if in series parallel circuit of part c if the bulb for r2 burns out what happens to r1 and r3

Learning Objectives

By the cease of this department, you will be able to:

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

Most circuits have more one component, called a resistor that limits the catamenia of charge in the excursion. A measure of this limit on charge flow is chosen resistance. The simplest combinations of resistors are the serial and parallel connections illustrated in Effigy ane. The full resistance of a combination of resistors depends on both their private values and how they are connected.

Effigy ane. (a) A serial connection of resistors. (b) A parallel connectedness of resistors.

Resistors in Serial

When are resistors in series? Resistors are in series whenever the menstruum of charge, called the current, must menstruum through devices sequentially. For case, if current flows through a person belongings a screwdriver and into the Earth, then R ₁ in Effigy ane(a) could be the resistance of the screwdriver'south shaft, R _ii the resistance of its handle, R ₃ the person's trunk resistance, and R ₄ the resistance of her shoes. Figure 2 shows resistors in series continued to a voltage source. Information technology 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 past wearing high-resistance prophylactic-soled shoes. Information technology 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 battery (left) and the equivalent single or series resistance (correct).

To verify that resistances in series do indeed add together, permit united states consider the loss of electrical power, called a voltage drop, in each resistor in Effigy 2. According to Ohm'south police, the voltage drop, V, across a resistor when a electric 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). Some other way to call up of this is that V is the voltage necessary to make a current I menstruation through a resistance R. So the voltage drib across R _ane is 5 ₁=IR _one, that beyond R ₂ is 5 _two=IR ₂, and that across R _iii is Five _iii=IR ₃. The sum of these voltages equals the voltage output of the source; that is,

V=Five ₁+V _ii+V ₃.

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

qV ₁+ qV _two+ qV ₃.

Making Connections: Conservation Laws

The derivations of the expressions for series and parallel resistance are based on the laws of conservation of energy and conservation of charge, which state that full charge and total free energy are constant in whatever process. These two laws are directly involved in all electric phenomena and volition exist invoked repeatedly to explain both specific effects and the general behavior of electricity.

These energies must be equal, considering there is no other source and no other destination for energy in the circuit. Thus, qV=qV ₁+qV _ii+qV ₃. The charge q cancels, yielding 5=5 _one+V ₂+V ₃, as stated. (Note that the same amount of charge passes through the battery and each resistor in a given amount of time, since at that place is no capacitance to shop charge, in that location is no place for charge to leak, and charge is conserved.) Now substituting the values for the individual voltages gives

5= IR ₁+ IR ₂ + IR ₃ =I(R ₁+R ₂+R ₃).

Annotation that for the equivalent unmarried series resistance R _s, nosotros have

Five = IR _south.

This implies that the total or equivalent series resistance R _southward of iii resistors is R _s=R _ane+R ₂+R ₃. This logic is valid in full general for whatever number of resistors in series; thus, the total resistance R _south of a series connexion is

R _s=R ₁+R ₂+R ₃+…,

equally proposed. Since all of the electric current must pass through each resistor, it experiences the resistance of each, and resistances in series just add up.

Example i. Calculating Resistance, Current, Voltage Drop, and Power Dissipation: Analysis of a Series Circuit

Suppose the voltage output of the bombardment in Figure ii is 12.0 V, and the resistances are R _one= 1.00 Ω, R _ii= 6.00 Ω, and R ₃= 13.0 Ω. (a) What is the total resistance? (b) Notice the current. (c) Summate the voltage drop in each resistor, and show these add to equal the voltage output of the source. (d) Summate the power dissipated by each resistor. (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors.

Strategy and Solution for (a)

The total resistance is simply the sum of the individual resistances, as given by this equation:

[latex]\begin{array}{lll}{R}_{\text{south}}& =& {R}_{1}+{R}_{2}+{R}_{3}\\ & =& 1.00\text{ }\Omega + six.00\text{ }\Omega + 13.0\text{ }\Omega\\ & =& 20.0\text{ }\Omega\end{array}\\[/latex].

