resistance is lower.

More complex connections of resistors are sometimes just combinations of series and parallel. These are commonly encountered, especially when wire resistance is considered. In that case, wire resistance is in series with other resistances that are in parallel. Combinations of series and parallel can be reduced to a single equivalent resistance using the technique illustrated in Figure 4. Various parts are identified as either series or parallel, reduced to their equivalents, and further reduced until a single resistance is left. The process is more time consuming than difficult.

The simplest combination of series and parallel resistance, shown in Figure 4, is also the most instructive, since it is found in many applications. For example, R₁ could be the resistance of wires from a car battery to its electrical devices, which are in parallel. R₂ and R₃ could be the starter motor and a passenger compartment light. We have previously assumed that wire resistance is negligible, but, when it is not, it has important effects, as the next example indicates.

Example 3. Calculating Resistance, IR Drop, Current, and Power Dissipation: Combining Series and Parallel Circuits

Figure 5 shows the resistors from the previous two examples wired in a different way—a combination of series and parallel. We can consider R₁ to be the resistance of wires leading to R₂ and R₃. (a) Find the total resistance. (b) What is the IR drop in R₁? (c) Find the current I₂ through R₂. (d) What power is dissipated by R₂?

Strategy and Solution for (a)

To find the total resistance, we note that R₂ and R₃ are in parallel and their combination R_p is in series with R₁. Thus the total (equivalent) resistance of this combination is

R_tot = R₁ + R_p.

First, we find R_p using the equation for resistors in parallel and entering known values:

[latex]\frac{1}{{R}_{\text{p}}}=\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}=\frac{1}{6\text{.}\text{00}\text{ }\Omega }+\frac{1}{\text{13}\text{.}0\text{ }\Omega }=\frac{0.2436}{\text{ }\Omega}\\[/latex].

Inverting gives

[latex]{R}_{\text{p}}=\frac{1}{0.2436}\text{ }\Omega =4.11\text{ }\Omega\\[/latex].

So the total resistance is

R_tot = R₁ + R_p = 1.00 Ω + 4.11 Ω = 5.11 Ω.

Discussion for (a)

The total resistance of this combination is intermediate between the pure series and pure parallel values (20.0 Ω and 0.804 Ω, respectively) found for the same resistors in the two previous examples.

Strategy and Solution for (b)

To find the IR drop in R₁, we note that the full current I flows through R₁. Thus its IR drop is

V₁ = IR₁

We must find I before we can calculate V₁. The total current I is found using Ohm’s law for the circuit. That is,

[latex]I=\frac{V}{{R}_{\text{tot}}}=\frac{\text{12.0}\text{ V}}{5.11\text{ }\Omega}=2.35\text{ A}\\[/latex].

Entering this into the expression above, we get

V₁ = IR₁ = (2.35 A)(1.00 Ω) = 2.35 V.

Discussion for (b)

The voltage applied to R₂ and R₃ is less than the total voltage by an amount V₁. When wire resistance is large, it can significantly affect the operation of the devices represented by R₂ and R₃.

Strategy and Solution for (c)

To find the current through R₂, we must first find the voltage applied to it. We call this voltage V_p, because it is applied to a parallel combination of resistors. The voltage applied to both R₂ and R₃ is reduced by the amount V₁, and so it is

V_p = V − V₁ = 12.0 V − 2.35 V = 9.65 V.

Now the current I₂ through resistance R₂ is found using Ohm’s law:

[latex]{I}_{2}=\frac{{V}_{\text{p}}}{{R}_{2}}=\frac{9.65\text{ V}}{6.00\text{ }\Omega}=1.61\text{A}\\[/latex].

Discussion for (c)

The current is less than the 2.00 A that flowed through R₂ when it was connected in parallel to the battery in the previous parallel circuit example.

Strategy and Solution for (d)

The power dissipated by R₂ is given by

P₂ = (I₂)²R₂ = (1.61 A)²(6.00 Ω) = 15.5 W

Discussion for (d)

The power is less than the 24.0 W this resistor dissipated when connected in parallel to the 12.0-V source.

One implication of this last example is that resistance in wires reduces the current and power delivered to a resistor. If wire resistance is relatively large, as in a worn (or a very long) extension cord, then this loss can be significant. If a large current is drawn, the IR drop in the wires can also be significant.

For example, when you are rummaging in the refrigerator and the motor comes on, the refrigerator light dims momentarily. Similarly, you can see the passenger compartment light dim when you start the engine of your car (although this may be due to resistance inside the battery itself).

What is happening in these high-current situations is illustrated in Figure 6. The device represented by R₃ has a very low resistance, and so when it is switched on, a large current flows. This increased current causes a larger IR drop in the wires represented by R₁, reducing the voltage across the light bulb (which is R₂), which then dims noticeably.

Glossary

series:

a sequence of resistors or other components wired into a circuit one after the other

resistor:

a component that provides resistance to the current flowing through an electrical circuit

resistance:

causing a loss of electrical power in a circuit

Ohm’s law:

the relationship between current, voltage, and resistance within an electrical circuit: V = IR

voltage:

the electrical potential energy per unit charge; electric pressure created by a power source, such as a battery

voltage drop:

the loss of electrical power as a current travels through a resistor, wire or other component

current:

the flow of charge through an electric circuit past a given point of measurement

Joule’s law:

the relationship between potential electrical power, voltage, and resistance in an electrical circuit, given by: [latex]{P}_{e}=\text{IV}[/latex]

parallel:

the wiring of resistors or other components in an electrical circuit such that each component receives an equal voltage from the power source; often pictured in a ladder-shaped diagram, with each component on a rung of the ladder