How Do You Get the Current if You Know Resistance in a Series Cuircit

Learning Objectives

Past the end of this section, y'all volition be able to:

• Draw a circuit with resistors in parallel and in serial.
• Calculate the voltage drop of a current across a resistor using Ohm's law.
• Contrast the way total resistance is calculated for resistors in series and in parallel.
• Explain why full resistance of a parallel excursion is less than the smallest resistance of any of the resistors in that circuit.
• Calculate total resistance of a circuit that contains a mixture of resistors connected in serial and in parallel.

About circuits have more than one component, chosen a resistor that limits the catamenia of accuse in the circuit. A measure of this limit on accuse flow is called resistance. The simplest combinations of resistors are the series and parallel connections illustrated in Figure 1. The total resistance of a combination of resistors depends on both their private values and how they are connected.

Effigy 1. (a) A series connection of resistors. (b) A parallel connection of resistors.

Resistors in Serial

When are resistors in series? Resistors are in serial whenever the flow of charge, called the current, must menstruum through devices sequentially. For example, if electric current flows through a person holding a screwdriver and into the Earth, then R ₁ in Figure 1(a) could be the resistance of the screwdriver's shaft, R _two the resistance of its handle, R ₃ the person'due south trunk resistance, and R ₄ the resistance of her shoes. Figure two 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 electric current has to laissez passer through each resistor in sequence. (This fact would be an reward to a person wishing to avoid an electric shock, who could reduce the electric current past wearing high-resistance prophylactic-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.)

Figure 2. Three resistors connected in serial to a bombardment (left) and the equivalent single or series resistance (correct).

To verify that resistances in serial do indeed add, let u.s. consider the loss of electric power, called a voltage drop, in each resistor in Effigy ii. Co-ordinate to Ohm'southward police, the voltage drop, Five, across a resistor when a electric current flows through it is calculated using the equation Five = IR, where I equals the current in amps (A) and R is the resistance in ohms (Ω). Some other mode to think of this is that V is the voltage necessary to brand a electric current I flow through a resistance R. So the voltage drop across R _ane is V _one=IR ₁, that across R ₂ is 5 ₂=IR ₂, and that across R _three is V ₃=IR ₃. The sum of these voltages equals the voltage output of the source; that is,

Five=V _one+V ₂+5 ₃.

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

qV ₁+ qV ₂+ qV ₃.

Making Connections: Conservation Laws

The derivations of the expressions for serial and parallel resistance are based on the laws of conservation of energy and conservation of charge, which land that total charge and full energy are constant in any process. These ii laws are directly involved in all electrical phenomena and volition be invoked repeatedly to explain both specific effects and the general behavior of electricity.

These energies must be equal, because in that location is no other source and no other destination for energy in the circuit. Thus, qV=qV ₁+qV ₂+qV _three. The accuse q cancels, yielding 5=Five ₁+5 ₂+V ₃, as stated. (Note that the aforementioned amount of accuse passes through the bombardment and each resistor in a given corporeality of time, since there is no capacitance to store accuse, there is no place for charge to leak, and charge is conserved.) Now substituting the values for the private voltages gives

V= IR _one+ IR _two + IR _iii =I(R ₁+R _ii+R ₃).

Note that for the equivalent single series resistance R _s, we have

V = IR _s.

This implies that the total or equivalent series resistance R _{due south} of three resistors is R _s=R ₁+R ₂+R ₃. This logic is valid in general for any number of resistors in series; thus, the total resistance R _s of a serial connection is

R _s=R ₁+R ₂+R ₃+…,

as proposed. Since all of the current must pass through each resistor, it experiences the resistance of each, and resistances in series simply add together up.

Example 1. Computing Resistance, Current, Voltage Driblet, and Power Dissipation: Analysis of a Series Circuit

Suppose the voltage output of the battery in Effigy 2 is 12.0 Five, and the resistances are R ₁= 1.00 Ω, R _two= 6.00 Ω, and R ₃= 13.0 Ω. (a) What is the total resistance? (b) Find the current. (c) Calculate the voltage drop in each resistor, and testify these add together to equal the voltage output of the source. (d) Calculate the power dissipated by each resistor. (e) Detect the power output of the source, and show that it equals the total ability dissipated by the resistors.

