Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 21: Circuits and DC Instruments

40 problems · 10 figures

21-1 §21.1 source ↗

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

(a) What is the resistance of ten 275-Ω resistors connected in series? (b) In parallel?

Answer

(a) 2.75 kΩ

(b) 27.5Ω

21-3 §21.1 source ↗

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

What are the largest and smallest resistances you can obtain by connecting a 36.0-Ω, a 50.0-Ω, and a 700-Ω resistor together?

Answer

(a) 786Ω

(b) 20.3Ω

21-5 §21.1 source ↗

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

Your car’s 30.0-W headlight and 2.40-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 circuit and any change in resistance in the two devices.)

Answer

29 . 6 W

21-7 §21.1 source ↗

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

Referring to the example combining series and parallel circuits and Figure 21.6, calculate I3 in the following two different ways: (a) from the known values of I and I2; (b) using Ohm’s law for R3. In both parts explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors.

Figure 21.6 in the book ↗
These three resistors are connected to a voltage source so that R2 and R3 are in parallel with one another and that combination is in series with R1.

Answer

(a) 0.74 A

(b) 0.742 A

21-9 §21.1 source ↗

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

Refer to Figure 21.7 and the discussion of lights dimming when a heavy appliance comes on. (a) Given the voltage source is 120 V, the wire resistance is 0.400Ω, and the bulb is nominally 75.0 W, what power will the bulb dissipate if a total of 15.0 A passes through the wires when the motor comes on? Assume negligible change in bulb resistance. (b) What power is consumed by the motor?

Figure 21.7 in the book ↗
Why do lights dim when a large appliance is switched on? The answer is that the large current the appliance motor draws causes a significant IR drop in the wires and reduces the voltage across the light.

Answer

(a) 67.7 W

(b) 1.64 kW

21-11 §21.1 source ↗

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

Show that if two resistors R1 and R2 are combined and one is much greater than the other (R1>>R2): (a) Their series resistance is very nearly equal to the greater resistance R1. (b) Their parallel resistance is very nearly equal to smaller resistance R2.

Answer

(a) Rs=R1+R2⇒Rs≈R1R1>>R2

(b) 1Rp= 1R1+ 1R2= R1+R2R1R2,

so that

R p = R 1 R 2 R 1 + R 2 ≈ R 1 R 2 R 1 = R 2 R 1 >> R 2 .

21-13 §21.1 source ↗

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

Unreasonable Results

Two 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 second resistance? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

Answer

(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 any of the resistors.

21-14 §21.2 source ↗

Standard automobile batteries have six lead-acid cells in series, creating a total emf of 12.0 V. What is the emf of an individual lead-acid cell?

Answer

2.00 V

21-16 §21.2 source ↗

What is the output voltage of a 3.0000-V lithium cell in a digital wristwatch that draws 0.300 mA, if the cell’s internal resistance is 2.00Ω?

Answer

2.9994 V

21-18 §21.2 source ↗

What is the internal resistance of an automobile battery that has an emf of 12.0 V and a terminal voltage of 15.0 V while a current of 8.00 A is charging it?

Answer

0 . 375 Ω

21-21 §21.2 source ↗

The hot resistance of a flashlight bulb is 2.30Ω, and it is run by a 1.58-V alkaline cell having a 0.100-Ω internal resistance. (a) What current flows? (b) Calculate the power supplied to the bulb using I2Rbulb. (c) Is this power the same as calculated using V2Rbulb?

Answer

(a) 0.658 A

(b) 0.997 W

(c) 0.997 W; yes

21-23 §21.2 source ↗

An automobile starter motor has an equivalent resistance of 0.0500Ω and is supplied by a 12.0-V battery with a 0.0100-Ω internal resistance. (a) What is the current to the motor? (b) What voltage is applied to it? (c) What power is supplied to the motor? (d) Repeat these calculations for when the battery connections are corroded and add 0.0900Ω to the circuit. (Significant problems are caused by even small amounts of unwanted resistance in low-voltage, high-current applications.)

