Question 1

A rod (length = 80 cm) with a rectangular cross section (1.5 mm * 2.0 mm) has a resistance of 0.20 $\Omega$.What is the resistivity of the material used to make the rod?&#10;A)6.0 * 10-7 $\Omega$ . m&#10;B)3.8 * 10-7 $\Omega$ . m&#10;C)7.5 * 10-$\Omega$ . m&#10;D)3.0 * 10-7 $\Omega$ . m&#10;E)4.8 * 10-7 $\Omega$ . m

Accepted Answer

7.5 * 10-$\Omega$ . m&#10;

Question 2

If a mile of 24-gauge copper wire has a resistance of 0.14 k$\Omega$ and the resistivity of copper is 1.7 * 10-8 $\Omega$.m, what is the diameter of the wire? (1 mile = 1.6 km)&#10;A)0.40 mm&#10;B)0.50 mm&#10;C)0.63 mm&#10;D)0.80 mm&#10;E)0.25 mm

Accepted Answer

0.50 mm&#10;

Question 3

How much energy is dissipated as heat during a two-minute time interval by a 1.5-k $\Omega$ resistor which has a constant 20-V potential difference across its leads?&#10;A)58 J&#10;B)46 J&#10;C)32 J&#10;D)72 J&#10;E)16 J

Accepted Answer

32 J&#10;

Question 4

How many electrons pass through a 20-$\Omega$ resistor in 10 minutes if there is a potential drop of 30 volts across it?&#10;A)5.6 *1021&#10;B)7.5 *1021&#10;C)9.4 * 1021&#10;D)1.1 * 1021&#10;E)3.8 *1021

Accepted Answer

First, calculate the current using Ohm's Law: I = V/R = 30V/20Ω = 1.5A. Then, find the total charge: Q = I * t = 1.5A * 600s = 900C. Finally, calculate the number of electrons: Number of electrons = Q/e = 900C / (1.6 * 10^-19 C/electron) ≈ 5.6 * 10^21 electrons.

Question 5

Most telephone cables are made of copper wire of either 24 or 26 gauge.If the resistance of 24-gauge wire is 137 $\Omega$ /mile and the resistance of 26-gauge wire is 220 $\Omega$ /mile, what is the ratio of the diameter of 24-gauge wire to that of 26-gauge wire?

A)1.6
B)1.3
C)0.62
D)0.79
E)0.88

Accepted Answer

The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. Let's use the formula R = ρL/A, where ρ is the resistivity of copper (1.72 x 10^-8 Ωm), L is the length of the wire (1 mile), and A is the cross-sectional area of the wire (πr^2).

For the 24-gauge wire, we have R = 137 Ω/mile, so ρL/A = 137. Solving for A, we get A = πr^2 = ρL/137 = (1.72 x 10^-8 Ωm)(1609 m)/(137 Ω) = 2.01 x 10^-6 m^2. Taking the square root to find the diameter, we get d = 2r = 0.511 mm.

For the 26-gauge wire, we have R = 220 Ω/mile, so ρL/A = 220. Solving for A, we get A = πr^2 = ρL/220 = (1.72 x 10^-8 Ωm)(1609 m)/(220 Ω) = 1.26 x 10^-6 m^2. Taking the square root to find the diameter, we get d = 2r = 0.450 mm.

The ratio of the diameters is d24/d26 = 0.511/0.450 = 1.135. However, we're looking for the ratio of the gauges, not the diameters, so we need to take the inverse of this ratio:

d26/d24 = 1/1.135 = 0.880

Therefore, the ratio of the diameter of 24-gauge wire to that of 26-gauge wire is 0.880, or approximately 1.3. Answer: B

Question 6

A 4.0-$\Omega$ resistor has a current of 3.0 A in it for 5.0 minutes.How many electrons pass through the resistor during this time interval?&#10;A)7.5 * 1021&#10;B)5.6 * 1021&#10;C)6.6 * 1021&#10;D)8.4 * 1021&#10;E)2.1 *1021

Accepted Answer

The answer of A 4.0-$\Omega$ resistor has a current of...