Strategy and Solution for (b)

The current is plant using Ohm'south law, Five = IR. Inbound the value of the applied voltage and the total resistance yields the current for the circuit:

[latex]I=\frac{V}{{R}_{\text{south}}}=\frac{12.0\text{ V}}{20.0\text{ }\Omega}=0.lx\text{ A}\\[/latex].

Strategy and Solution for (c)

The voltage—or IR drop—in a resistor is given by Ohm's law. Entering the electric current and the value of the start resistance yields

V ₁ = IR _one = (0.600A)(1.0 Ω) = 0.600 V.

Similarly,

Five ₂ = IR ₂ = (0.600A)(6.0 Ω) = iii.60 V

and

V3 = IR ₃ = (0.600A)(13.0 Ω) = seven.80 5.

Discussion for (c)

The three IR drops add to 12.0 V, as predicted:

V ₁ + 5 ₂ + V ₃ = (0.600 + 3.60 + 7.80)V = 12.0 V.

Strategy and Solution for (d)

The easiest way to calculate power in watts (W) prodigal by a resistor in a DC circuit is to utilise Joule'south police force, P = IV, where P is electric power. In this case, each resistor has the same full current flowing through it. Past substituting Ohm's law 5 = IR into Joule's law, we go the power dissipated by the first resistor as

P _ane = I ² R ₁ = ( 0 . 600 A ) ² ( 1 . 00 Ω ) = 0 . 360 West .

Similarly,

P ₂ = I ⁱⁱ R _ii = ( 0 . 600 A ) ^two (half-dozen . 00 Ω ) = two.xvi Westward .

and

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

Discussion for (d)

Power tin can also be calculated using either P = IV or [latex]P=\frac{{V}^{ii}}{R}\\[/latex], where V is the voltage drib across the resistor (not the total voltage of the source). The same values will be obtained.

Strategy and Solution for (e)

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

P = (0.600 A)(12.0 Five) = vii.20 Westward.

Discussion for (e)

Note, coincidentally, that the total power dissipated by the resistors is also 7.20 Westward, the aforementioned as the power put out by the source. That is,

P ₁ + P ₂ + P ₃ = (0.360 + 2.16 + 4.68) W = 7.xx W.

Power is energy per unit of measurement time (watts), and so conservation of free energy requires the power output of the source to be equal to the full power prodigal past the resistors.

Major Features of Resistors in Series

1. Series resistances add: R _{due south}=R _one+R ₂+R ₃+….
2. The aforementioned current flows through each resistor in series.
3. Individual resistors in series do non get the total source voltage, but separate information technology.

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 electric current it would if it alone were continued to the voltage source (provided the voltage source is not overloaded). For example, an automobile's headlights, radio, so on, are wired in parallel, and so that they utilize the full voltage of the source and can operate completely independently. The same is truthful in your business firm, or any building. (Run across Figure 3(b).)

Figure 3. (a) Three resistors connected in parallel to a bombardment and the equivalent unmarried or parallel resistance. (b) Electrical power setup in a house. (credit: Dmitry M, Wikimedia Eatables)

To detect an expression for the equivalent parallel resistance R _p, permit u.s.a. consider the currents that period and how they are related to resistance. Since each resistor in the circuit has the total voltage, the currents flowing through the individual resistors are [latex]{I}_{1}=\frac{V}{{R}_{1}}\\[/latex], [latex]{I}_{two}=\frac{V}{{R}_{ii}}\\[/latex], and [latex]{I}_{3}=\frac{V}{{R}_{three}}\\[/latex]. Conservation of charge implies that the total current I produced by the source is the sum of these currents:

I=I _ane+I ₂+I ₃.

Substituting the expressions for the private currents gives

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

Notation that Ohm's law for the equivalent unmarried resistance gives

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

The terms within the parentheses in the last two equations must exist equal. Generalizing to any number of resistors, the total resistance R _p of a parallel connection is related to the private resistances past

[latex]\frac{1}{{R}_{p}}=\frac{i}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{ane}{{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 instance.) When resistors are continued in parallel, more than electric current flows from the source than would flow for any of them individually, then the total resistance is lower.