Strategy and Solution for (a)

The full resistance is simply the sum of the individual resistances, equally given by this equation:

[latex]\begin{array}{lll}{R}_{\text{due south}}& =& {R}_{1}+{R}_{2}+{R}_{3}\\ & =& i.00\text{ }\Omega + half dozen.00\text{ }\Omega + thirteen.0\text{ }\Omega\\ & =& 20.0\text{ }\Omega\end{assortment}\\[/latex].

Strategy and Solution for (b)

The current is found using Ohm'southward constabulary, 5 = IR. Entering the value of the practical voltage and the total resistance yields the current for the circuit:

[latex]I=\frac{V}{{R}_{\text{s}}}=\frac{12.0\text{ V}}{xx.0\text{ }\Omega}=0.60\text{ A}\\[/latex].

Strategy and Solution for (c)

The voltage—or IR drib—in a resistor is given by Ohm'due south law. Inbound the current and the value of the get-go resistance yields

5 ₁ = IR ₁ = (0.600A)(ane.0 Ω) = 0.600 V.

Similarly,

V ₂ = IR _two = (0.600A)(6.0 Ω) = 3.60 V

and

V3 = IR _iii = (0.600A)(13.0 Ω) = 7.80 V.

Discussion for (c)

The three IR drops add together to 12.0 Five, as predicted:

V _one + V ₂ + V ₃ = (0.600 + 3.threescore + 7.80)Five = 12.0 V.

Strategy and Solution for (d)

The easiest fashion to summate ability in watts (W) dissipated by a resistor in a DC excursion is to use Joule'southward law, P = IV, where P is electric power. In this case, each resistor has the same total current flowing through information technology. By substituting Ohm's police force V = IR into Joule's police force, we get the ability dissipated by the first resistor equally

P ₁ = I ² R ₁ = ( 0 . 600 A ) ² ( 1 . 00 Ω ) = 0 . 360 Due west .

Similarly,

P _ii = I ^two R ₂ = ( 0 . 600 A ) ² (six . 00 Ω ) = 2.16 Westward .

and

P _three = I ⁱⁱ R ₃ = ( 0 . 600 A ) ² (13.0 Ω ) = 4.68 Westward .

Give-and-take for (d)

Power can also be calculated using either P = Four or [latex]P=\frac{{V}^{2}}{R}\\[/latex], where V is the voltage drop beyond the resistor (not the total voltage of the source). The same values volition be obtained.

Strategy and Solution for (e)

The easiest way to calculate power output of the source is to use P = IV, where Five is the source voltage. This gives

P = (0.600 A)(12.0 5) = vii.xx W.

Discussion for (e)

Notation, coincidentally, that the total power prodigal past the resistors is also vii.twenty West, the aforementioned equally the power put out by the source. That is,

P _ane + P ₂ + P _three = (0.360 + 2.xvi + 4.68) W = 7.xx W.

Power is energy per unit time (watts), and then conservation of energy requires the ability output of the source to exist equal to the total power dissipated past the resistors.

Major Features of Resistors in Serial

1. Series resistances add together: R _{due south}=R ₁+R _ii+R ₃+….
2. The same current flows through each resistor in series.
3. Individual resistors in serial practise not go the total source voltage, but divide it.

Resistors in Parallel

Figure three 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 lonely 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 tin operate completely independently. The same is true in your house, or any edifice. (Meet Effigy three(b).)

Figure 3. (a) Three resistors connected in parallel to a battery and the equivalent single or parallel resistance. (b) Electric ability setup in a house. (credit: Dmitry Thou, Wikimedia Eatables)

To observe an expression for the equivalent parallel resistance R _p, let u.s.a. consider the currents that menses and how they are related to resistance. Since each resistor in the excursion has the full voltage, the currents flowing through the individual resistors are [latex]{I}_{ane}=\frac{5}{{R}_{1}}\\[/latex], [latex]{I}_{2}=\frac{V}{{R}_{ii}}\\[/latex], and [latex]{I}_{3}=\frac{Five}{{R}_{iii}}\\[/latex]. Conservation of charge implies that the full current I produced by the source is the sum of these currents:

I=I ₁+I ₂+I _iii.