Question 7

What is the resistance of a wire made of a material with a resistivity of 3.2 *10-8 $\Omega$ . m if its length is 2.5 m and its diameter is 0.50 mm?&#10;A)0.16 $\Omega$&#10;B)0.10 $\Omega$&#10;C)1.28 $\Omega$&#10;D)0.41 $\Omega$&#10;E)0.81 $\Omega$

Accepted Answer

We can use the formula for resistance of a wire:

R = ρL/A

where: 
ρ = resistivity of the material (3.2 x 10^-8 Ω·m) 
L = length of the wire (2.5 m) 
A = cross-sectional area of the wire (πr^2, where r is the radius or half of the diameter)

First, we need to convert the diameter to radius and then to meters:

0.50 mm = 0.50 x 10^-3 m

Radius = diameter/2 = 0.25 x 10^-3 m

Area = π(0.25 x 10^-3 m)^2 = 1.96 x 10^-7 m^2

Now we can substitute into the formula:

R = (3.2 x 10^-8 Ω·m)(2.5 m)/(1.96 x 10^-7 m^2)

R = 0.41 Ω

Therefore, the correct answer is D) 0.41 Ω.

Question 8

A small bulb is rated at 7.50 W when operated at 125 V.The tungsten filament has a temperature coefficient of resistivity $\alpha$ = 4.50 *10-3 / $\degree$ C.When the filament is hot and glowing, its temperature is seven times room temperature (20 $\degree$ C).What is the resistance of the filament (in ohms) at room temperature?

A)234
B)1350
C)2080
D)4530
E)5630

Accepted Answer

The resistance at room temperature can be calculated using the formula $ R = \frac{V^2}{P} $ for the hot filament, and then adjusting for temperature using $ R_0 = \frac{R}{1 + \alpha \Delta T} $. Solving gives approximately 234 ohms.

Question 9

If 5.0 *1021 electrons pass through a 20-$\Omega$resistor in 10 minutes, what is the potential difference across the resistor?&#10;A)21 V&#10;B)32 V&#10;C)27 V&#10;D)37 V&#10;E)54 V

Accepted Answer

We can use Ohm's Law to find the potential difference across the resistor:
V = IR
where V is the potential difference, I is the current, and R is the resistance.

First, we need to find the current:
I = Q/t
where Q is the charge and t is the time.

We are given the charge, which is:
Q = ne
where n is the number of electrons and e is the charge of one electron.

Substituting the given values, we get:
Q = (5.0 × 10^21)(1.6 × 10^-19) = 8.0 × 10^2 C

The time is given as 10 minutes, which is equivalent to:
t = 10 × 60 = 600 s

Substituting the known values, we get:
I = 8.0 × 10^2 C / 600 s = 1.33 A

Now we can use Ohm's Law to find the potential difference:
V = IR = (1.33 A)(20 Ω) = 26.6 V

Therefore, the closest answer choice is C) 27 V.

Question 10

A nichrome wire and an aluminum wire, each with the same initial resistance, have the same change in resistance when heated separately.( $\rho$ Al = 2.82 * 10-8 $\Omega$ . m; $\alpha$ Al = 3.9 * 10-3 / $\degree$ C; $\alpha$ nichrome = 1.50 *10-6 $\Omega$ . m; $\alpha$ nichrome = 0.40* 10-3 / $\degree$ C.) The ratio of the temperature change of the nichrome wire to the temperature change of the aluminum wire is:

A)0.019.
B)0.10.
C)0.18.
D)9.8.
E)53.

Accepted Answer

The change in resistance is given by $\Delta R = R_0 \alpha \Delta T$. Since the initial resistances and changes in resistance are the same for both wires, $\alpha_{\text{nichrome}} \Delta T_{\text{nichrome}} = \alpha_{\text{Al}} \Delta T_{\text{Al}}$. Solving for the ratio $\frac{\Delta T_{\text{nichrome}}}{\Delta T_{\text{Al}}}$ gives $\frac{\alpha_{\text{Al}}}{\alpha_{\text{nichrome}}} = \frac{3.9 \times 10^{-3}}{0.40 \times 10^{-3}} = 9.8$.

Question 11

An electric heater is constructed by applying a potential difference of 110 V across a wire with a resistance of 5.0 $\Omega$.What is the power rating of the heater?&#10;A)2.0 kW&#10;B)2.4 kW&#10;C)1.7 kW&#10;D)1.5 kW&#10;E)60 kW

Accepted Answer

The answer of An electric heater is constructed by applying...

Question 12

A rod of 2.0-m length and a square (2.0 mm * 2.0 mm) cross section is made of a material with a resistivity of 6.0 * 10-8 $\Omega$ . m.If a potential difference of 0.50 V is placed across the ends of the rod, at what rate is heat generated in the rod?

A)3.0 W
B)5.3 W
C)8.3 W
D)1.3 W
E)17 W

Accepted Answer

The answer of A rod of 2.0-m length and a...

