Electric Current and Resistance Practice Questions — JAMB UTME

Question 1Question →

A uniform metal wire of length $20\,\text{m}$ and total mass $0.034\,\text{kg}$ is manufactured from a material of density $8.5 \times 10^3\,\text{kg/m}^3$ and electrical resistivity $1.7 \times 10^{-8}\,\Omega\cdot\text{m}$ . What is the electrical resistance of the wire in ohms?

Show answer & explanation

Answer: 1.7

Answer

The electrical resistance of the wire is

1.7\,\Omega

.

Combining the density relation

V = \frac{m}{d}

with the geometric expression

V = A \cdot L

gives

A = \frac{m}{d L}

. Substituting this into Pouillet's law

R = \frac{\rho L}{A}

yields

R = \frac{\rho d L^2}{m}

. Evaluating with the given values:

R = \frac{(1.7 \times 10^{-8})(8.5 \times 10^3)(20)^2}{0.034} = 1.7\,\Omega

.

Step-by-Step Solution

1

Calculate the volume of the wire using mass and density

V = 4.0 \times 10^{-6}\,\text{m}^3

Volume is related to mass and density by

V = \frac{m}{d}

.

2

Calculate the cross-sectional area of the wire

A = 2.0 \times 10^{-7}\,\text{m}^2

For a cylindrical wire of uniform cross-section,

V = A \cdot L

, so

A = \frac{V}{L}

.

3

Apply resistivity formula to find electrical resistance

R = 1.7\,\Omega

Resistance is given by

R = \frac{\rho L}{A}

.

Key Concept

Relationship between Resistance, Mass, Density, and Resistivity

Estimated Time:2m 0s

Question 2Question →

Two cylindrical metallic conductors, $X$ and $Y$ , are connected in series across a direct-current source. Wire $X$ has a diameter of $1.0\,\text{mm}$ and a free-electron density of $6.0 \times 10^{28}\,\text{m}^{-3}$ . Wire $Y$ has a diameter of $3.0\,\text{mm}$ and a free-electron density of $2.0 \times 10^{28}\,\text{m}^{-3}$ . What is the ratio of the drift velocity of free electrons in wire $X$ to that in wire $Y$ ?

3.0

0.33

1.0

9.0

Show answer & explanation

Answer: $3.0$

Answer

The ratio of the drift velocity in wire X to that in wire Y is 3.0.

Because the wires are connected in series, the same current passes through both (

I_X = I_Y

). Using the formula for electric current in terms of drift velocity

I = n A e v_d

, where cross-sectional area

A = \pi d^2 / 4

, we find

v_d \propto 1 / (n d^2)

. Taking the ratio yields

v_X / v_Y = (n_Y d_Y^2) / (n_X d_X^2) = (2.0 \times 10^{28} \times 3.0^2) / (6.0 \times 10^{28} \times 1.0^2) = 18 / 6 = 3.0

.

Step-by-Step Solution

1

Relate electric current to drift velocity and conductor geometry.

I = n A e v_d = n \left( \frac{\pi d^2}{4} \right) e v_d

Electric current

I

depends on free-electron density

n

, cross-sectional area

A

, elementary charge

e

, and electron drift velocity

v_d

.

2

Apply the series connection constraint.

I_X = I_Y \implies n_X d_X^2 v_X = n_Y d_Y^2 v_Y

In a series circuit, the steady current flowing through every conductor is identical.

3

Rearrange to solve for the drift velocity ratio

v_X / v_Y

.

\frac{v_X}{v_Y} = \frac{n_Y d_Y^2}{n_X d_X^2}

Isolating

v_X / v_Y

demonstrates inverse proportionality to electron density and the square of conductor diameter.

4

Substitute the given values into the ratio expression.

\frac{v_X}{v_Y} = \frac{(2.0 \times 10^{28}\,\text{m}^{-3}) \times (3.0\,\text{mm})^2}{(6.0 \times 10^{28}\,\text{m}^{-3}) \times (1.0\,\text{mm})^2} = \frac{2.0 \times 9.0}{6.0 \times 1.0} = \frac{18.0}{6.0} = 3.0

Numerical calculation yields the simplified dimensionless ratio.