Substituting the expressions for the individual currents gives

[latex]I=\frac{V}{{R}_{one}}+\frac{V}{{R}_{two}}+\frac{5}{{R}_{iii}}=V\left(\frac{1}{{R}_{i}}+\frac{1}{{R}_{two}}+\frac{1}{{R}_{3}}\right)\\[/latex].

Notation that Ohm's police force for the equivalent single resistance gives

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

The terms within the parentheses in the final two equations must be equal. Generalizing to any number of resistors, the total resistance R _p of a parallel connexion is related to the individual resistances past

[latex]\frac{1}{{R}_{p}}=\frac{i}{{R}_{ane}}+\frac{i}{{R}_{ii}}+\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 adjacent example.) When resistors are connected in parallel, more than current flows from the source than would menstruum for any of them individually, and and then the full resistance is lower.

Case ii. Calculating Resistance, Current, Power Dissipation, and Power Output: Analysis of a Parallel Circuit

Let the voltage output of the bombardment and resistances in the parallel connection in Figure 3 exist the same as the previously considered series connection: Five = 12.0 5, R ₁= 1.00 Ω, R _two= 6.00 Ω, and R ₃= 13.0 Ω. (a) What is the total resistance? (b) Find the total electric current. (c) Calculate the currents in each resistor, and show these add together to equal the full electric current output of the source. (d) Summate the ability prodigal by each resistor. (e) Find the power output of the source, and prove that information technology equals the total ability dissipated by the resistors.

Strategy and Solution for (a)

The total resistance for a parallel combination of resistors is institute using the equation below. Entering known values gives

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

Thus,

[latex]\frac{1}{{R}_{p}}=\frac{ane.00}{\text{ }\Omega }+\frac{0\text{.}\text{1667}}{\text{ }\Omega }+\frac{0\text{.}\text{07692}}{\text{ }\Omega }=\frac{ane\text{.}\text{2436}}{\text{ }\Omega }\\[/latex].

(Note that in these calculations, each intermediate answer is shown with an extra digit.) We must invert this to find the total resistance R _p. This yields

[latex]{R}_{\text{p}}=\frac{one}{i\text{.}\text{2436}}\text{ }\Omega =0\text{.}\text{8041}\text{ }\Omega\\[/latex].

The total resistance with the correct number of pregnant digits is R _p= 0.804 Ω

Discussion for (a)

R _p is, every bit predicted, less than the smallest individual resistance.

Strategy and Solution for (b)

The total current tin exist found from Ohm'southward law, substituting R _p for the total resistance. This gives

[latex]I=\frac{V}{{R}_{\text{p}}}=\frac{\text{12.0 V}}{0.8041\text{ }\Omega }=\text{fourteen}\text{.}\text{92 A}\\[/latex].

Discussion for (b)

Electric current I for each device is much larger than for the same devices connected in series (run across the previous case). A circuit with parallel connections has a smaller total resistance than the resistors connected in series.

Strategy and Solution for (c)

The individual currents are easily calculated from Ohm's police force, since each resistor gets the full voltage. Thus,

[latex]{I}_{1}=\frac{V}{{R}_{1}}=\frac{12.0\text{ V}}{1.00\text{ }\Omega}=12.0\text{ A}\\[/latex].

Similarly,

[latex]{I}_{2}=\frac{V}{{R}_{2}}=\frac{12.0\text{ V}}{6.00\text{ }\Omega}=2\text{.}\text{00}\text{ A}\\[/latex]

and

[latex]{I}_{3}=\frac{5}{{R}_{3}}=\frac{\text{12}\text{.}0\text{ 5}}{\text{13}\text{.}\text{0}\text{ }\Omega }=0\text{.}\text{92}\text{ A}\\[/latex].

Discussion for (c)

The total current is the sum of the individual currents:

I ₁+I ₂+I ₃= 14.92 A.

This is consistent with conservation of charge.

Strategy and Solution for (d)

The power dissipated by each resistor tin can be found using whatever of the equations relating ability to current, voltage, and resistance, since all three are known. Let united states use [latex]P=\frac{{V}^{2}}{R}\\[/latex], since each resistor gets full voltage. Thus,

[latex]{P}_{1}=\frac{{V}^{ii}}{{R}_{1}}=\frac{(12.0\text{ 5})^{2}}{1.00\text{ }\Omega}=144\text{ Due west}\\[/latex].