Answer

(a) 200 A

(b) 10.0 V

(c) 2.00 kW

(d) 0.1000Ω; 80.0 A, 4.0 V, 320 W

21-25 §21.2 source ↗

(a) What is the internal resistance of a voltage source if its terminal voltage drops by 2.00 V when the current supplied increases by 5.00 A? (b) Can the emf of the voltage source be found with the information supplied?

Answer

(a) 0.400 Ω

(b) No, there is only one independent equation, so only r can be found.

21-29 §21.2 source ↗

Unreasonable Results

A 1.58-V alkaline cell with a 0.200-Ω internal resistance is supplying 8.50 A to a load. (a) What is its terminal voltage? (b) What is the value of the load resistance? (c) What is unreasonable about these results? (d) Which assumptions are unreasonable or inconsistent?

Answer

(a) –0.120 V

(b) -1.41×10−2Ω

(c) Negative terminal voltage; negative load resistance.

(d) The assumption that such a cell could provide 8.50 A is inconsistent with its internal resistance.

21-31 §21.3 source ↗

Apply the loop rule to loop abcdefgha in Figure 21.25.
Figure 21.25 in the book ↗
This circuit is similar to that in Figure 21.21, but the resistances and emfs are specified. (Each emf is denoted by script E.) The currents in each branch are labeled and assumed to move in the directions shown. This example uses Kirchhoff’s rules to find the currents.

Answer

− I 2 R 2 + E 1 − I 2 r 1 + I 3 R 3 + I 3 r 2 - E 2 = 0

21-35 §21.3 source ↗

Apply the junction rule at point a in Figure 21.50.
Figure 21.50 in the book ↗

Answer

I 3 = I 1 + I 2

21-37 §21.3 source ↗

Apply the loop rule to loop akledcba in Figure 21.50.
Figure 21.50 in the book ↗

Answer

emf 2 - I 2 r 2 - I 2 R 2 + I 1 R 5 + I 1 r 1 - emf 1 + I 1 R 1 = 0

21-39 §21.3 source ↗

Solve Example 21.5, but use loop abcdefgha instead of loop akledcba. Explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors.
Example 21.5 Calculating Current: Using Kirchhoff’s Rules in the book ↗

Answer

(a) I1=4.75 A

(b) I2 =-3.5 A

(c) I3=8.25 A

21-41 §21.3 source ↗

Unreasonable Results

Consider the circuit in Figure 21.51, and suppose that the emfs are unknown and the currents are given to be I1=5.00 A, I2=3.0 A, and I3=–2.00 A. (a) Could you find the emfs? (b) What is wrong with the assumptions?

Figure 21.51 in the book ↗

Answer

(a) No, you would get inconsistent equations to solve.

(b) I1≠I2+I3. The assumed currents violate the junction rule.

21-42 §21.4 source ↗

What is the sensitivity of the galvanometer (that is, what current gives a full-scale deflection) inside a voltmeter that has a 1.00-MΩ resistance on its 30.0-V scale?

Answer

30 μA

21-44 §21.4 source ↗

Find the resistance that must be placed in series with a 25.0-Ω galvanometer having a 50.0-μA sensitivity (the same as the one discussed in the text) to allow it to be used as a voltmeter with a 0.100-V full-scale reading.

Answer

1 . 98 k Ω

21-46 §21.4 source ↗

Find the resistance that must be placed in parallel with a 25.0-Ω galvanometer having a 50.0-μA sensitivity (the same as the one discussed in the text) to allow it to be used as an ammeter with a 10.0-A full-scale reading. Include a circuit diagram with your solution.

Answer

1 . 25 × 10 - 4 Ω

21-48 §21.4 source ↗

Find the resistance that must be placed in series with a 10.0-Ω galvanometer having a 100-μA sensitivity to allow it to be used as a voltmeter with: (a) a 300-V full-scale reading, and (b) a 0.300-V full-scale reading.

