Question 1

A light bulb operating at a dc voltage of 120 V has a resistance of 200 Ω. How much power is dissipated in this bulb?
A)100 W
B)60 W
C)72 W
D)14 mW
E)7.2 W

Accepted Answer

Using the formula P = V^2/R, we get P = (120 V)^2 / 200 Ω which equals 72 W. Therefore, the correct answer is C.

Question 2

When 5.00 A is flowing through an 10.0-Ω device, how much power is being dissipated in the device?
A)50.0 W
B)250 W
C)500 W
D)2.50 kW

Accepted Answer

The answer of When 5.00 A is flowing through an...

Question 3

When a potential difference is applied across a piece of wire made of metal A, a $5.0 - \mathrm { mA }$ current flows. If the metal-A wire is replaced with a wire made of metal B having twice the diameter of the metal-A wire, how much current will flow through the metal-B wire? The lengths of both wires are the same, and the voltage difference remains unchanged. The resistivity of metal A is 1.68 × 10^-8 ? ? m, and the resistivity of metal B is 1.59 × 10^-8 ? ? m. A)21 mA B)19 mA C)11 mA D)5.3 mA

Accepted Answer

The current through a wire is given by the equation:$$I = \frac{V}{R}$$where:* I is the current in amps* V is the voltage in volts* 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, the voltage and length of the wire are the same for both wires. Therefore, the only factor that will affect the current is the resistivity of the material and the cross-sectional area of the wire.The resistivity of metal B is slightly less than the resistivity of metal A. This means that metal B will have a lower resistance than metal A.The cross-sectional area of the metal-B wire is twice the cross-sectional area of the metal-A wire. This means that the metal-B wire will have a lower resistance than the metal-A wire.Therefore, the current through the metal-B wire will be greater than the current through the metal-A wire.The correct answer is A: 21 mA.

Question 4

How much current will flow through a $40.0 - \mathrm { m }$ length of metal wire with a radius of $0.0 \mathrm {~mm}$ if it is connected to a power source supplying $16.0 \mathrm {~V} ?$ The resistivity of the metal is 1.68 × 10^-8 ? ? m. A)0 A B)0 x 10⁸ A C)0 nA D)0 A.

Accepted Answer

The answer of How much current will flow through a...

Question 5

A light bulb operating at a dc voltage of 120 V has a power rating of 60 W. How much current is flowing through this bulb?
A)1.5 A
B)2.5 A
C)0.50 A
D)2.0 A
E)1.0 A

Accepted Answer

Using the formula: power = voltage x current, we can rearrange it to solve for the current.
Current = power / voltage
Current = 60 W / 120 V
Current = 0.5 A

Question 6

The power rating of a 400-? resistor is 0.800 W.
(a)What is the maximum safe voltage across this resistor?
(b)What is the maximum current the resistor can safely draw?

Accepted Answer

The answer of The power rating of a 400-? resistor...

Question 7

A flashlight draws 0.133 A from a 3.0-V battery pack. In 2.0 minutes (a)how much charge flows from the battery, (b)how much energy does the battery supply, and (c)how many electrons have passed any point in the circuit every second?

Accepted Answer

The answer of A flashlight draws 0.133 A from a...

Question 8

The resistivity of a 1.0 m long copper wire is 1.72 × 10^-8 Ω ∙ m and its cross sectional area is 2.0 × 10^-6 m². If the wire carries a current of 0.20 A, what is the voltage across the wire? A)17 mV B)0.90 mV C)1.7 mV D)10 mV E)90 mV

Accepted Answer

The formula to calculate voltage (V) is:

V = IR

where I is the current and R is the resistance.

The resistance of the wire can be calculated as:

R = ρL/A

where ρ is the resistivity, L is the length of the wire, and A is the cross-sectional area.

Substituting the given values, we get:

R = (1.72 × 10^-8 Ω ∙ m)(1.0 m)/(2.0 × 10^-6 m^2)
R = 8.6 × 10^-3 Ω

Now, substituting the values for I and R in the formula for voltage, we get:

V = (0.20 A)(8.6 × 10^-3 Ω)
V = 1.72 × 10^-3 V or 1.7 mV

Therefore, the voltage across the wire is 1.7 mV, which is closest to option C.

