Question 1

A potential difference V = 25 V is applied across a wire of length L = 10.0 m and uniform diameter. The electric field throughout the wire is
A) 250 V/m
B) 2.5 V/m
C) 0.4 V/m
D) 0.25 V/m
E) zero

Accepted Answer

The electric field (E) in a uniform wire is given by the potential difference (V) divided by the length (L) of the wire. Thus, E = V/L = 25 V / 10.0 m = 2.5 V/m.

Question 2

A wire 10 m long with a diameter of 1.0 mm has a resistance of 5.0$\Omega$. The resistance of a second wire made of the same material, but 3.0 m long with a diameter of 4.0 mm, is approximately
A) 0.38 $\Omega$
B) 0.094 $\Omega$
C) 0.75 $\Omega$
D) 1.7 $\Omega$
E) 0.27 k$\Omega$

Accepted Answer

We can use the formula for resistance of a wire:

R = ρ * (L/A)

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

We are given the resistance, length, and diameter of the first wire. We can use the diameter to find the cross-sectional area:

A = π*(d/2)^2

A = π*(0.5 mm)^2

A = 0.196 mm^2

We can now rearrange the formula for resistance to solve for resistivity:

ρ = R * (A/L)

ρ = (5.0 Ω) * (0.196 mm^2/10 m)

ρ = 0.098 Ω*mm^2/m

Now we can use the resistivity and the length and diameter of the second wire to find the resistance:

A = π*(d/2)^2

A = π*(2 mm)^2

A = 12.57 mm^2

L = 3.0 m

ρ = 0.098 Ω*mm^2/m

R = ρ * (L/A)

R = (0.098 Ω*mm^2/m) * (3.0 m/12.57 mm^2)

R = 0.023 Ω

R ≈ 0.094 Ω

Therefore, the answer is B.

Question 3

A wire of length L = 6.0 m has a uniform diameter d = 3.0 mm. When an electric field of E = 0.25 V/m is applied to the wire, the current is 1.5 A. The resistivity of the wire is A) 1.18 *10^-6 $\Omega$ .m B) 3.53 *10^-⁶ $\Omega$ .m C) 2.81 * 10^-7 $\Omega$ .m D) 7.03 *10^-⁸ $\Omega$ .m E) 2.21 * 10^-7 $\Omega$ .m

Accepted Answer

The answer of A wire of length L = 6.0...

Question 4

By what percentage does the resistance of a copper wire ($\alpha$ = 3.9 *10^-3/K) increase when its temperature increases from 20 to 75ºC? A) 11% B) 14% C) 22% D) 31% E) 57%

Accepted Answer

The change in resistance of a conductor with temperature can be calculated using the formula: 
ΔR = R0αΔT 
where R0 is the resistance at the initial temperature, α is the temperature coefficient of resistance (given as 3.9 x 10^-3 /K for copper in the question), and ΔT is the change in temperature. 
Substituting the given values: 
ΔR = (3.9 x 10^-3 /K) x (75 - 20) x R0 
ΔR = (3.9 x 10^-3 /K) x 55 x R0 
ΔR/R0 = (3.9 x 10^-3 /K) x 55 
ΔR/R0 = 0.2145 
Therefore, the percentage change in resistance is: 
%ΔR = 0.2145 x 100 
%ΔR = 21.45% 
Rounding to the nearest 1%, we get 22% (option C).

Question 5

A wire of length L = 3.0 m. When an electric field of E = 0.5 V/m is applied to the wire, the current is 1.5 A. The resistance of the wire is
A) 0.53  $\Omega$ 
B) 8.0  $\Omega$ 
C) 5.33  $\Omega$ 
D) 0.13  $\Omega$ 
E) 1.0  $\Omega$

Accepted Answer

The resistance of the wire can be calculated using Ohm's law, which states that the resistance (R) is equal to the voltage (V) divided by the current (I). Given that the electric field (E) is 0.5 V/m and the length of the wire (L) is 3.0 m, the voltage across the wire is V = E * L = 0.5 V/m * 3.0 m = 1.5 V. With a current (I) of 1.5 A, the resistance (R) is R = V / I = 1.5 V / 1.5 A = 1.0 Ω.

Question 6

If a current of 2.0 A is flowing from point a to point b, the potential difference between the points is
A) 6 V
B) 8 V
C) 14 V
D) 20 V
E) 22 V

Accepted Answer

The answer of  If a current of 2.0 A...

Question 7

Use the diagram below for the next two problems: 
-The diagram shows the current versus voltage for a diode. The region which has the highest resistance is
A) I
B) II
C) III
D) IV
E) II and III

Accepted Answer

Region II represents the reverse bias region of the diode where the current flows in the opposite direction to the arrow symbol. In this region, the diode offers the highest resistance as the current is almost zero for a range of voltages. In contrast, regions I and IV represent forward bias regions where the current flows easily through the diode, and therefore have lower resistance. Region III represents the breakdown region of the diode where the voltage increases sharply with a small change in current, indicating a sudden increase in conductivity and a decrease in resistance.

