Question 1

A simple circuit has a total resistance of 30 Ω. If a 2.0-A current is maintained in this circuit, how much energy is dissipated in this circuit in 4.0 seconds?
A) 24 J
B) 4.8 J
C) 48 J
D) 480 J
E) 6.0 J

Accepted Answer

We can use the formula for electrical energy:

E = I²Rt

where E is the energy, I is the current, R is the resistance, and t is time.

Plugging in the given values, we get:

E = (2.0 A)² (30 Ω) (4.0 s)

E = 480 J

Therefore, the energy dissipated in the circuit is 480 J. The answer is D.

Question 2

A battery is rated at 12 V and 160 A-h. How much energy does this battery store?
A) 1.9 kJ
B) 6.0 kJ
C) 1.9 MJ
D) 6.9 MJ

Accepted Answer

The answer of A battery is rated at 12 V...

Question 3

A 400-W computer (including its monitor) is turned on for 8.0 hours per day. If electricity costs 10¢ per kWh, how much does it cost to run the computer annually for a 365-day year?
A) $120
B) $1200
C) $15
D) $150

Accepted Answer

The answer of A 400-W computer (including its monitor) is...

Question 4

A 120-Ω laboratory resistor is rated at 0.25 W. How much current can safely flow through the resistor?
A) 30 A
B) 2.1 mA
C) 22 mA
D) 46 mA

Accepted Answer

The answer of A 120-Ω laboratory resistor is rated at...

Question 5

How much energy does a 100-W light bulb use in 8.0 hours?
A) 0.0080 kWh
B) 0.80 kWh
C) 13 kWh
D) 800 kWh

Accepted Answer

To find the energy consumed, we use the formula E = P × t, where E is energy, P is power, and t is time. Substituting the values given, we get E = 100 W × 8.0 hours = 800 Wh or 0.8 kWh (since 1 kWh = 1000 Wh). Therefore, the answer is B, 0.80 kWh.

Question 6

A toaster is rated at 800 W when operating at 120 V dc. What is the resistance of its heating element?
A) 16 Ω
B) 18 Ω
C) 6.7 Ω
D) 0.15 Ω

Accepted Answer

The answer of A toaster is rated at 800 W...

Question 7

What is the resistance of a 0.100-kW light bulb designed to be used in a 120-V circuit dc?
A) 12.0 Ω
B) 144 Ω
C) 1.2 Ω
D) 0.83 Ω

Accepted Answer

The resistance can be calculated using the formula $P = \frac{V^2}{R}$, where $P$ is the power in watts, $V$ is the voltage in volts, and $R$ is the resistance in ohms. Rearranging the formula to solve for $R$ gives $R = \frac{V^2}{P}$. Substituting the given values ($P = 100$ watts, $V = 120$ volts) into the formula yields $R = \frac{120^2}{100} = 144$ ohms.

Question 8

How much does it cost to operate a 25-W soldering iron for 8.0 hours if energy costs 8.0¢/kWh?
A) $1.50
B) 25¢
C) 16¢
D) 1.6¢

Accepted Answer

The answer of How much does it cost to operate...

Question 9

The voltage drop across a metal bar is $5.0 \mathrm {~V}$ while a current of $10.0 \mathrm {~mA}$ flows through it. How much power does this bar dissipate?
A) 50 mW
B) 500 μW
C) 2.5 kW
D) 0.50 kW

Accepted Answer

Power dissipated is calculated using the formula $ P = VI $. Substituting the given values, $ P = 5.0 \, \text{V} \times 10.0 \, \text{mA} = 50 \, \text{mW} $.

Question 10

A 200-Ω resistor is rated at 1/4 W. What is the maximum voltage that can safely be connected across it?
A) 0.71 V
B) 7.1 V
C) 50 V
D) 0.25 V

Accepted Answer

The maximum power rating of the resistor is 1/4 W or 0.25 W. Using the power formula P = V^2/R, where P is power, V is voltage, and R is resistance, we can solve for V: V = sqrt(P*R) = sqrt(0.25W * 200Ω) = sqrt(50) = 7.07 V, which is approximately 7.1 V.

Question 11

A 9.0-V battery costs $1.49 and will run a portable CD player for 6.0 hours before running down and needing to be replaced. If this battery supplies a current of 25 mA to the player, what is the cost of the energy provided by the battery in dollars per kWh?
A) $11/kWh
B) $110/kWh
C) $1100/kWh
D) $11,000/kWh

Accepted Answer

We can start by finding the total energy supplied by the battery:
E = VIt = (9.0 V)(25 mA)(6.0 hours) = 1.35 Wh
To convert this to kWh, we need to divide by 1000:
E = 1.35 Wh ÷ 1000 = 0.00135 kWh
The cost of the energy provided by the battery is the cost of the battery divided by the energy it supplies:
cost/kWh = $1.49 ÷ 0.00135 kWh = $1100/kWh
Therefore, the answer is C, $1100/kWh.