Similarly,

[latex]{P}_{ii}=\frac{{5}^{2}}{{R}_{two}}=\frac{(12.0\text{ V})^{2}}{6.00\text{ }\Omega}=24.0\text{ West}\\[/latex].

and

[latex]{P}_{iii}=\frac{{V}^{two}}{{R}_{3}}=\frac{(12.0\text{ 5})^{2}}{xiii.0\text{ }\Omega}=eleven.1\text{ W}\\[/latex].

Give-and-take for (d)

The power prodigal by each resistor is considerably higher in parallel than when connected in series to the same voltage source.

Strategy and Solution for (e)

The total power can also exist calculated in several means. Choosing P = 4, and entering the full current, yields

P = IV= (xiv.92 A)(12.0 5) = 179 W.

Discussion for (east)

Total power dissipated past the resistors is also 179 W:

P ₁+P ₂+P ₃= 144 West + 24.0 W + 11.1 W = 179 W.

This is consistent with the police force of conservation of energy.

Overall Discussion

Note that both the currents and powers in parallel connections are greater than for the same devices in series.

Major Features of Resistors in Parallel

1. Parallel resistance is found from [latex]\frac{i}{{R}_{\text{p}}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{three}}+\text{…}\\[/latex], and it is smaller than whatever private resistance in the combination.
2. Each resistor in parallel has the aforementioned full voltage of the source applied to it. (Power distribution systems nearly oftentimes utilize parallel connections to supply the myriad devices served with the same voltage and to permit them to operate independently.)
3. Parallel resistors practise not each get the total current; they divide it.

Combinations of Series and Parallel

More complex connections of resistors are sometimes just combinations of series and parallel. These are commonly encountered, peculiarly 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 unmarried resistance is left. The process is more than time consuming than difficult.

Effigy iv. This combination of seven resistors has both series and parallel parts. Each is identified and reduced to an equivalent resistance, and these are further reduced until a single equivalent resistance is reached.

The simplest combination of series and parallel resistance, shown in Effigy four, is too the most instructive, since information technology is institute in many applications. For example, R _i could exist the resistance of wires from a auto battery to its electrical devices, which are in parallel. R ₂ and R ₃ could be the starter motor and a rider compartment lite. We take previously assumed that wire resistance is negligible, simply, when it is not, it has important furnishings, as the adjacent example indicates.

Example 3. Computing Resistance, IR Drib, Current, and Ability Dissipation: Combining Series and Parallel Circuits

Figure five shows the resistors from the previous ii examples wired in a different mode—a combination of series and parallel. We can consider R ₁ to be the resistance of wires leading to R ₂ and R ₃. (a) Observe the full resistance. (b) What is the IR drib in R _one? (c) Find the electric current I ₂ through R ₂. (d) What power is dissipated by R ₂?

Effigy 5. These three resistors are connected to a voltage source so that R ₂ and R _three are in parallel with one another and that combination is in series with R _i.

Strategy and Solution for (a)

To find the full resistance, we note that R ₂ and R ₃ are in parallel and their combination R _p is in serial with R ₁. Thus the total (equivalent) resistance of this combination is

R _tot=R ₁+R _p.

Commencement, we find R _p using the equation for resistors in parallel and inbound known values:

[latex]\frac{i}{{R}_{\text{p}}}=\frac{ane}{{R}_{2}}+\frac{1}{{R}_{3}}=\frac{i}{half-dozen\text{.}\text{00}\text{ }\Omega }+\frac{i}{\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].

And so the total resistance is

R _tot= R ₁+ R _p= 1.00 Ω + iv.11 Ω = 5.11 Ω.

Discussion for (a)

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

Strategy and Solution for (b)

To find the IRdrop in R _one, nosotros annotation that the full current I flows through R _one. Thus itsIR driblet is

V ₁ = IR _i

We must observe I before we can summate V ₁. The total current I is institute using Ohm'southward police force for the circuit. That is,

[latex]I=\frac{5}{{R}_{\text{tot}}}=\frac{\text{12.0}\text{ V}}{v.11\text{ }\Omega}=2.35\text{ A}\\[/latex].

Entering this into the expression above, nosotros get

V ₁ = IR ₁ = ( 2 . 35 A ) ( i . 00 Ω ) = 2 . 35 5 .