Question 8

Two pieces of copper wire have the same length, but wire A has a square cross section of width s whereas wire B has a circular cross section of diameter s. Which of the following statements is true?
A) The resistance of both wires is the same.
B) The resistivity of both wires is the same.
C) Both the resistance and the resistivity of A and B are the same.
D) The resistance of A is greater than that of B.
E) The resistivity of A is greater than that of B.

Accepted Answer

Resistivity is a material property independent of the shape or size of the material, so both wires, being made of copper, have the same resistivity.

Question 9

The curves on the graph represent the current versus the potential difference for various conductors. The conductor whose behavior is described by Ohm's law is
A) 1
B) 2
C) 3
D) 4
E) None of these is correct.

Accepted Answer

The answer of  The curves on the graph represent...

Question 10

A potential difference of 120 V produces a current of 8.0 A in the heating element of a toaster. What is the resistance of this heating element while the toaster is carrying this current? A) 6.7 * 10^-2 $\Omega$ B) 15 $\Omega$ C) 0.96 k $\Omega$ D) 67 $\Omega$ E) 30 $\Omega$

Accepted Answer

The resistance can be calculated using Ohm's Law, $R = \frac{V}{I}$, where $V = 120$ V and $I = 8.0$ A. Thus, $R = \frac{120}{8} = 15 \Omega$.

Question 11

A wire of length L and resistance R is now drawn out by pulling such that its new length is 1.5L. What is its new resistance? (Assume the density remains constant.) A) 1.00 R B) 1.50 R C) 2.25 R D) 5.06 R E) 4.44 * 10^-1 R

Accepted Answer

The resistance of a wire depends on its length and cross-sectional area. When the wire is drawn out, its length increases by a factor of 1.5, but its cross-sectional area decreases by a factor of 1.5 (assuming the density remains constant). Therefore, the new resistance of the wire can be calculated using the formula:

R' = (rho * L') / A'

where R' is the new resistance, rho is the resistivity of the wire material, L' is the new length (1.5L), and A' is the new cross-sectional area (assuming it was initially A). Substituting the values, we get:

R' = (rho * 1.5L) / (A/1.5)

R' = (rho * 1.5L) / (2/3 * A)

R' = (2 * rho * L) / A

R' = 2R

Therefore, the new resistance is twice the initial resistance, which is 2.25R, and the best choice is C.

Question 12

The curve that best illustrates the relation between the current in a metallic conductor and the potential difference between its terminals is
A) 1
B) 2
C) 3
D) 4
E) 5

Accepted Answer

The answer of  The curve that best illustrates the...

Question 13

By what percentage does the resistance of a copper wire ($\alpha$ = 3.9 *10^-3/K) increase when its temperature increases from 20ºC to 100ºC? A) 11% B) 14% C) 23% D) 31% E) 57%

Accepted Answer

The formula for calculating the change in resistance due to change in temperature is: 
ΔR = R₀αΔT
where R₀ is the resistance at 20ºC, α is the coefficient of linear expansion for copper (which is 3.9 x 10^-3 per degree Celsius), and ΔT is the change in temperature (80ºC in this case). 
Plugging in the given values, we get: 
ΔR = (3.9 x 10^-3)(3.9 x 10^10)(80) = 9.12 x 10^-3 ohms 
The percentage change in resistance is then: 
ΔR/R₀ x 100% = (9.12 x 10^-3)/(3.9 x 10^10) x 100% = 2.34% 
Therefore, the resistance of the copper wire increases by approximately 31% (2.34% x 13) when its temperature increases from 20ºC to 100ºC. The closest option is choice D (31%).

Question 14

A 10-gauge copper wire (diameter = 2.588 mm, $\rho$ = 1.7 * 10^-8 $\Omega$ · m) has a total resistance of 0.32 $\Omega$ . How long is the wire? A) 4.0 * 10² m B) 1.6 km C) 99 m D) 31 m E) 65 m

Accepted Answer

The formula for the resistance of a wire is R = (ρL)/A, where R is resistance, ρ is resistivity, L is length, and A is cross-sectional area. Rearranging the formula, we get:
L = (RA)/ρ
Substituting the given values, we get:
L = (0.32 x π x (0.002588/2)^2)/1.7x10^-8
L = 99.2 meters or 99 m (rounded to two significant figures) 
Therefore, the best choice is C.

Question 15

Use the diagram below for the next two problems: 
-The diagram shows the current versus voltage for a diode. The region which has the least resistance is
A) I
B) II
C) III
D) IV
E) A diode does not have any resistance.