Question 13

A light bulb is rated at 30 W when operated at 120 V.How much charge enters (and leaves) the light bulb in 1.0 minute?&#10;A)17 C&#10;B)15 C&#10;C)14 C&#10;D)13 C&#10;E)60 C

Accepted Answer

The answer of A light bulb is rated at 30...

Question 14

An electric device, which heats water by immersing a resistance wire in the water, generates 50 cal of heat per second when an electric potential difference of 12 V is placed across its leads.What is the resistance of the heater wire? (Note: 1 cal = 4.186 J)

A)0.94

\Omega

B)0.81

\Omega

C)0.58

\Omega

D)0.69

\Omega

E)1.5

\Omega

Accepted Answer

The answer of An electric device, which heats water by...

Question 15

In the Drude model of electrical conduction, the current density is directly proportional to:&#10;A)the average time interval between successive collisions.&#10;B)the number of charge carriers per unit volume.&#10;C)the square of the electron charge.&#10;D)the electric field present in the wire.&#10;E)the product of all four quantities listed above.

Accepted Answer

The answer of In the Drude model of electrical conduction,...

Question 16

The temperature coefficient of resistivity of iron is 5.0 * 10-3 / $\degree$ C and that of carbon is -0.50 * 10-3 / $\degree$ C.When an iron wire and a carbon rod, each having the same 10 $\Omega$ resistance at 20 $\degree$ C, are cooled from that temperature to -80 $\degree$ C, the new ratio of the resistance of the carbon rod to the resistance of the iron wire at the lower temperature is:

A)(-0.10.)
B)+1.9.
C)+2.1.
D)(-10.)
E)+10.

Accepted Answer

The answer of The temperature coefficient of resistivity of iron...

Question 17

A conductor of radius r, length and resistivity $\rho$ has resistance R.What is the new resistance if it is stretched to 4 times its original length?  &#10;A)R  &#10;B)R  &#10;C)R&#10;D)4R&#10;E)16R

Accepted Answer

The answer of A conductor of radius r, length and...

Question 18

What maximum power can be generated from an 18-V emf using any combination of a 6.0- $\Omega$ resistor and a 9.0- $\Omega$resistor?&#10;A)54 W&#10;B)71 W&#10;C)90 W&#10;D)80 W&#10;E)22 W

Accepted Answer

The answer of What maximum power can be generated from...

Question 19

A wire (length = 2.0 m, diameter = 1.0 mm) has a resistance of 0.45$\Omega$.What is the resistivity of the material used to make the wire?&#10;A)5.6 * 10-7 $\Omega$ .m&#10;B)1.2 * 10-7 $\Omega$ . m&#10;C)1.8 * 10-7 $\Omega$ .m&#10;D)2.3 *10-7 $\Omega$ .m&#10;E)7.1 *10-7$\Omega$ . m

Accepted Answer

The answer of A wire (length = 2.0 m, diameter...

Question 20

The electron density in copper is 8.49 *1028 electrons/m3.The electron charge is e = -1.60 *10-19 C.When a 1.00 A current is present in a copper wire with a 0.40 cm2 cross-section, the electron drift velocity, in m/s, with direction defined relative to the current density, is:

A)(-1.84 *10-6.)
B)+1.84 * 10-6.
C)(-1.84.)
D)(-5.43 * 105.)
E)+5.43 * 105.

Accepted Answer

The answer of The electron density in copper is 8.49...

Question 21

A cook plugs a 500 W crockpot and a 1000 W kettle into a 240 V power supply, all operating on direct current.When we compare the two, we find that: A)Icrockpot < Ikettle and Rcrockpot < Rkettle. B)Icrockpot < Ikettle and Rcrockpot > Rkettle. C)Icrockpot = Ikettle and Rcrockpot = Rkettle. D)Icrockpot > Ikettle and Rcrockpot < Rkettle. E)Icrockpot > Ikettle and Rcrockpot > Rkettle.

Accepted Answer

The answer of A cook plugs a 500 W crockpot...

Question 22

To increase the current density in a wire of length and diameter D, you can:  &#10;A)decrease the potential difference between the two ends of the wire.&#10;B)increase the potential difference between the two ends of the wire.&#10;C)decrease the magnitude of the electric field in the wire.&#10;D)heat the wire to a higher temperature.&#10;E)combine both (b) and (d).

Accepted Answer

The answer of To increase the current density in a...

Question 23

Jadeen says that you can increase the resistance of a copper wire by hammering the wire to make it narrower and longer.Arnell says that you can increase its resistance by heating the wire.Which one, if either, is correct, and why?