Example 2. Calculating Resistance, Current, Power Dissipation, and Power Output: Analysis of a Parallel Circuit

Let the voltage output of the battery and resistances in the parallel connection in Figure 3 be the same as the previously considered series connection: 5 = 12.0 V, 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 current. (c) Calculate the currents in each resistor, and evidence these add to equal the total current output of the source. (d) Calculate the ability dissipated by each resistor. (e) Find the ability output of the source, and show that it equals the total ability dissipated past the resistors.

Strategy and Solution for (a)

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

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

Thus,

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

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

[latex]{R}_{\text{p}}=\frac{1}{1\text{.}\text{2436}}\text{ }\Omega =0\text{.}\text{8041}\text{ }\Omega\\[/latex].

The total resistance with the right number of significant digits is R _p= 0.804 Ω

Discussion for (a)

R _p is, as predicted, less than the smallest private resistance.

Strategy and Solution for (b)

The total current tin be found from Ohm'due south police, substituting R _p for the total resistance. This gives

[latex]I=\frac{Five}{{R}_{\text{p}}}=\frac{\text{12.0 5}}{0.8041\text{ }\Omega }=\text{14}\text{.}\text{92 A}\\[/latex].

Give-and-take for (b)

Current I for each device is much larger than for the same devices connected in series (run across the previous instance). A circuit with parallel connections has a smaller full resistance than the resistors continued in series.

Strategy and Solution for (c)

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

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

Similarly,

[latex]{I}_{2}=\frac{5}{{R}_{ii}}=\frac{12.0\text{ 5}}{vi.00\text{ }\Omega}=2\text{.}\text{00}\text{ A}\\[/latex]

and

[latex]{I}_{3}=\frac{V}{{R}_{3}}=\frac{\text{12}\text{.}0\text{ V}}{\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 _i+I _ii+I ₃= 14.92 A.

This is consistent with conservation of charge.

Strategy and Solution for (d)

The ability prodigal by each resistor can be found using whatsoever of the equations relating power to electric current, voltage, and resistance, since all three are known. Permit us utilize [latex]P=\frac{{V}^{2}}{R}\\[/latex], since each resistor gets full voltage. Thus,

[latex]{P}_{i}=\frac{{V}^{2}}{{R}_{i}}=\frac{(12.0\text{ V})^{2}}{one.00\text{ }\Omega}=144\text{ W}\\[/latex].

Similarly,

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

and

[latex]{P}_{3}=\frac{{V}^{2}}{{R}_{3}}=\frac{(12.0\text{ V})^{two}}{13.0\text{ }\Omega}=eleven.1\text{ W}\\[/latex].

Discussion for (d)

The ability dissipated by each resistor is considerably college in parallel than when connected in series to the same voltage source.

Strategy and Solution for (e)

The total power tin also be calculated in several ways. Choosing P = 4, and inbound the total electric current, yields

P = Four= (14.92 A)(12.0 V) = 179 W.

Discussion for (e)

Total power dissipated by the resistors is also 179 West:

P _ane+P ₂+P ₃= 144 W + 24.0 Westward + eleven.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{1}{{R}_{\text{p}}}=\frac{1}{{R}_{one}}+\frac{1}{{R}_{two}}+\frac{1}{{R}_{3}}+\text{…}\\[/latex], and information technology is smaller than any private resistance in the combination.
2. Each resistor in parallel has the aforementioned full voltage of the source practical to it. (Power distribution systems most often utilise parallel connections to supply the myriad devices served with the aforementioned voltage and to allow them to operate independently.)
3. Parallel resistors do non each become 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, especially when wire resistance is considered. In that instance, wire resistance is in serial with other resistances that are in parallel. Combinations of series and parallel can exist reduced to a single equivalent resistance using the technique illustrated in Effigy 4. Diverse parts are identified equally either series or parallel, reduced to their equivalents, and further reduced until a single resistance is left. The process is more than time consuming than difficult.