Discussion for (b)

The voltage applied to R ₂ and R ₃ is less than the full voltage by an amount V ₁. When wire resistance is large, it can significantly touch on the operation of the devices represented by R _ii and R ₃.

Strategy and Solution for (c)

To find the current through R ₂, we must offset find the voltage applied to it. We call this voltage 5 _p, because it is applied to a parallel combination of resistors. The voltage applied to both R ₂ and R ₃ is reduced past the amount V ₁, and and so information technology is

V _p= V− V ₁= 12.0 5 − 2.35 V = 9.65 V.

Now the current I ₂ through resistance R _ii is establish using Ohm'southward law:

[latex]{I}_{2}=\frac{{V}_{\text{p}}}{{R}_{two}}=\frac{9.65\text{ Five}}{six.00\text{ }\Omega}=ane.61\text{A}\\[/latex].

Discussion for (c)

The current is less than the 2.00 A that flowed through R ₂ when it was continued in parallel to the bombardment in the previous parallel circuit case.

Strategy and Solution for (d)

The power dissipated by R _two is given past

P _two= (I ₂)^two R ₂= (one.61 A)²(6.00 Ω) = fifteen.v Due west

Discussion for (d)

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

Practical Implications

1 implication of this final example is that resistance in wires reduces the current and power delivered to a resistor. If wire resistance is relatively big, as in a worn (or a very long) extension string, then this loss tin be pregnant. If a large electric current is drawn, the IR drop in the wires can also be pregnant.

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

What is happening in these loftier-current situations is illustrated in Figure 6. The device represented past R ₃ has a very depression resistance, and and so when it is switched on, a large electric current flows. This increased electric current causes a larger IR drop in the wires represented past R ₁, reducing the voltage across the low-cal seedling (which is R _two), which then dims noticeably.

Figure 6. Why do lights dim when a large apparatus is switched on? The reply is that the large current the apparatus motor draws causes a significant drop in the wires and reduces the voltage across the light.

Check Your Agreement

Can whatsoever capricious combination of resistors be broken downward into series and parallel combinations? See if you can draw a circuit diagram of resistors that cannot exist broken downwards into combinations of series and parallel.

Solution

No, there are many ways to connect resistors that are not combinations of series and parallel, including loops and junctions. In such cases Kirchhoff'due south rules, to be introduced in Kirchhoff'south Rules, will allow you to analyze the circuit.

Problem-Solving Strategies for Series and Parallel Resistors

1. Draw a articulate excursion diagram, labeling all resistors and voltage sources. This pace includes a listing of the knowns for the problem, since they are labeled in your circuit diagram.
2. Identify exactly what needs to be adamant in the trouble (identify the unknowns). A written list is useful.
3. Determine whether resistors are in serial, parallel, or a combination of both serial and parallel. Examine the circuit diagram to make this assessment. Resistors are in series if the same current must pass sequentially through them.
4. Utilise the appropriate list of major features for series or parallel connections to solve for the unknowns. There is one listing for series and another for parallel. If your trouble has a combination of series and parallel, reduce information technology in steps by considering private groups of series or parallel connections, every bit done in this module and the examples. Special note: When finding R , the reciprocal must be taken with care.
5. Cheque to see whether the answers are reasonable and consistent. Units and numerical results must be reasonable. Total series resistance should be greater, whereas total parallel resistance should be smaller, for example. Ability should exist greater for the same devices in parallel compared with series, and so on.

Section Summary

• The total resistance of an electrical circuit with resistors wired in a series is the sum of the individual resistances:R _{due south}= R ₁+ R ₂+ R ₃+….
• Each resistor in a serial circuit has the same amount of electric current flowing through it.
• The voltage drop, or power dissipation, across each individual resistor in a series is different, and their combined total adds up to the power source input.
• The full resistance of an electrical circuit with resistors wired in parallel is less than the lowest resistance of whatsoever of the components and can be adamant using the formula:

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

• Each resistor in a parallel circuit has the same full voltage of the source applied to it.
• The current flowing through each resistor in a parallel circuit is different, depending on the resistance.
• If a more complex connexion of resistors is a combination of series and parallel, it tin can be reduced to a unmarried equivalent resistance past identifying its diverse parts equally serial or parallel, reducing each to its equivalent, and standing until a unmarried resistance is eventually reached.