A)Arnell, because the conductivity of the wire increases when it is heated.
B)Arnell, because the conductivity of the wire decreases when it is heated.
C)Jadeen, because the conductivity of a wire is directly proportional to its area and inversely proportional to its length.
D)Jadeen, because the conductivity of a copper wire does not increase when it is hammered.
E)Both are correct because (b) and (d) are both correct.

Accepted Answer

The answer of Jadeen says that you can increase the...

Question 24

The heating coil of a hot water heater has a resistance of 20 ohms and operates at 210 V.If electrical energy costs 5.5 cents per kW-hr, what does it cost to raise the 200 kg of water in the tank from 15$\degree$C to 80$\degree$C? (The specific heat of water is 4186 J/kg$\degree$C)

Accepted Answer

The answer of The heating coil of a hot water...

Question 25

What is the resistance of 1000 m of 4-mm diameter copper wire?&#10;($\rho$Cu = 1.7 *10-8  $\Omega$ .m)

Accepted Answer

The answer of What is the resistance of 1000 m...

Question 26

A conductor of radius r, length and resistivity$\rho$ has resistance R.It is melted down and formed into a new conductor, also cylindrical, with one fourth the length of the original conductor.The resistance of the new conductor is:  &#10;A)R  &#10;B)R  &#10;C)R&#10;D)4R&#10;E)16R

Accepted Answer

The answer of A conductor of radius r, length and...

Question 27

A copper cable is to be designed to carry a current of 300 A with a power loss of only 2.0 watts per meter.What is the required radius of the copper cable? (The resistivity of copper is 1.7 * 10-8  $\Omega$.m.)

Accepted Answer

The answer of A copper cable is to be designed...

Question 28

An aluminium wire of cross-sectional area 4.0 mm2 is carrying a current of 8.0 A.The density of aluminium is 2.7 g/cm3, and its molar mass is 27
g.Assuming one free electron per aluminium atom, what is the drift speed of the electrons in this wire?

Accepted Answer

The answer of An aluminium wire of cross-sectional area 4.0...

Question 29

A high-resistance material is used as an insulator between the conductors of a length of coaxial cable.The resistance material, which forms a hollow tube, has an inner radius a and an outer radius b, and the insulator provides a resistance R between the conductors.If a second insulator, made of the same material and having the same length, is made with double both the inner radius and the outer radius of the first, what resistance would it provide between the conductors?

A)R
B)2 R
C)4 R
D)(ln2)R
E)R/(ln2)

Accepted Answer

The answer of A high-resistance material is used as an...

Question 30

Jadeen says that you can increase the resistance of a copper wire by hammering the wire to make it narrower and longer.Arnell says that you can increase its resistance by heating the wire.Which one, if either, is correct, and why?

A)Arnell, because the conductivity of the wire increases when it is heated.
B)Arnell, because the conductivity of the wire decreases when it is heated.
C)Jadeen, because the conductivity of a wire is directly proportional to its area and inversely proportional to its length.
D)Jadeen, because the conductivity of a copper wire does not increase when it is hammered.
E)Both are correct because (b) and (d) are both correct.

Accepted Answer

The answer of Jadeen says that you can increase the...

Question 31

A high-voltage transmission line carries 1000 A at 700 kV for a distance of 100 miles.If the resistance per length in the wire is 0.5  $\Omega$/mile, what is the power loss due to resistive losses?

Accepted Answer

The answer of A high-voltage transmission line carries 1000 A...

Question 32

Light bulb A is rated at 60 W and light bulb B is rated at 100 W.Both are designed to operate at 110 V.Which statement is correct?&#10;A)The 60 W bulb has a greater resistance and greater current than the 100 W bulb.&#10;B)The 60 W bulb has a greater resistance and smaller current than the 100 W bulb.&#10;C)The 60 W bulb has a smaller resistance and smaller current than the 100 W bulb.&#10;D)The 60 W bulb has a smaller resistance and greater current than the 100 W bulb.&#10;E)We need to know the resistivities of the filaments to answer this question.

Accepted Answer

The answer of Light bulb A is rated at 60...

Question 33

In the Drude model of electrical conduction, the current density is NOT directly proportional to:&#10;A)the average time interval between successive collisions.&#10;B)the number of charge carriers per unit volume.&#10;C)the square of the electron charge.&#10;D)the electric field present in the wire.&#10;E)the resistivity of the wire.

Accepted Answer

The answer of In the Drude model of electrical conduction,...

Deck 5: Current and Resistance