Figure 4. 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 Figure 4, is also the about instructive, since it is found in many applications. For instance, R _one could be the resistance of wires from a machine bombardment to its electrical devices, which are in parallel. R ₂ and R ₃ could be the starter motor and a rider compartment light. Nosotros have previously assumed that wire resistance is negligible, simply, when it is non, it has important effects, as the next instance indicates.

Example iii. Computing Resistance, IR Driblet, Current, and Ability Dissipation: Combining Series and Parallel Circuits

Effigy five shows the resistors from the previous two examples wired in a different way—a combination of serial and parallel. Nosotros tin can consider R ₁ to be the resistance of wires leading to R ₂ and R _iii. (a) Find the total resistance. (b) What is the IR drop in R ₁? (c) Find the electric current I ₂ through R ₂. (d) What ability is dissipated by R ₂?

Figure v. These iii resistors are continued to a voltage source then that R _two and R ₃ are in parallel with one another and that combination is in serial with R _ane.

Strategy and Solution for (a)

To observe the total resistance, we annotation that R _two 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 _one+R _p.

Kickoff, 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}_{iii}}=\frac{i}{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{one}{0.2436}\text{ }\Omega =4.eleven\text{ }\Omega\\[/latex].

So the total resistance is

R _tot= R _one+ R _p= 1.00 Ω + 4.11 Ω = 5.xi Ω.

Discussion for (a)

The total resistance of this combination is intermediate between the pure serial 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 notice the IRdrop in R _ane, we notation that the full current I flows through R _ane. Thus itsIR drop is

Five ₁ = IR ₁

Nosotros must find I earlier nosotros can summate V _one. The total current I is establish using Ohm's law for the circuit. That is,

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

Entering this into the expression to a higher place, we get

5 ₁ = IR ₁ = ( 2 . 35 A ) ( 1 . 00 Ω ) = ii . 35 Five .

Word for (b)

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

Strategy and Solution for (c)

To find the electric current through R ₂, we must kickoff find the voltage applied to it. We call this voltage V _p, because it is applied to a parallel combination of resistors. The voltage practical to both R ₂ and R _three is reduced past the corporeality V ₁, and so it is

V _p= V− Five ₁= 12.0 V − 2.35 V = ix.65 V.

Now the current I ₂ through resistance R _ii is found using Ohm's police:

[latex]{I}_{2}=\frac{{V}_{\text{p}}}{{R}_{2}}=\frac{9.65\text{ V}}{6.00\text{ }\Omega}=one.61\text{A}\\[/latex].

Word for (c)

The current is less than the ii.00 A that flowed through R ₂ when it was connected in parallel to the battery in the previous parallel circuit example.

Strategy and Solution for (d)

The ability dissipated by R _ii is given by

P ₂= (I ₂)² R _two= (1.61 A)²(6.00 Ω) = 15.five W

Discussion for (d)

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

Applied Implications

One implication of this final case 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 string, then this loss can be significant. If a large electric current is drawn, the IR driblet in the wires can likewise be significant.

For example, when you are rummaging in the refrigerator and the motor comes on, the fridge low-cal dims momentarily. Similarly, yous tin run into the passenger compartment light dim when you start the engine of your car (although this may be due to resistance inside the bombardment itself).

What is happening in these high-current situations is illustrated in Effigy six. The device represented past R _iii has a very low resistance, and and then when information technology is switched on, a large current flows. This increased electric current causes a larger IR drop in the wires represented by R ₁, reducing the voltage across the low-cal bulb (which is R _ii), which and then dims noticeably.

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

Check Your Understanding

Can any arbitrary combination of resistors be broken down into series and parallel combinations? See if you can draw a circuit diagram of resistors that cannot be broken down 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's rules, to be introduced in Kirchhoff's Rules, volition allow you to analyze the excursion.