Question 9

A resistor operated at 120 V dc is rated at 1.40 kW.
(a)What is the normal operating current through the resistor?
(b)What is the resistance of the resistor?

Accepted Answer

The answer of A resistor operated at 120 V dc...

Question 10

If the power rating of a 400-Ω resistor is 0.800 W, what is the maximum voltage that can safely be connected across the resistor?
A)110 V
B)17.9 V
C)170 V
D)1.80 V

Accepted Answer

The maximum voltage that can be safely connected across the resistor can be found using the formula P=V^2/R, where P is the power rating of the resistor, V is the voltage, and R is the resistance. Solving for V, we get V=sqrt(P*R)=sqrt(0.8*400)=17.9 V. Therefore, the best choice is B) 17.9 V.

Question 11

A 200-W light bulb is connected across a 110-V dc power supply. What current will flow through this bulb?
A)0.90 A
B)0.36 A
C)0 A
D)0.60 A
E)1.8 A

Accepted Answer

The answer of A 200-W light bulb is connected across...

Question 12

The power rating of a 400-Ω resistor is 0.25 W. What is the maximum voltage you can safely connect across its ends?
A)10 V
B)20 V
C)30 V
D)40 V
E)50 V

Accepted Answer

The maximum voltage can be calculated using the power 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 $V$, we get $V = \sqrt{P \times R}$. Substituting $P = 0.25$ W and $R = 400$ Ω, we get $V = \sqrt{0.25 \times 400} = \sqrt{100} = 10$ V.

Question 13

If the power rating of a 400-Ω resistor is 0.80 W, what is the maximum current it can safely draw?
A)45 mA
B)18 mA
C)4.4 mA
D)2.0 mA
E)320 mA

Accepted Answer

The answer of If the power rating of a 400-Ω...

Question 14

A 150-W light bulb is designed to operate at 110 V dc. How much current does it draw?
A)0.73 A
B)1.4 A
C)2.0 A
D)15 A

Accepted Answer

Using the formula P = VI, where P is power, V is voltage, and I is current, we can solve for I:

150 W = 110 V x I

I = 150 W / 110 V

I = 1.36 A

The closest answer choice is B, 1.4 A.

Question 15

A tube of mercury with resistivity 9.84 × 10^-7 Ω ∙ m has a uniform electric field of 23 N/C inside the mercury. How much current is flowing in the tube, if the radius of the tube is 0.495 mm? A)18 A B)180 A C)29 A D)280 A

Accepted Answer

The answer of A tube of mercury with resistivity 9.84...

Question 16

A 100-W resistance heater is connected to a 110-V dc source. What current flows through the heater?
A)1.1 A
B)2.2 A
C)3.3 A
D)4.4 A
E)0.91 A

Accepted Answer

Using Ohm's Law, we can find the current:

I = V/R

where V is the voltage (110 V) and R is the resistance (100 ohms):

I = 110/100

I = 1.1 A

Therefore, the answer is (E) 0.91 A.

Question 17

A resistance heater is rated at 1200 W when operating at 110 V dc. What current will it draw?
A)0.090 A
B)1.0 A
C)11 A
D)12 A

Accepted Answer

We can use Ohm's Law: V = IR, where V is voltage, I is current, and R is resistance. We can rearrange this equation to solve for I: I = V/R. 
The resistance of the heater can be found by using the power formula: P = IV, where P is power, I is current, and V is voltage. Rearranging to solve for resistance, we get: R = V/I. 
Substituting in the given values: 
Resistance = V^2/P = (110 V)^2 / 1200 W = 10.17 Ω 
Current = V/R = 110 V / 10.17 Ω = 10.8 A 
However, this calculation assumes that the heater operates at 100% efficiency, which is not realistic. In reality, some of the electrical energy is converted to heat, while some is lost to other forms of energy (such as radiation). Therefore, the actual current drawn by the heater will be slightly less than 10.8 A, but still close to this value. The closest choice is C) 11 A.