Conceptual Questions

1. A switch has a variable resistance that is nearly nada when airtight and extremely large when open up, and information technology is placed in series with the device it controls. Explain the effect the switch in Figure 7 has on current when open up and when closed.

Figure 7. A switch is ordinarily in serial with a resistance and voltage source. Ideally, the switch has nearly nothing resistance when closed but has an extremely big resistance when open. (Note that in this diagram, the script Eastward represents the voltage (or electromotive force) of the battery.)

2. What is the voltage across the open switch in Effigy 7?

3. There is a voltage across an open switch, such equally in Figure vii. Why, then, is the power prodigal by the open switch small?

4. Why is the power dissipated by a closed switch, such as in Figure 7, small?

5. A student in a physics lab mistakenly wired a calorie-free bulb, bombardment, and switch as shown in Figure viii. Explain why the bulb is on when the switch is open, and off when the switch is closed. (Exercise non try this—it is hard on the battery!)

Figure 8. A wiring mistake put this switch in parallel with the device represented by [latex]R[/latex] . (Notation that in this diagram, the script Due east represents the voltage (or electromotive force) of the battery.)

six. Knowing that the severity of a shock depends on the magnitude of the electric current through your torso, would you adopt to be in serial or parallel with a resistance, such as the heating element of a toaster, if shocked past it? Explain.

7. Would your headlights dim when you lot start your car'due south engine if the wires in your motorcar were superconductors? (Exercise not neglect the battery's internal resistance.) Explain.

8. Some strings of holiday lights are wired in series to save wiring costs. An old version utilized bulbs that pause the electrical connection, like an open switch, when they fire out. If one such bulb burns out, what happens to the others? If such a string operates on 120 5 and has 40 identical bulbs, what is the normal operating voltage of each? Newer versions use bulbs that curt circuit, like a closed switch, when they burn out. If one such seedling burns out, what happens to the others? If such a string operates on 120 V and has 39 remaining identical bulbs, what is then the operating voltage of each?

ix. If 2 household lightbulbs rated 60 Westward and 100 Westward are connected in series to household power, which volition be brighter? Explain.

x. Suppose you are doing a physics lab that asks you to put a resistor into a circuit, just all the resistors supplied have a larger resistance than the requested value. How would yous connect the available resistances to attempt to become the smaller value asked for?

11. Earlier Globe War 2, some radios got power through a "resistance string" that had a significant resistance. Such a resistance cord reduces the voltage to a desired level for the radio's tubes and the similar, and it saves the expense of a transformer. Explain why resistance cords go warm and waste energy when the radio is on.

12. Some lite bulbs have iii power settings (not including zip), obtained from multiple filaments that are individually switched and wired in parallel. What is the minimum number of filaments needed for three power settings?

Problems & Exercises

Note: Data taken from figures can be assumed to be accurate to three pregnant digits.

1. (a) What is the resistance of ten 275-Ω resistors continued in serial? (b) In parallel?

2. (a) What is the resistance of a 1.00 × 10²-Ω, a 2.fifty-kΩ, and a four.00-kΩ resistor continued in series? (b) In parallel?

iii. What are the largest and smallest resistances you can obtain by connecting a 36.0-Ω, a l.0-Ω, and a 700-Ω resistor together?

4. An 1800-W toaster, a 1400-Westward electric frying pan, and a 75-W lamp are plugged into the aforementioned outlet in a fifteen-A, 120-V circuit. (The three devices are in parallel when plugged into the aforementioned socket.). (a) What current is drawn by each device? (b) Will this combination blow the 15-A fuse?

5. Your car'south 30.0-Westward headlight and 2.twoscore-kW starter are ordinarily connected in parallel in a 12.0-V system. What power would one headlight and the starter consume if connected in series to a 12.0-V battery? (Neglect any other resistance in the excursion and any change in resistance in the two devices.)

6. (a) Given a 48.0-V battery and 24.0-Ω and 96.0-Ω resistors, observe the current and ability for each when connected in series. (b) Echo when the resistances are in parallel.

seven. Referring to the example combining series and parallel circuits and Figure 5, summate I ₃ in the following two unlike ways: (a) from the known values of IandI _two ; (b) using Ohm's law for R _three. In both parts explicitly bear witness how yous follow the steps in the Problem-Solving Strategies for Serial and Parallel Resistors to a higher place.