Question 12

When a 14.0-A current flows through an 8.00-Ω device for 24.0 hours, how much does this cost if energy costs $0.0900 per kW ∙ h?
A) $0.24
B) $1.04
C) $2.16
D) $3.39

Accepted Answer

First, calculate the energy used in kilowatt-hours (kWh). The power (P) in watts (W) is given by P = IV, where I is the current in amperes (A) and V is the voltage (V). Since V = IR (Ohm's Law), where R is the resistance in ohms (Ω), we find P = I^2R. For a 14.0-A current through an 8.00-Ω device, P = (14.0)^2 * 8.00 = 1568 W = 1.568 kW. The energy used over 24.0 hours is E = P * time = 1.568 kW * 24.0 h = 37.632 kWh. At $0.0900 per kWh, the cost is 37.632 kWh * $0.0900/kWh = $3.39.

Question 13

A 200-Ω resistor is rated at 1/4 W. What is the maximum current it can safely draw?
A) 0.035 A
B) 0.35 A
C) 50 A
D) 0.25 A

Accepted Answer

The maximum current can be calculated using the power formula P = I^2 * R, where P is power in watts, I is current in amperes, and R is resistance in ohms. Rearranging the formula to solve for I gives I = sqrt(P/R). Substituting the given values (P = 1/4 W = 0.25 W, R = 200 Ω) yields I = sqrt(0.25 / 200) = sqrt(0.00125) = 0.035 A.

Question 14

The nichrome heating element in an electric drier operates on 240 V and generates heat at the rate of 2.0 kW. The heating element shorts out and, in repairing it, the repairman shortens the nichrome wire by 10%. (Assume the temperature is unchanged. In reality, the resistivity of the wire will depend on its temperature.) What effect will the repair have on the power dissipated in the heating element?
A) The power is still 2.0 kW.
B) The power increases to 2.2 kW.
C) The power decreases to 1.8 kW.
D) None of the given answers is correct.

Accepted Answer

Shortening the wire by 10% reduces its resistance, since resistance is directly proportional to length. With a lower resistance, the power dissipated (P = V^2/R) increases, making 2.2 kW the correct answer due to the increase in power with decreased resistance.

Question 15

An electronic component with a 17-Ω resistor is rated for use at power levels not exceeding 14$ {~W}$ How much current can safely flow through the component?
A) 0.91 A
B) 1.21 A
C) 238 A
D) 0.22 A

Accepted Answer

The correct answer is A because the current through the resistor can be calculated using Ohm's law: I = V/R. Since the power is given by P = I^2 * R, we can solve for the current: I = sqrt(P/R) = sqrt(14 W / 17 Ω) = 0.91 A.

Question 16

A 1500-W heater is connected to a 120-V line for 2.0 hours. How much heat energy is produced?
A) 1.5 kJ
B) 3.0 kJ
C) 0.18 MJ
D) 11 MJ

Accepted Answer

The heat energy produced can be calculated using the formula: energy (in joules) = power (in watts) × time (in seconds). First, convert 2.0 hours to seconds: 2.0 hours × 3600 seconds/hour = 7200 seconds. Then, calculate the energy: 1500 W × 7200 s = 10,800,000 joules or 10.8 MJ, which is closest to option D, 11 MJ.

Question 17

A 25-W soldering iron runs on 110 V dc. What is its resistance?
A) 0.0020 Ω
B) 4.4 Ω
C) 0.48 kΩ
D) 2.8 kΩ

Accepted Answer

We can use the formula P = V^2/R, where P is the power (25 W), V is the voltage (110 V), and R is the resistance.

Rearranging the formula to solve for R, we get:

R = V^2/P = (110 V)^2 / 25 W = 4840/25 Ω = 193.6 Ω

However, this is the resistance at room temperature. As the soldering iron heats up, its resistance will increase due to the increase in temperature. So, the actual resistance of the soldering iron when in use may be slightly higher than 193.6 Ω.

Among the given choices, the closest answer to 193.6 Ω is C, 0.48 kΩ.

Question 18

A 100-W driveway light bulb is on 10 hours per day. If the power company charges 10¢ for each kilowatt-hour of electricity used, estimate the yearly cost to operate the bulb if it is used every day for a 365-day year.
A) $3.65
B) $7.30
C) $37
D) $73

Accepted Answer

First, we need to convert the wattage of the bulb to kilowatts:

100 W = 0.1 kW

Next, we can calculate the daily energy usage:

Energy = power x time

Energy = 0.1 kW x 10 hours = 1 kWh

Since the bulb is used every day for a year, we can calculate the yearly energy usage:

Yearly energy usage = daily energy usage x number of days

Yearly energy usage = 1 kWh x 365 days = 365 kWh

Finally, we can calculate the cost of operating the bulb for a year:

Cost = energy usage x cost per kWh

Cost = 365 kWh x $0.10/kWh = $36.50

Therefore, the answer is C) $37.