Answer

(a) 3.00 MΩ

(b) 2.99 kΩ

21-50 §21.4 source ↗

Suppose you measure the terminal voltage of a 1.585-V alkaline cell having an internal resistance of 0.100Ω by placing a 1.00-kΩ voltmeter across its terminals. (See Figure 21.52.) (a) What current flows? (b) Find the terminal voltage. (c) To see how close the measured terminal voltage is to the emf, calculate their ratio.
Figure 21.52 in the book ↗

Answer

(a) 1.58 mA

(b) 1.5848 V (need four digits to see the difference)

(c) 0.99990 (need five digits to see the difference from unity)

21-52 §21.4 source ↗

A certain ammeter has a resistance of 5.00×10−5Ω on its 3.00-A scale and contains a 10.0-Ω galvanometer. What is the sensitivity of the galvanometer?

Answer

15 . 0 μA

21-54 §21.4 source ↗

A 0.0200-Ω ammeter is placed in series with a 10.00-Ω resistor in a circuit. (a) Draw a circuit diagram of the connection. (b) Calculate the resistance of the combination. (c) If the voltage is kept the same across the combination as it was through the 10.00-Ω resistor alone, what is the percent decrease in current? (d) If the current is kept the same through the combination as it was through the 10.00-Ω resistor alone, what is the percent increase in voltage? (e) Are the changes found in parts (c) and (d) significant? Discuss.

Answer

(a)

The figure shows part of a circuit that includes an ammeter with internal resistance r connected in series with a load resistance R.

(b) 10.02 Ω

(c) 0.9980, or a 2.0×10–1 percent decrease

(d) 1.002, or a 2.0×10–1 percent increase

(e) Not significant.

21-56 §21.4 source ↗

Unreasonable Results

(a) What resistance would you put in parallel with a 40.0-Ω galvanometer having a 25.0-μA sensitivity to allow it to be used as an ammeter that has a full-scale deflection for 10.0-μA? (b) What is unreasonable about this result? (c) Which assumptions are responsible?

Answer

(a) −66.7Ω

(b) You can’t have negative resistance.

(c) It is unreasonable that IG is greater than Itot (see Figure 21.30). You cannot achieve a full-scale deflection using a current less than the sensitivity of the galvanometer.

Figure 21.30 in the book ↗
A small shunt resistance R placed in parallel with a galvanometer G produces an ammeter, the full-scale deflection of which depends on the choice of R. The larger the current to be measured, the smaller R must be. Most of the current (I) flowing through the meter is shunted through R to protect the galvanometer. (Note that r represents the internal resistance of the galvanometer.) Ammeters may also have multiple scales for greater flexibility in application. The various scales are achieved by switching various shunt resistances in parallel with the galvanometer—the greater the maximum current to be measured, the smaller the shunt resistance must be.

21-57 §21.5 source ↗

What is the emfx of a cell being measured in a potentiometer, if the standard cell’s emf is 12.0 V and the potentiometer balances for Rx=5.000Ω and Rs=2.500Ω?

Answer

24.0 V

21-59 §21.5 source ↗

When an unknown resistance Rx is placed in a Wheatstone bridge, it is possible to balance the bridge by adjusting R3 to be 2500Ω. What is Rx if R2R1=0.625?

Answer

1 . 56 k Ω

21-61 §21.5 source ↗

(a) What is the unknown emfx in a potentiometer that balances when Rx is 10.0Ω, and balances when Rs is 15.0Ω for a standard 3.000-V emf? (b) The same emfx is placed in the same potentiometer, which now balances when Rs is 15.0Ω for a standard emf of 3.100 V. At what resistance Rx will the potentiometer balance?

Answer

(a) 2.00 V

(b) 9.68 Ω

21-62 §21.5 source ↗

Suppose you want to measure resistances in the range from 10.0Ω to 10.0 kΩ using a Wheatstone bridge that has R2R1=2.000. Over what range should R3 be adjustable?

Answer

Range = 5 . 00 Ω to 5 . 00 k Ω

21-63 §21.6 source ↗

The timing device in an automobile’s intermittent wiper system is based on an RC time constant and utilizes a 0.500-μF capacitor and a variable resistor. Over what range must R be made to vary to achieve time constants from 2.00 to 15.0 s?

Answer

range 4 . 00 to 30 . 0 M Ω

21-65 §21.6 source ↗

The duration of a photographic flash is related to an RC time constant, which is 0.100 μs for a certain camera. (a) If the resistance of the flash lamp is 0.0400Ω during discharge, what is the size of the capacitor supplying its energy? (b) What is the time constant for charging the capacitor, if the charging resistance is 800 kΩ?

