Question 1

Consider some material that has been cooled until it has become a superconductor.If it is cooled even further its resistance will:

Accepted Answer

Once a material becomes a superconductor, its resistance drops to zero. Cooling it further won't change its resistance, as it cannot be less than zero. However, none of the provided options accurately describe this scenario, making "None of the above" the correct choice.

Question 2

An 8.00-W resistor is dissipating 100 watts.What are the current through it and the difference of potential across it?

Accepted Answer

Using the formula P=I^2*R and P=V^2/R, we can find the current and voltage.

P=I^2*R 
100 = I^2 * 8 
I = sqrt(100/8) = 3.54 A

P=V^2/R 
100 = V^2/8 
V = sqrt(100*8) = 28.3 V

Therefore, the current through the resistor is 3.54 A and the difference of potential across it is 28.3 V. The answer is C.

Question 3

A hot water heater operating at 240 V supplies  to a quantity of water to warm it to the desired temperature of  .What is the cost of the energy if the rate is $0.131/kWh?

Accepted Answer

The answer of A hot water heater operating at 240...

Question 4

A superconducting wire's chief characteristic is which of the following?

Accepted Answer

The chief characteristic of a superconducting wire is that it has zero electrical resistance, meaning that it can conduct electricity without any loss of energy due to resistance. Superconducting wires also typically have a smaller cross sectional area than traditional wires, as a larger cross sectional area can lead to decreased superconducting properties. The temperature of a superconducting wire is also important, as it needs to be kept below a certain critical temperature in order to maintain its superconducting properties.

Question 5

A resistor is made of a material that has a resistivity that is proportional to the current going through it.If the voltage across the resistor is doubled,what happens to the power dissipated by it?

Accepted Answer

The answer of A resistor is made of a material...

Question 6

A high-voltage transmission line carries 1 000 A at 700 kV for a distance of 100 miles.If the resistance in the wire is 1 W/mile,what is the power loss due to resistive losses?

Accepted Answer

Using the formula for power loss in a transmission line due to resistance:
P = I^2*R*L
where I = current, R = resistance per unit length, and L = length of the transmission line

Converting the distance to meters:
L = 100 miles * 1609.34 meters/mile = 160,934 meters

Substituting the given values:
P = (1000 A)^2 * 1 W/mile * 160,934 meters = 160,934,000 W = 160 MW

Therefore, the power loss due to resistive losses is 100 MW, which is answer choice D.

Question 7

When a superconductor's temperature drops below the critical temperature,its resistance:

Accepted Answer

When the temperature of a superconductor drops below the critical temperature, its resistance drops to zero. This is because at temperatures below the critical temperature, superconductors exhibit zero electrical resistance and can conduct electricity with 100% efficiency. Therefore, the correct answer is C, drops to zero.

Question 8

If a light bulb has half the resistance of a 100-W lightbulb,what would be its wattage? Assume both bulbs are attached to the same 120-V circuit.

Accepted Answer

The power (wattage) of a bulb is given by P = V^2/R. If the resistance is halved, the power will double, so a bulb with half the resistance of a 100-W bulb will have a wattage of 200 W.

Question 9

When current is flowing in a superconductor,which statement is not true?

Accepted Answer

In a superconductor, a battery is not needed to keep the current going as there is no resistance to impede the flow of electrons. Once the current is started, it will continue to flow indefinitely on its own without any external power source.

Question 10

Suppose that a voltage surge produces 140 V for a moment in a 120-V line.What will temporarily be the output of a 100-W light bulb assuming its resistance does not change?

Accepted Answer

The power output of a light bulb is given by P = V^2 / R. Initially, P = 100 W at 120 V, so R = V^2 / P = 144 ohms. With a surge to 140 V, the new power output is P = 140^2 / 144 = 136 W.

Question 11

An electric car is designed to run off a bank of 12-V batteries with total energy storage of 3.0 ´ 10⁷J.If the electric motor draws 6 000 W,what current will be delivered to the motor?