Question 19

The resistivity of gold is $2.44 \times 10 ^{- 8} \Omega \cdot \mathrm { m }$ at a temperature of 20°C. A gold wire, 1.8 mm in diameter and 14 cm long, carries a current of 480 mA. What power is dissipated in this wire at 20°C? A) 0.31 mW B) 0.077mW C) 0.14 mW D) 0.19 mW E) 0.25 mW

Accepted Answer

The power dissipated in a wire is given by the equation:$$P = I^2R$$where:* P is the power in watts* I is the current in amperes* R is the resistance in ohmsThe resistance of a wire is given by the equation:$$R = \rho \frac{L}{A}$$where:* R is the resistance in ohms* ρ is the resistivity of the material in ohm-meters* L is the length of the wire in meters* A is the cross-sectional area of the wire in square metersIn this problem, we are given the following information:* ρ = $2.44 \times 10 ^{- 8} \Omega \cdot \mathrm { m }$* L = 14 cm = 0.14 m* d = 1.8 mm = 0.0018 m* I = 480 mA = 0.48 AFirst, we need to calculate the cross-sectional area of the wire:$$A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.0018 \mathrm { m}}{2}\right)^2 = 2.54 \times 10^{-6} \mathrm { m}^2$$Now we can calculate the resistance of the wire:$$R = \rho \frac{L}{A} = (2.44 \times 10^{-8} \Omega \cdot \mathrm { m}) \frac{0.14 \mathrm { m}}{2.54 \times 10^{-6} \mathrm { m}^2} = 0.14 \Omega$$Finally, we can calculate the power dissipated in the wire:$$P = I^2R = (0.48 \mathrm { A})^2 (0.14 \Omega) = 0.031 \mathrm { W} = 31 \mathrm { mW}$$Therefore, the correct answer is A) 0.31 mW.

Question 20

A battery is rated such that it provides $1.0 \mathrm {~mW}$ of power at $9.0 \mathrm {~V}$ when fully charged. How much current can it deliver?
A) 0.11 mA
B) 81 kA
C) 0.33 mA
D) 9.1 kA

Accepted Answer

The answer of A battery is rated such that it...

Quiz 22: Current and Resistance

A simple circuit has a total resistance of 30 Ω. If a 2.0-A current is maintained in this circuit, how much energy is dissipated in this circuit in 4.0 seconds?

A battery is rated at 12 V and 160 A-h. How much energy does this battery store?

A 400-W computer (including its monitor) is turned on for 8.0 hours per day. If electricity costs 10¢ per kWh, how much does it cost to run the computer annually for a 365-day year?

A 120-Ω laboratory resistor is rated at 0.25 W. How much current can safely flow through the resistor?

How much energy does a 100-W light bulb use in 8.0 hours?

A toaster is rated at 800 W when operating at 120 V dc. What is the resistance of its heating element?

What is the resistance of a 0.100-kW light bulb designed to be used in a 120-V circuit dc?

How much does it cost to operate a 25-W soldering iron for 8.0 hours if energy costs 8.0¢/kWh?

The voltage drop across a metal bar is $5.0 \mathrm {~V}$ while a current of $10.0 \mathrm {~mA}$ flows through it. How much power does this bar dissipate?

A 200-Ω resistor is rated at 1/4 W. What is the maximum voltage that can safely be connected across it?

A 9.0-V battery costs $1.49 and will run a portable CD player for 6.0 hours before running down and needing to be replaced. If this battery supplies a current of 25 mA to the player, what is the cost of the energy provided by the battery in dollars per kWh?

When a 14.0-A current flows through an 8.00-Ω device for 24.0 hours, how much does this cost if energy costs $0.0900 per kW ∙ h?

A 200-Ω resistor is rated at 1/4 W. What is the maximum current it can safely draw?

An electronic component with a 17-Ω resistor is rated for use at power levels not exceeding 14 ${~W}$ How much current can safely flow through the component?

A 1500-W heater is connected to a 120-V line for 2.0 hours. How much heat energy is produced?

A 25-W soldering iron runs on 110 V dc. What is its resistance?

A 100-W driveway light bulb is on 10 hours per day. If the power company charges 10¢ for each kilowatt-hour of electricity used, estimate the yearly cost to operate the bulb if it is used every day for a 365-day year.

The resistivity of gold is $2.44 \times 10 ^{- 8} \Omega \cdot \mathrm { m }$ at a temperature of 20°C. A gold wire, 1.8 mm in diameter and 14 cm long, carries a current of 480 mA. What power is dissipated in this wire at 20°C?

A battery is rated such that it provides $1.0 \mathrm {~mW}$ of power at $9.0 \mathrm {~V}$ when fully charged. How much current can it deliver?