Answer

(a) 2.50 μF

(b) 2.00 s

21-67 §21.6 source ↗

After two time constants, what percentage of the final voltage, emf, is on an initially uncharged capacitor C, charged through a resistance R?

Answer

86.5%

21-69 §21.6 source ↗

A heart defibrillator being used on a patient has an RC time constant of 10.0 ms due to the resistance of the patient and the capacitance of the defibrillator. (a) If the defibrillator has an 8.00-μF capacitance, what is the resistance of the path through the patient? (You may neglect the capacitance of the patient and the resistance of the defibrillator.) (b) If the initial voltage is 12.0 kV, how long does it take to decline to 6.00×102V?

Answer

(a) 1.25 kΩ

(b) 30.0 ms

21-71 §21.6 source ↗

Figure 21.53 shows how a bleeder resistor is used to discharge a capacitor after an electronic device is shut off, allowing a person to work on the electronics with less risk of shock. (a) What is the time constant? (b) How long will it take to reduce the voltage on the capacitor to 0.250% (5% of 5%) of its full value once discharge begins? (c) If the capacitor is charged to a voltage V0 through a 100-Ω resistance, calculate the time it takes to rise to 0.865V0 (This is about two time constants.)
Figure 21.53 in the book ↗

Answer

(a) 20.0 s

(b) 120 s

(c) 16.0 ms

21-73 §21.6 source ↗

Using the exact exponential treatment, find how much time is required to charge an initially uncharged 100-pF capacitor through a 75.0-MΩ resistor to 90.0% of its final voltage.

Answer

1 . 73 × 10 − 2 s

21-74 §21.6 source ↗

Integrated Concepts

If you wish to take a picture of a bullet traveling at 500 m/s, then a very brief flash of light produced by an RC discharge through a flash tube can limit blurring. Assuming 1.00 mm of motion during one RC constant is acceptable, and given that the flash is driven by a 600-μF capacitor, what is the resistance in the flash tube?

Answer

3 . 33 × 10 − 3 Ω

21-76 §21.6 source ↗

Integrated Concepts

A 160-μF capacitor charged to 450 V is discharged through a 31.2-kΩ resistor. (a) Find the time constant. (b) Calculate the temperature increase of the resistor, given that its mass is 2.50 g and its specific heat is 1.67kJkg⋅ºC, noting that most of the thermal energy is retained in the short time of the discharge. (c) Calculate the new resistance, assuming it is pure carbon. (d) Does this change in resistance seem significant?

Answer

(a) 4.99 s

(b) 3.87ºC

(c) 31.1 kΩ

(d) No

21-80 §21.6 source ↗

Critical Thinking A circuit has a voltage source producing 1.00×102V and a 2.50×103Ω resistor. (a) If the circuit contains nothing else, how much power is dissipated by the resistor? (b) A second 2.50×103Ω resistor is added in parallel to the first resistor. Now how much power is dissipated by the resistors? (c) Now place the two resistors in series. How much power is dissipated by the resistors? (d) Why would a parallel arrangement of resistors dissipate more power than a series arrangement of resistors of the same value? (e) Which delivers more energy in a given time period, the parallel combination of resistors, the series combination of resistors, or is the energy delivered in a given time period the same?

Answer

(a) P= V 2 R = (1.00× 10 2 ) 2 2.50× 10 3 w= 4.00 W

(b) 1 R eq = 2 R 1 R eq = R 1 2 =1.25× 10 3 Ω P= V 2 R = (1.00× 10 2 ) 2 1.25× 10 3  W=8W

(c) R eq = R 1 + R 2 =2 R 1 P= V 2 R = (1.00× 10 2 ) 2 5.00× 10 3 W=2W

(d) In the parallel case current passes through each branch, so more current is passed through the resistor system. In the series, the higher resistance of the circuit allows less current to flow. P=IV is proportional to the current when the voltage does not change.

(e) The parallel combination delivers more energy in a given time period.

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Problems and answers are adapted from College Physics 2e by Paul Peter Urone, Roger Hinrichs, Kim Dirks and Manjula Sharma, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.