Question 1

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

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Question 2

A 2.0 mm diameter wire of length 20 m has a resistance of 0.25 Ω. What is the resistivity of the wire?

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Question 3

If an all electric house uses 2400 kWh in a month, what will the amount of the bill for electricity be at a rate of 9.5 cents per kWh?

Accepted Answer

The bill is calculated by multiplying the total kWh used by the rate per kWh: 2400 kWh * $0.095 = $228.

Question 4

A 110-V hair dryer is rated at 1200 W. What current will it draw?

Accepted Answer

The current can be calculated using the formula: Power (P) = Voltage (V) x Current (I). Rearranging gives I = P/V. Substituting the given values, I = 1200 W / 110 V = 10.91 A, which rounds to 11 A.

Question 5

A Nichrome wire is used as a heating element in a toaster. From the moment the toaster is first turned on to the moment the wire reaches it maximum temperature, the current in the wire drops by 20% from its initial value. What is the temperature change in the wire? The temperature coefficient for Nichrome is α = 0.0004 (°C)^-1.

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Question 6

A 24-kW electric furnace is connected to a 240-V line. What is the resistance of the furnace?

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Question 7

When a 60-W lightbulb is operating on a 120-V household line, what current does it draw?

Accepted Answer

The power (P) of the lightbulb is 60 W and the voltage (V) is 120 V. Using the formula P = IV, where I is the current, we can solve for I: I = P/V = 60 W / 120 V = 0.5 A.

Question 8

A 100-W driveway light bulb is on 10 hours per day. Assuming the power company charges 10 cents for each kilowatt-hour of electricity used, estimate the annual cost to operate the bulb.

Accepted Answer

The bulb uses 100 watts or 0.1 kilowatts. Over 10 hours, it uses 1 kWh per day. At 10 cents per kWh, it costs $0.10 per day. Over 365 days, the annual cost is $36.50.

Question 9

A light bulb operating at a voltage of 120 V has a resistance of 200 Ω. What is the power?

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Question 10

A 100-W light bulb is connected to a 110-V source. What is the resistance of this bulb?

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Question 11

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

Accepted Answer

A 100-W light bulb uses 0.1 kW. Over 8 hours, it uses 0.1 kW * 8 hours = 0.80 kWh.

Question 12

A platinum wire is used to determine the melting point of indium. The resistance of the platinum wire is 2.000 Ω at 20°C and increases to 3.072 Ω as indium starts to melt. What is the melting point of indium? (The temperature coefficient of resistivity for platinum is 3.9 × 10^-3/C°.)

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Question 13

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

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Question 14

Calculate the current through a 10.0-m long 22 gauge (the radius is 0.321 mm) nichrome wire if it is connected to a 12.0-V battery. (The resistivity of nichrome is 100 × 10^-⁸ Ω∙m.)

Accepted Answer

The resistance of the wire is:$$R = ho \frac{L}{A} = ho \frac{4L}{\pi d^2}$$$$= (100 	imes 10^{-8} \Omega \cdot m) \frac{4(10.0 \ m)}{\pi (0.321 	imes 10^{-3} \ m)^2}$$$$= 0.388 \Omega$$The current through the wire is:$$I = \frac{V}{R} = \frac{12.0 \ V}{0.388 \Omega} = 30.9 \ A$$

Question 15

The resistivity of a 1.0 m long wire is 1.72 × 10^-8 Ωm 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 resistance of the wire can be calculated using the formula:

R = ρL/A

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

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

Using Ohm's Law (V = IR), we can find the voltage across the wire:

V = IR = (0.20 A)(8.6 Ω) = 1.72 V

However, the question asks for the voltage in millivolts (mV), so we need to convert:

1.72 V = 1720 mV

Therefore, the answer is 1.7 mV, which is choice C.

Question 16

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

Accepted Answer

The energy stored in a battery can be calculated using the formula: Energy (in watt-hours) = Voltage (V) × Capacity (A-h). For this battery, Energy = 12 V × 160 A-h = 1920 watt-hours. Converting watt-hours to joules (1 watt-hour = 3600 joules), the energy is 1920 × 3600 = 6,912,000 joules, or 6.9 MJ.

Question 17

A carbon resistor has a resistance of 18 Ω at a temperature of 20°C. What is its resistance at a temperature of 120°C? (The temperature coefficient of resistivity for carbon is -5.0 × 10^-⁴ /C°.)

Accepted Answer

The resistance of a carbon resistor changes with temperature according to the following equation:$$R_t = R_0 [1 + \alpha (t - t_0)]$$where:* $$R_t$$ is the resistance at temperature t* $$R_0$$ is the resistance at temperature $$t_0$$* $$\alpha$$ is the temperature coefficient of resistivityIn this problem, we are given that $$R_0 = 18 \Omega$$, $$t_0 = 20°C$$, $$t = 120°C$$, and $$\alpha = -5.0 × 10^{-4} /C°$$. Plugging these values into the equation, we get:$$R_t = 18 \Omega [1 + (-5.0 × 10^{-4} /C°) (120°C - 20°C)]$$$$R_t = 18 \Omega [1 + (-5.0 × 10^{-4} /C°) (100°C)]$$$$R_t = 18 \Omega [1 - 0.05]$$$$R_t = 17.1 \Omega$$Therefore, the resistance of the carbon resistor at a temperature of 120°C is 17.1 Ω, which is closest to choice B) 17 Ω.

Question 18

The temperature coefficient of resistivity for copper is 0.0068°C^-1. If a copper wire has a resistance of 104 Ω at 20°C, what is its resistance 80°C?

Accepted Answer

The resistance of a conductor at a temperature $T_2$ can be calculated using the formula $R_2 = R_1[1 + \alpha(T_2 - T_1)]$, where $R_1$ is the initial resistance, $R_2$ is the final resistance, $\alpha$ is the temperature coefficient of resistivity, $T_1$ is the initial temperature, and $T_2$ is the final temperature. Plugging in the given values: $R_2 = 104[1 + 0.0068(80 - 20)] = 104[1 + 0.0068(60)] = 104[1 + 0.408] = 104 \times 1.408 = 146.432 \Omega$. Rounding to the nearest whole number gives 146 Ω.

Question 19

A 120-m long copper wire (resistivity 1.68 × 10^-⁸ Ω∙m) has resistance 6.0 Ω. What is the diameter of the wire?

Accepted Answer

The resistance R of a wire is given by R = ρL/A, where ρ is resistivity, L is length, and A is cross-sectional area. Solving for A gives A = ρL/R. The area A of a circle is π(d/2)^2, where d is the diameter. Solving for d gives d = 2√(A/π). Substituting the given values, we find d ≈ 0.65 mm.

Question 20

The resistance of a 100-cm wire of cross sectional area 2 × 10^-6 m² is 400 Ω. What is the resistivity of the material of this wire?

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Quiz 25: Electric Currents and Resistance