Figure 5. These 3 resistors are connected to a voltage source so that R _ii and R ₃ are in parallel with one another and that combination is in series with R _ane.

8. Referring to Effigy 5: (a) CalculateP ₃ and annotation how information technology compares withP _iii institute in the beginning ii example problems in this module. (b) Find the total power supplied by the source and compare it with the sum of the powers dissipated by the resistors.

9. Refer to Figure 6 and the word of lights dimming when a heavy appliance comes on. (a) Given the voltage source is 120 V, the wire resistance is 0.400 Ω, and the bulb is nominally 75.0 W, what power will the seedling dissipate if a total of xv.0 A passes through the wires when the motor comes on? Assume negligible alter in bulb resistance. (b) What ability is consumed by the motor?

Effigy six. Why exercise lights dim when a large apparatus is switched on? The answer is that the big current the apparatus motor draws causes a pregnant drop in the wires and reduces the voltage across the low-cal.

10. A 240-kV power transmission line carrying 5.00 × 10² is hung from grounded metallic towers by ceramic insulators, each having a 1.00 × 10⁹-Ω resistance (Figure ix(a)). What is the resistance to ground of 100 of these insulators? (b) Calculate the power dissipated by 100 of them. (c) What fraction of the ability carried by the line is this? Explicitly bear witness how y'all follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors in a higher place.

Effigy 9. Loftier-voltage (240-kV) transmission line conveying five.00 × 10ⁱⁱ is hung from a grounded metal manual belfry. The row of ceramic insulators provide 1.00 × 10^ix Ω of resistance each.

11. Show that if two resistorsR _one andR ₂ are combined and ane is much greater than the other (R ₁>> R ₂): (a) Their series resistance is very nigh equal to the greater resistanceR ₁ . (b) Their parallel resistance is very virtually equal to smaller resistanceR ₂.

12. Unreasonable ResultsTwo resistors, one having a resistance of 145 Ω, are continued in parallel to produce a total resistance of 150 Ω. (a) What is the value of the 2d resistance? (b) What is unreasonable nearly this result? (c) Which assumptions are unreasonable or inconsistent?

13. Unreasonable ResultsTwo resistors, one having a resistance of 900 kΩ, are connected in serial to produce a full resistance of 0.500 MΩ. (a) What is the value of the 2nd resistance? (b) What is unreasonable about this issue? (c) Which assumptions are unreasonable or inconsistent?

Glossary

serial:

a sequence of resistors or other components wired into a excursion i 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 inside an electrical circuit:V = IR

voltage:

the electric potential free energy per unit accuse; electric pressure created by a power source, such as a bombardment

voltage drop:

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

current:

the menses of charge through an electric excursion past a given point of measurement

Joule'due south law:

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

parallel:

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

Selected Solutions to Problems & Exercises

1. (a) 2.75 kΩ (b) 27.5 Ω

iii.(a) 786 Ω (b) 20.3 Ω

5. 29.vi W

7. (a) 0.74 A (b) 0.742 A

9. (a) 60.8 W (b) 3.18 kW

11. (a) [latex]\begin{array}{}{R}_{\text{south}}={R}_{1}+{R}_{2}\\ \Rightarrow {R}_{\text{south}}\approx {R}_{1}\left({R}_{ane}\text{>>}{R}_{2}\right)\end{array}\\[/latex]

(b) [latex]\frac{1}{{R}_{p}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}=\frac{{R}_{1}+{R}_{2}}{{R}_{1}{R}_{ii}}\\[/latex] ,

so that

[latex]\begin{assortment}{}{R}_{p}=\frac{{R}_{1}{R}_{2}}{{R}_{one}+{R}_{2}}\approx \frac{{R}_{1}{R}_{ii}}{{R}_{1}}={R}_{2}\left({R}_{1}\text{>>}{R}_{2}\right)\text{.}\end{array}\\[/latex]

13.(a) −400 kΩ (b) Resistance cannot be negative. (c) Series resistance is said to be less than 1 of the resistors, but information technology must be greater than whatsoever of the resistors.

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Source: https://courses.lumenlearning.com/physics/chapter/21-1-resistors-in-series-and-parallel/

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