Key Concept

Drift Velocity and Current Density in Series Conductors

Question 3Question →

A steady electric current of $2.5\,\text{A}$ flows through a conductor for $4.0\,\text{minutes}$ . What is the total electric charge that passes through any cross-section of the conductor during this period?

10\,\text{C}

96\,\text{C}

600\,\text{C}

150\,\text{C}

Show answer & explanation

Answer: $600\,\text{C}$

Answer

600\,\text{C}

Electric current (

I

) is defined as the rate of charge flow (

Q

) per unit time (

t

), expressed as

Q = I \times t

. Converting

4.0\,\text{minutes}

into seconds gives

4.0 \times 60 = 240\,\text{s}

. Multiplying the current of

2.5\,\text{A}

by

240\,\text{s}

yields

600\,\text{C}

.

Step-by-Step Solution

1

Convert the time duration from minutes to seconds

t = 4.0\,\text{minutes} = 4.0 \times 60\,\text{s} = 240\,\text{seconds}

The standard SI unit for time in electromagnetism formulas is seconds.

2

Apply the formula relating electric charge, current, and time (

Q = I \times t

)

Q = 2.5\,\text{A} \times 240\,\text{s} = 600\,\text{C}

Electric current is defined as the rate of flow of electric charge (

I = \frac{Q}{t}

).

Key Concept

Relationship between Electric Current, Charge, and Time

Question 4Question →

A metallic wire has a resistance of $10.0\,\Omega$ at $0^\circ\text{C}$ and a temperature coefficient of resistance $\alpha = 4.0 \times 10^{-3}\,\text{K}^{-1}$ . The wire is uniformly stretched at constant temperature until its length is doubled while maintaining constant mass and volume. It is subsequently heated to $50^\circ\text{C}$ . What is the electric current, in Amperes, that flows through the wire when a potential difference of $120.0\,\text{V}$ is applied across its ends?

Show answer & explanation

Answer: 2.5

Answer

The electric current passing through the heated, stretched wire is 2.5 A.

When a wire of initial resistance 10.0 ohms is stretched to twice its original length, conservation of volume requires its cross-sectional area to halve, which quadruples its resistance to 40.0 ohms at 0 °C. Heating the wire by 50 K increases its resistance by a factor of (1 + 0.004 * 50) = 1.2, producing a final resistance of 48.0 ohms. Applying a 120.0 V potential difference across 48.0 ohms results in a current of 2.5 A.

Step-by-Step Solution

1

Determine resistance change due to wire stretching

R_0' = 40.0 ohms at 0 °C

Uniform stretching conserves total volume (V = A * L). Doubling length halves area, making resistance increase by a factor of 2^2 = 4.

2

Calculate resistance at 50 °C using temperature coefficient

R(50 °C) = 48.0 ohms

Resistance increases linearly with temperature: R(T) = R_0'(1 + alpha * Delta T).

3

Apply Ohm's law to find current

I = 2.5 A

Current is given by potential difference divided by total resistance at the operational temperature.

Key Concept

Resistance variation with geometric stretching and temperature coefficient of resistance

Question 5Question →

A uniform metallic wire of resistance $R$ is stretched uniformly until its radius decreases by $20\%$ . The stretched wire is subsequently cut into two equal halves, which are then connected in parallel across a constant potential difference $V$ . What is the ratio of the total electrical power dissipated in this parallel combination to the power dissipated by the original unstretched wire under the same potential difference?

1.64

0.61

2.56

0.41

Show answer & explanation

Answer: 1.64

Answer

The ratio of the total power dissipated in the parallel combination to the original power is 1.64

When a wire of initial resistance

R

is stretched so that its radius decreases by

20\%

, its new radius is

0.8r

. Because volume is conserved (

A_1 L_1 = A_2 L_2

), reducing the cross-sectional area to

0.64A

causes the length to increase to

L/0.64

. Consequently, the resistance scales inversely with the fourth power of the radius:

R_{\text{stretched}} = R / (0.8)^4 = R / 0.4096 = 2.4414R

. Cutting this wire into two equal pieces gives two resistors of

1.2207R

each. Connecting them in parallel yields an equivalent resistance

R_{\text{eq}} = 1.2207R / 2 = 0.61035R

. Power at constant voltage is

P = V^2/R

, so the new power is

P_{\text{new}} = V^2 / (0.61035R) = 1.64 (V^2/R) = 1.64 P_{\text{orig}}

.