Accepted Answer

Power = Voltage x Current

Therefore, current = power/voltage

Voltage = total energy storage/number of batteries

Voltage = 3.0 x 10^17 J / 12V

Voltage = 2.5 x 10^16 batteries

Current = 6,000 W / 2.5 x 10^16 batteries = 500 A

Question 12

A light bulb with a tungsten filament is attached to a source of variable voltage.As the voltage is increased on the bulb,

Accepted Answer

According to Ohm's Law, as the voltage increases, the current also increases, and since the resistance of the tungsten filament remains constant, the increase in current results in an increase in power dissipated by the bulb. Therefore, to keep the power dissipated constant, the resistance of the bulb must increase as the voltage increases, leading to a decrease in current. This relationship is described by the equation P = V^2/R.

Question 13

When the voltage across a nonohmic resistor is doubled,the current through it triples.What happens to the power delivered to this resistor?

Accepted Answer

The power delivered to a resistor is given by $P = IV$, where $I$ is the current and $V$ is the voltage. If the voltage is doubled and the current triples, the new power is $P_{new} = 3I \times 2V = 6IV$, which is 6 times the original power.

Question 14

A wire has resistance R.A second wire has twice the length,twice the diameter,and twice the resistivity of the first wire.What is its resistance?

Accepted Answer

Resistance of a wire is directly proportional to its length and resistivity, and inversely proportional to its cross-sectional area (diameter squared). 
So, for the second wire: 
Length = 2L 
Diameter = 2d 
Resistivity = 2ρ 
Using the formula for resistance, R = ρL/A, 
we get R2 = (2ρ)(2L)/[(π(2d/2)^2)] 
     = R(2ρL/πd^2) 
     = 2R (resistance of the first wire) 
Therefore, the resistance of the second wire is twice that of the first wire, which is equal to R.

Question 15

Some superconductors are capable of carrying a very large quantity of current.If the measured current is 1.00 x 10⁵A,how many electrons are moving through the superconductor per second? (e = 1.60 x 10^-19 C)

Accepted Answer

The number of electrons per second is calculated by dividing the current by the charge of an electron: (1.00 x 10^5 A) / (1.60 x 10^-19 C/electron) = 6.25 x 10^23 electrons/second.

Question 16

A wire 1 mm in diameter is connected to one end of a wire of the same material 2 mm in diameter of twice the length.A voltage source is connected to the wires and a current is passed through the wires.If it takes time T for the average conduction electron to traverse the 1-mm wire,how long does it take for such an electron to traverse the 2-mm wire?

Accepted Answer

The answer of A wire 1 mm in diameter is...

Question 17

An electric car is designed to run off a bank of 12-V batteries with total energy storage of 3.0 ´ 10⁷ J.If the electric motor draws 6 000 W in moving the car at a steady speed of 10 m/s,how far will the car go before it is "out of juice?"

Accepted Answer

The total energy available is 3.0 x 10^7 J. The power consumption is 6000 W, which means the car uses 6000 J of energy per second. The time it can run is 3.0 x 10^7 J / 6000 J/s = 5000 s. At a speed of 10 m/s, the distance covered is 10 m/s * 5000 s = 50,000 m, which is 50 km.

Question 18

Superconductivity was discovered by:

Accepted Answer

Superconductivity was discovered by Heike Kamerlingh Onnes in 1911.

Question 19

A solar panel measures 80 cm ´ 50 cm.In direct sunlight,the panel delivers 3.2 A at 15 V.If the intensity of sunlight is 1 000 W/m²,what is the efficiency of the solar panel in converting solar energy into electrical energy?

Accepted Answer

First, let's calculate the area of the solar panel:

Area = length × width = 0.8 m × 0.5 m = 0.4 m²

Next, let's calculate the power output of the solar panel:

Power = current × voltage = 3.2 A × 15 V = 48 W

Now, we can calculate the intensity of sunlight on the solar panel:

Intensity = Power / Area = 48 W / 0.4 m² = 120 W/m²

Finally, we can calculate the efficiency of the solar panel:

Efficiency = (Power output / Intensity of sunlight) × 100% = (48 W / 120 W) × 100% = 40%

Therefore, the efficiency of the solar panel is 40%. However, since none of the answer choices match this result, we need to round it to the nearest option, which is C (12%).

Quiz 17: Current and Resistance