Question 18

An instrument is rated at 250 W if it is connected across a 120-V dc power supply.
(a)What current does it draw under normal operation?
(b)What is its resistance?
(c)How many kilowatt-hours does it use in a day if it is left on all the time?

Accepted Answer

The answer of An instrument is rated at 250 W...

Question 19

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 20

A 100-W light bulb is operated by a 110-V dc source. What is the resistance of this bulb? A)100 Ω B)8.0 × 10^-3 Ω C)6.0 × 10^-3 Ω D)120 Ω E)240 Ω

Accepted Answer

The answer of A 100-W light bulb is operated by...

Quiz 22: Current and Resistance

A light bulb operating at a dc voltage of 120 V has a resistance of 200 Ω. How much power is dissipated in this bulb?

When 5.00 A is flowing through an 10.0-Ω device, how much power is being dissipated in the device?

How much current will flow through a 40.0−m40.0 - \mathrm { m }40.0−m length of metal wire with a radius of 0.0 mm0.0 \mathrm {~mm}0.0 mm if it is connected to a power source supplying 16.0 V?16.0 \mathrm {~V} ?16.0 V? The resistivity of the metal is 1.68 × 10-8 ? ? m.

A light bulb operating at a dc voltage of 120 V has a power rating of 60 W. How much current is flowing through this bulb?

The power rating of a 400-? resistor is 0.800 W. (a)What is the maximum safe voltage across this resistor? (b)What is the maximum current the resistor can safely draw?

A flashlight draws 0.133 A from a 3.0-V battery pack. In 2.0 minutes (a)how much charge flows from the battery, (b)how much energy does the battery supply, and (c)how many electrons have passed any point in the circuit every second?

The resistivity of a 1.0 m long copper wire is 1.72 × 10-8 Ω ∙ m and its cross sectional area is 2.0 × 10-6 m2. If the wire carries a current of 0.20 A, what is the voltage across the wire?

A resistor operated at 120 V dc is rated at 1.40 kW. (a)What is the normal operating current through the resistor? (b)What is the resistance of the resistor?

If the power rating of a 400-Ω resistor is 0.800 W, what is the maximum voltage that can safely be connected across the resistor?

A 200-W light bulb is connected across a 110-V dc power supply. What current will flow through this bulb?

The power rating of a 400-Ω resistor is 0.25 W. What is the maximum voltage you can safely connect across its ends?

If the power rating of a 400-Ω resistor is 0.80 W, what is the maximum current it can safely draw?

A 150-W light bulb is designed to operate at 110 V dc. How much current does it draw?

A tube of mercury with resistivity 9.84 × 10-7 Ω ∙ m has a uniform electric field of 23 N/C inside the mercury. How much current is flowing in the tube, if the radius of the tube is 0.495 mm?

A 100-W resistance heater is connected to a 110-V dc source. What current flows through the heater?

A resistance heater is rated at 1200 W when operating at 110 V dc. What current will it draw?

An instrument is rated at 250 W if it is connected across a 120-V dc power supply. (a)What current does it draw under normal operation? (b)What is its resistance? (c)How many kilowatt-hours does it use in a day if it is left on all the time?

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

A 100-W light bulb is operated by a 110-V dc source. What is the resistance of this bulb?

How much current will flow through a $40.0 - \mathrm { m }$ length of metal wire with a radius of $0.0 \mathrm {~mm}$ if it is connected to a power source supplying $16.0 \mathrm {~V} ?$ The resistivity of the metal is 1.68 × 10^-8 ? ? m.

The resistivity of a 1.0 m long copper wire is 1.72 × 10^-8 Ω ∙ m and its cross sectional area is 2.0 × 10^-6 m². If the wire carries a current of 0.20 A, what is the voltage across the wire?

A tube of mercury with resistivity 9.84 × 10^-7 Ω ∙ m has a uniform electric field of 23 N/C inside the mercury. How much current is flowing in the tube, if the radius of the tube is 0.495 mm?