Step-by-Step Solution

1

Determine the new resistance of the wire after stretching

R_{\text{stretched}} = \frac{R}{(0.8)^4} = \frac{R}{0.4096} \approx 2.4414 R

Since mass and density remain constant, volume

V_{\text{vol}} = A \cdot L

is conserved. Decreasing radius to

r_2 = 0.8 r_1

reduces area to

A_2 = 0.64 A_1

and increases length to

L_2 = L_1 / 0.64

. Resistance

R = \rho L / A \propto 1/r^4

.

2

Calculate the equivalent resistance of the two equal halves connected in parallel

R_{\text{eq}} = \frac{1}{4} R_{\text{stretched}} = \frac{2.4414 R}{4} \approx 0.61035 R

Cutting the stretched wire in half gives two pieces each of resistance

R_{\text{half}} = R_{\text{stretched}} / 2

. Connecting two identical resistors in parallel yields an equivalent resistance

R_{\text{eq}} = R_{\text{half}} / 2 = R_{\text{stretched}} / 4

.

3

Calculate the ratio of power dissipated across a constant potential difference V

\frac{P_{\text{new}}}{P_{\text{orig}}} = \frac{V^2 / R_{\text{eq}}}{V^2 / R} = \frac{R}{R_{\text{eq}}} = \frac{1}{0.61035} \approx 1.64

Electrical power dissipated at constant voltage is given by

P = V^2 / R

, which means power is inversely proportional to equivalent resistance.

Key Concept

Dependence of electrical resistance on conductor geometry under volume conservation, and power dissipation in parallel circuits.

Question 6Question →

A uniform conductor of length $50\,\text{m}$ and cross-sectional area $2.0 \times 10^{-6}\,\text{m}^2$ is made of a material with a resistivity of $4.0 \times 10^{-7}\,\Omega\cdot\text{m}$ . What is the electrical resistance of the conductor in ohms?

Show answer & explanation

Answer: 10

Answer

The resistance of the conductor is

10\,\Omega

.

The resistance of a uniform conductor is given by

R = \frac{\rho L}{A}

. Substituting the values

L = 50\,\text{m}

,

A = 2.0 \times 10^{-6}\,\text{m}^2

, and

\rho = 4.0 \times 10^{-7}\,\Omega\cdot\text{m}

yields

R = \frac{(4.0 \times 10^{-7})(50)}{2.0 \times 10^{-6}} = 10\,\Omega

.

Step-by-Step Solution

1

Identify given physical quantities

L = 50\,\text{m}

,

A = 2.0 \times 10^{-6}\,\text{m}^2

,

\rho = 4.0 \times 10^{-7}\,\Omega\cdot\text{m}

Extracting given parameter values clearly sets up the mathematical relationship.

2

Apply the resistivity formula for resistance

R = \frac{\rho L}{A}

Resistance varies directly with length and resistivity, and inversely with cross-sectional area.

3

Perform the calculation

R = \frac{(4.0 \times 10^{-7}\,\Omega\cdot\text{m})(50\,\text{m})}{2.0 \times 10^{-6}\,\text{m}^2} = 10\,\Omega

Multiplying the numerator gives

2.0 \times 10^{-5}\,\Omega\cdot\text{m}^2

; dividing by

2.0 \times 10^{-6}\,\text{m}^2

yields

10\,\Omega

.

Key Concept

Direct computation of electrical resistance using resistivity, length, and cross-sectional area

Estimated Time:45s

Question 7Question →

An electric heating element made of a wire with resistivity $1.0 \times 10^{-6}\,\Omega\cdot\text{m}$ and a uniform cross-sectional area of $5.0 \times 10^{-7}\,\text{m}^2$ carries a steady current of $4.0\,\text{A}$ when connected to a $240\,\text{V}$ direct-current supply. What is the length of the wire?

15\,\text{m}

30\,\text{m}

60\,\text{m}

120\,\text{m}

Show answer & explanation

Answer: $30\,\text{m}$

Answer

30\,\text{m}

By applying Ohm's Law (

R = \frac{V}{I}

), the total resistance of the heating element is found to be

60\,\Omega

. Substituting this along with resistivity

\rho = 1.0 \times 10^{-6}\,\Omega\cdot\text{m}

and cross-sectional area

A = 5.0 \times 10^{-7}\,\text{m}^2

into the resistivity equation

R = \frac{\rho L}{A}

yields

L = \frac{R A}{\rho} = 30\,\text{m}

.