Problem-Solving Strategies for Series and Parallel Resistors

1. Draw a clear circuit diagram, labeling all resistors and voltage sources. This pace includes a list of the knowns for the problem, since they are labeled in your circuit diagram.
2. Identify exactly what needs to be determined in the problem (place the unknowns). A written list is useful.
3. Determine whether resistors are in series, parallel, or a combination of both series and parallel. Examine the circuit diagram to make this assessment. Resistors are in serial if the same current must pass sequentially through them.
4. Use the appropriate list of major features for series or parallel connections to solve for the unknowns. In that location is one list for series and another for parallel. If your problem has a combination of series and parallel, reduce it in steps by considering private groups of series or parallel connections, as done in this module and the examples. Special note: When finding R , the reciprocal must be taken with care.
5. Cheque to run into 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. Power should be greater for the same devices in parallel compared with serial, 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 _s= R _i+ R ₂+ R ₃+….
• Each resistor in a serial circuit has the aforementioned amount of current flowing through it.
• The voltage drib, or ability dissipation, across each individual resistor in a series is dissimilar, and their combined total adds up to the power source input.
• The total resistance of an electrical excursion with resistors wired in parallel is less than the lowest resistance of any of the components and can exist adamant using the formula:

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

• Each resistor in a parallel circuit has the aforementioned total voltage of the source applied to information technology.
• The current flowing through each resistor in a parallel excursion is different, depending on the resistance.
• If a more than complex connection of resistors is a combination of serial and parallel, information technology tin be reduced to a single equivalent resistance by identifying its various parts as series or parallel, reducing each to its equivalent, and continuing until a unmarried resistance is eventually reached.

Conceptual Questions

1. A switch has a variable resistance that is about goose egg when airtight and extremely big when open, and it is placed in series with the device it controls. Explain the effect the switch in Figure 7 has on current when open and when closed.

Figure 7. A switch is ordinarily in series with a resistance and voltage source. Ideally, the switch has nearly zero resistance when closed but has an extremely large resistance when open. (Note that in this diagram, the script E represents the voltage (or electromotive forcefulness) of the bombardment.)

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

3. There is a voltage across an open switch, such equally in Figure 7. Why, and then, is the power dissipated by the open switch pocket-size?

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

5. A student in a physics lab mistakenly wired a light bulb, battery, and switch every bit shown in Figure 8. Explain why the bulb is on when the switch is open, and off when the switch is closed. (Do not try this—it is difficult on the battery!)

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

half dozen. Knowing that the severity of a daze depends on the magnitude of the current through your body, would you prefer to be in series or parallel with a resistance, such as the heating chemical element of a toaster, if shocked by it? Explicate.

vii. Would your headlights dim when you commencement your automobile's engine if the wires in your automobile were superconductors? (Practise 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 break the electrical connection, like an open up switch, when they burn out. If ane such bulb burns out, what happens to the others? If such a string operates on 120 V and has 40 identical bulbs, what is the normal operating voltage of each? Newer versions apply bulbs that short circuit, similar a closed switch, when they burn out. If one such bulb burns out, what happens to the others? If such a cord operates on 120 5 and has 39 remaining identical bulbs, what is then the operating voltage of each?

9. If 2 household lightbulbs rated 60 W and 100 W are connected in series to household power, which will exist brighter? Explain.

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

eleven. Before Globe War Two, some radios got power through a "resistance cord" that had a pregnant resistance. Such a resistance cord reduces the voltage to a desired level for the radio's tubes and the like, and it saves the expense of a transformer. Explicate why resistance cords become warm and waste product free energy when the radio is on.

12. Some light bulbs have iii power settings (not including aught), obtained from multiple filaments that are individually switched and wired in parallel. What is the minimum number of filaments needed for 3 ability settings?

Problems & Exercises

Annotation: Information taken from figures tin exist causeless to exist accurate to three pregnant digits.

ane. (a) What is the resistance of x 275-Ω resistors connected in serial? (b) In parallel?