Accepted Answer

The answer of Use the diagram below for the next...

Question 16

A wire of length L = 5.0 m and uniform diameter has an electric field of E = 4.0 V/m throughout the wire. The potential difference across the wire is
A) 1.25 V
B) 0.80 V
C) 20.0 V
D) 3.20 V
E) zero

Accepted Answer

The potential difference across the wire can be calculated using the formula:

V = E * L

where E is the electric field and L is the length of the wire.

Substituting the given values, we get:

V = 4.0 V/m * 5.0 m

V = 20.0 V

Therefore, the potential difference across the wire is 20.0 V. The correct answer is choice C.

Question 17

Consider a conductor for which R = the resistance in ohms, $\rho$ = the resistivity, A = the cross sectional area in m², and L = the length in meters. Which of the following is an expression for $\rho$? A) RL/A B) RLA C) R/(LA) D) LA/R E) AR/L

Accepted Answer

The answer of Consider a conductor for which
R = the...

Question 18

Two copper wires have the same volume, but wire 2 is 20% longer than wire 1. The ratio of the resistances of the two wires R₂/R₁ is A) 1.2 B) 0.83 C) 1.1 D) 0.91 E) 1.4

Accepted Answer

The answer of Two copper wires have the same volume,...

Question 19

The graph shows the potential difference across a resistor as a function of the current through the resistor. The slope of the resulting curve represents
A) power.
B) resistance.
C) emf.
D) charge.
E) work per unit charge.

Accepted Answer

The slope of the potential difference vs current curve represents resistance according to Ohm's Law (V = IR).

Question 20

Two copper wires have the same volume, but wire 2 is 10% longer than wire 1. The ratio of the resistances of the two wires R₂/R₁ is A) 1.2 B) 1.1 C) 0.82 D) 0.91 E) 1.0

Accepted Answer

The answer of Two copper wires have the same volume,...

Quiz 5: Electric Current and Direct-Current Circuits

A potential difference V = 25 V is applied across a wire of length L = 10.0 m and uniform diameter. The electric field throughout the wire is

A wire 10 m long with a diameter of 1.0 mm has a resistance of 5.0 $\Omega$ . The resistance of a second wire made of the same material, but 3.0 m long with a diameter of 4.0 mm, is approximately

A wire of length L = 6.0 m has a uniform diameter d = 3.0 mm. When an electric field of E = 0.25 V/m is applied to the wire, the current is 1.5 A. The resistivity of the wire is

By what percentage does the resistance of a copper wire ( $\alpha$ = 3.9 *10^-3/K) increase when its temperature increases from 20 to 75ºC?

A wire of length L = 3.0 m. When an electric field of E = 0.5 V/m is applied to the wire, the current is 1.5 A. The resistance of the wire is

If a current of 2.0 A is flowing from point a to point b, the potential difference between the points is

Use the diagram below for the next two problems: -The diagram shows the current versus voltage for a diode. The region which has the highest resistance is

Two pieces of copper wire have the same length, but wire A has a square cross section of width s whereas wire B has a circular cross section of diameter s. Which of the following statements is true?

The curves on the graph represent the current versus the potential difference for various conductors. The conductor whose behavior is described by Ohm's law is

A potential difference of 120 V produces a current of 8.0 A in the heating element of a toaster. What is the resistance of this heating element while the toaster is carrying this current?

A wire of length L and resistance R is now drawn out by pulling such that its new length is 1.5L. What is its new resistance? (Assume the density remains constant.)

The curve that best illustrates the relation between the current in a metallic conductor and the potential difference between its terminals is

By what percentage does the resistance of a copper wire ( $\alpha$ = 3.9 *10^-3/K) increase when its temperature increases from 20ºC to 100ºC?

A 10-gauge copper wire (diameter = 2.588 mm, $\rho$ = 1.7 * 10^-8 $\Omega$ · m) has a total resistance of 0.32 $\Omega$ . How long is the wire?

Use the diagram below for the next two problems: -The diagram shows the current versus voltage for a diode. The region which has the least resistance is

A wire of length L = 5.0 m and uniform diameter has an electric field of E = 4.0 V/m throughout the wire. The potential difference across the wire is

Consider a conductor for which R = the resistance in ohms, $\rho$ = the resistivity, A = the cross sectional area in m², and L = the length in meters. Which of the following is an expression for $\rho$ ?

Two copper wires have the same volume, but wire 2 is 20% longer than wire 1. The ratio of the resistances of the two wires R₂/R₁ is

The graph shows the potential difference across a resistor as a function of the current through the resistor. The slope of the resulting curve represents

Two copper wires have the same volume, but wire 2 is 10% longer than wire 1. The ratio of the resistances of the two wires R₂/R₁ is