Step-by-Step Solution

1

Calculate the electrical resistance of the wire using Ohm's Law.

R = \frac{V}{I} = \frac{240\,\text{V}}{4.0\,\text{A}} = 60\,\Omega

The resistance must be determined from the operational potential difference and current before finding the geometric dimensions.

2

Relate resistance to length, cross-sectional area, and resistivity using

R = \frac{\rho L}{A}

.

60\,\Omega = \frac{(1.0 \times 10^{-6}\,\Omega\cdot\text{m}) \times L}{5.0 \times 10^{-7}\,\text{m}^2}

The resistance of a uniform conductor is directly proportional to its length and inversely proportional to its cross-sectional area.

3

Solve the equation for the wire length

L

.

L = \frac{60 \times 5.0 \times 10^{-7}}{1.0 \times 10^{-6}} = 30\,\text{m}

Rearranging the formula gives

L = \frac{R A}{\rho}

to isolate the required length.

Key Concept

Relationship between potential difference, current, resistance, and wire dimensions (

R = \frac{V}{I} = \frac{\rho L}{A}

)

Question 8Question →

A resistance thermometer has a resistance of $5.0\,\Omega$ at $0^\circ\text{C}$ and $5.8\,\Omega$ at $40^\circ\text{C}$ . What is the temperature coefficient of resistance of the material in $\text{K}^{-1}$ ?

Show answer & explanation

Answer: 0.004

Answer

The temperature coefficient of resistance of the material is

0.004\,\text{K}^{-1}

(or

4.0 \times 10^{-3}\,\text{K}^{-1}

).

The temperature coefficient of resistance is calculated using

\alpha = \frac{R_T - R_0}{R_0 \Delta T}

. Substituting

R_0 = 5.0\,\Omega

,

R_T = 5.8\,\Omega

, and

\Delta T = 40\,\text{K}

yields

\alpha = \frac{0.8}{200} = 0.004\,\text{K}^{-1}

.

Step-by-Step Solution

1

Identify the relationship between resistance and temperature.

R_T = R_0(1 + \alpha \Delta T)

, with

R_0 = 5.0\,\Omega

,

R_T = 5.8\,\Omega

, and

\Delta T = 40^\circ\text{C}

.

The resistance of metallic conductors varies linearly with temperature for moderate temperature changes.

2

Rearrange the expression to isolate the temperature coefficient

\alpha

.

\alpha = \frac{R_T - R_0}{R_0 \Delta T}

Isolating

\alpha

allows direct computation from the given resistance values and temperature interval.

3

Calculate the numerical value of

\alpha

.

\alpha = \frac{5.8 - 5.0}{5.0 \times 40} = \frac{0.8}{200} = 0.004\,\text{K}^{-1}

Dividing the change in resistance by the product of initial resistance and temperature change gives the fractional resistance change per unit temperature change.

Key Concept

Temperature dependence of electrical resistance and temperature coefficient of resistance.

Question 9Question →

A uniform cylindrical wire of length $2.0\,\text{m}$ and radius $1.0\,\text{mm}$ is connected to a direct-current source. When a potential difference of $12\,\text{V}$ is applied across its ends, a steady current of $4.0\,\text{A}$ flows through it. Taking $\pi \approx 3.142$ , what is the resistivity of the material of the wire?

4.71 \times 10^{-6}\,\Omega\cdot\text{m}

4.71 \times 10^{-3}\,\Omega\cdot\text{m}

1.89 \times 10^{-5}\,\Omega\cdot\text{m}

9.43 \times 10^{-6}\,\Omega\cdot\text{m}

Show answer & explanation

Answer: $4.71 \times 10^{-6}\,\Omega\cdot\text{m}$

Answer

The resistivity of the material of the wire is

4.71 \times 10^{-6}\,\Omega\cdot\text{m}

.

