Question 1

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?

Accepted Answer

Using the formula P = V^2 / R, we can solve for the maximum voltage:

0.800 W = V^2 / 400 Ω
V^2 = 320
V = 17.9 V (rounded to one decimal place)

Therefore, the maximum voltage that can safely be connected across the resistor is 17.9 V. Choice B is correct.

Question 2

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

Accepted Answer

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

Question 3

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

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 4

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

Accepted Answer

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

Question 5

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?

Accepted Answer

We can use the formula P = V^2 / R to calculate the power dissipated by the bulb.

P = (120 V)^2 / 200 Ω
P = 14,400 / 200
P = 72 W

Therefore, the power dissipated by the bulb is 72 W.

Question 6

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?

Accepted Answer

The current flowing in the tube can be calculated using the formula:I = (E * π * r^2) / ρwhere:I is the current in amperes (A)E is the electric field strength in volts per meter (V/m)r is the radius of the tube in meters (m)ρ is the resistivity of the material in ohm-meters (Ω∙m)Plugging in the given values, we get:I = (23 V/m * π * (0.495 mm)^2) / (9.84 × 10^-7 Ω∙m)I ≈ 18 ATherefore, the correct answer is A) 18 A.

Question 7

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

Accepted Answer

The current can be calculated using Ohm's Law, which states I = P/V, where I is the current, P is the power (100 W), and V is the voltage (110 V). Thus, I = 100 W / 110 V = 0.91 A.

Question 8

When a potential difference is applied across a piece of wire made of metal A,a 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 Ω ∙ m,and the resistivity of metal B is 1.59 × 10^-8 Ω ∙ m.

Accepted Answer

The answer of When a potential difference is applied across...

Question 9

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

Accepted Answer

Using the formula P = VI, we can rearrange it to solve for I:
I = P / V

Plugging in the given values, we get:

I = 1200 W / 110 V 
I = 10.91 A

However, since the question asks for the current to the nearest hundredth of an ampere, we round off the answer to two decimal places:

I = 11 A

Therefore, the correct choice is C.

Question 10

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

Accepted Answer

The power rating of a resistor is given by the equation P = I^2 * R, where P is the power in watts, I is the current in amps, and R is the resistance in ohms. In this case, we have a 400-Ω resistor with a power rating of 0.80 W. To find the maximum current it can safely draw, we can rearrange the equation to solve for I:I = sqrt(P / R) = sqrt(0.80 W / 400 Ω) = 0.045 A = 45 mATherefore, the maximum current the resistor can safely draw is 45 mA.

Question 11

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

Accepted Answer

The energy used by the soldering iron is:$$E = Pt = (25 	ext{ W})(8.0 	ext{ h}) = 200 	ext{ Wh} = 0.200 	ext{ kWh}$$The cost of this energy is:$$C = E 	imes r = (0.200 	ext{ kWh})(8.0¢/kWh) = 1.6¢$$

Question 12

How much current will flow through a 40.0mm. Length of metal wire with a radius of 0.0mm. If it is connected to a power source supplying 16.0V? The resistivity of the metal is 1.68 × 10^-8 Ω ∙ m.

Accepted Answer

The radius of the wire is given as 0.0 mm, which means the wire effectively has no cross-sectional area through which current can flow. Regardless of the voltage applied or the material's resistivity, if there's no pathway for the current (due to the absence of a cross-sectional area), no current can flow.

Question 13

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?

Accepted Answer

Using the formula P=VI, we can rearrange to solve for current, I = P/V. Plugging in the given values, we get I = 60/120 = 0.50 A.

Question 14

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

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 15

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

Accepted Answer

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

Question 16

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

Accepted Answer

Using the formula P = VI, we can rearrange it to solve for current:
I = P/V
Substituting the given values, we get:
I = 150 W / 110 V = 1.36 A
Therefore, the closest answer choice is B) 1.4 A.

Question 17

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?

Accepted Answer

The voltage across the wire can be calculated using Ohm's Law, $V = IR$, where $I$ is the current and $R$ is the resistance. The resistance $R$ can be found using the formula $R = \rho \frac{L}{A}$, where $\rho$ is the resistivity, $L$ is the length, and $A$ is the cross-sectional area. Substituting the given values, $R = 1.72 \times 10^{-8} \Omega \cdot m \times \frac{1.0 m}{2.0 \times 10^{-6} m^2} = 8.6 \times 10^{-3} \Omega$. Then, $V = 0.20 A \times 8.6 \times 10^{-3} \Omega = 1.72 \times 10^{-3} V = 1.7 mV$.

Question 18

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

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 19

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

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 20

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

Accepted Answer

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

Quiz 21: Electric Current and Direct-Current Circuits