2. (a) What is the resistance of a one.00 × x²-Ω, a ii.50-kΩ, and a iv.00-kΩ resistor connected in series? (b) In parallel?

3. What are the largest and smallest resistances you tin can obtain past connecting a 36.0-Ω, a fifty.0-Ω, and a 700-Ω resistor together?

4. An 1800-W toaster, a 1400-Westward electrical frying pan, and a 75-W lamp are plugged into the aforementioned outlet in a 15-A, 120-Five circuit. (The 3 devices are in parallel when plugged into the same socket.). (a) What current is fatigued by each device? (b) Will this combination accident the xv-A fuse?

5. Your car's 30.0-W headlight and 2.40-kW starter are ordinarily connected in parallel in a 12.0-V organization. What power would one headlight and the starter consume if connected in series to a 12.0-Five bombardment? (Fail whatever other resistance in the circuit and any change in resistance in the two devices.)

half-dozen. (a) Given a 48.0-5 battery and 24.0-Ω and 96.0-Ω resistors, find the current and ability for each when continued in series. (b) Repeat when the resistances are in parallel.

7. Referring to the instance combining series and parallel circuits and Figure five, calculate I _iii in the following two dissimilar ways: (a) from the known values of IandI _ii ; (b) using Ohm's law for R ₃. In both parts explicitly bear witness how yous follow the steps in the Trouble-Solving Strategies for Serial and Parallel Resistors above.

Figure five. These 3 resistors are connected to a voltage source so that R ₂ and R _iii are in parallel with one another and that combination is in series with R _one.

8. Referring to Figure v: (a) SummateP ₃ and note how information technology compares withP _three found in the first two case problems in this module. (b) Notice the total ability supplied by the source and compare it with the sum of the powers prodigal by the resistors.

9. Refer to Figure vi and the discussion 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 seedling is nominally 75.0 W, what ability will the bulb dissipate if a total of 15.0 A passes through the wires when the motor comes on? Assume negligible alter in bulb resistance. (b) What power is consumed by the motor?

Figure 6. Why practise lights dim when a large appliance is switched on? The answer is that the big electric current the apparatus motor draws causes a significant drop in the wires and reduces the voltage across the lite.

10. A 240-kV power manual line carrying v.00 × 10² is hung from grounded metal towers by ceramic insulators, each having a i.00 × ten⁹-Ω resistance (Effigy ix(a)). What is the resistance to ground of 100 of these insulators? (b) Calculate the ability dissipated past 100 of them. (c) What fraction of the power carried by the line is this? Explicitly testify how you follow the steps in the Problem-Solving Strategies for Serial and Parallel Resistors in a higher place.

Figure nine. Loftier-voltage (240-kV) manual line carrying 5.00 × 10ⁱⁱ is hung from a grounded metallic transmission tower. The row of ceramic insulators provide one.00 × 10^ix Ω of resistance each.

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

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

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

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 electric power in a circuit

Ohm's law:

the human relationship betwixt current, voltage, and resistance inside an electrical circuit:5 = IR

voltage:

the electrical potential energy per unit charge; electrical force per unit area created by a ability source, such as a bombardment

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 electrical excursion past a given betoken of measurement

Joule's law:

the relationship betwixt potential electric 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; oftentimes 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.v Ω

3.(a) 786 Ω (b) 20.3 Ω

5. 29.six W

7. (a) 0.74 A (b) 0.742 A

9. (a) threescore.8 W (b) iii.18 kW

xi. (a) [latex]\brainstorm{assortment}{}{R}_{\text{s}}={R}_{ane}+{R}_{2}\\ \Rightarrow {R}_{\text{s}}\approx {R}_{ane}\left({R}_{1}\text{>>}{R}_{2}\right)\end{array}\\[/latex]

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

so that

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

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

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Source: https://courses.lumenlearning.com/austincc-physics2/chapter/21-1-resistors-in-series-and-parallel/

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