Using Ohm's law (

R = V/I

), the wire's resistance is

3.0\,\Omega

. Substituting this resistance, the wire length (

2.0\,\text{m}

), and the circular area (

A = \pi r^2 = 3.142 \times 10^{-6}\,\text{m}^2

) into

\rho = R A / L

yields

4.71 \times 10^{-6}\,\Omega\cdot\text{m}

.

Step-by-Step Solution

1

Calculate the resistance of the wire using Ohm's Law.

R = \frac{V}{I} = \frac{12\,\text{V}}{4.0\,\text{A}} = 3.0\,\Omega

Ohm's Law relates potential difference, current, and resistance.

2

Calculate the cross-sectional area of the wire from its radius.

A = \pi r^2 = 3.142 \times (1.0 \times 10^{-3}\,\text{m})^2 = 3.142 \times 10^{-6}\,\text{m}^2

The wire has a circular cross-section.

3

Rearrange the resistance formula

R = \frac{\rho L}{A}

to solve for resistivity

\rho

.

\rho = \frac{R A}{L} = \frac{3.0\,\Omega \times 3.142 \times 10^{-6}\,\text{m}^2}{2.0\,\text{m}} = 4.713 \times 10^{-6}\,\Omega\cdot\text{m} \approx 4.71 \times 10^{-6}\,\Omega\cdot\text{m}

Resistivity is an intrinsic property derived from resistance, length, and cross-sectional area.

Key Concept

Relationship between resistance, potential difference, current, and material resistivity.

Question 10Question →

A potential difference of $16\,\text{V}$ is applied across a uniform conductor of length $4.0\,\text{m}$ and cross-sectional area $1.5 \times 10^{-6}\,\text{m}^2$ . If the resistivity of the conductor material is $3.0 \times 10^{-7}\,\Omega\cdot\text{m}$ , what is the electric current, in amperes, flowing through the conductor?

Show answer & explanation

Answer: 20

Answer

The electric current flowing through the conductor is

20\,\text{A}

.

The electrical resistance of the wire is first determined using the formula

R = \frac{\rho L}{A} = \frac{(3.0 \times 10^{-7})(4.0)}{1.5 \times 10^{-6}} = 0.8\,\Omega

. Then, by applying Ohm's law (

I = \frac{V}{R}

), the current is computed as

I = \frac{16}{0.8} = 20\,\text{A}

.

Step-by-Step Solution

1

Calculate the electrical resistance of the conductor from its physical dimensions and resistivity.

R = 0.8\,\Omega

Substitute

\rho = 3.0 \times 10^{-7}\,\Omega\cdot\text{m}

,

L = 4.0\,\text{m}

, and

A = 1.5 \times 10^{-6}\,\text{m}^2

into

R = \frac{\rho L}{A}

.

2

Apply Ohm's law to calculate the current flowing through the conductor.

I = 20\,\text{A}

Substitute potential difference

V = 16\,\text{V}

and calculated resistance

R = 0.8\,\Omega

into

I = \frac{V}{R}

.

Key Concept

Relationship between resistivity, resistance, potential difference, and electric current

Question 11Question →

A steady electric current of $3.2\,\text{A}$ flows through a conductor for $5.0\,\text{minutes}$ . Given that the elementary charge is $1.6 \times 10^{-19}\,\text{C}$ , how many electrons pass through a cross-section of the conductor during this period?

1.0 \times 10^{20}

3.6 \times 10^{20}

6.0 \times 10^{21}

1.536 \times 10^{-16}

Show answer & explanation

Answer: $6.0 \times 10^{21}$

Answer

The number of electrons passing through the cross-section of the conductor is

6.0 \times 10^{21}

.

Electric current

I

is related to total electric charge

Q

and time

t

by

Q = I \times t

. Converting time into seconds gives

t = 5.0 \times 60 = 300\,\text{s}

. The total charge passed is

Q = 3.2\,\text{A} \times 300\,\text{s} = 960\,\text{C}

. Using the charge quantization formula

Q = n \cdot e

, the number of electrons

n

is given by

n = \frac{960\,\text{C}}{1.6 \times 10^{-19}\,\text{C}} = 6.0 \times 10^{21}

.

Step-by-Step Solution

1

Convert the time from minutes into seconds.

t = 5.0\,\text{minutes} = 5.0 \times 60\,\text{s} = 300\,\text{s}

SI units require time to be in seconds when calculating electric charge.

2

Calculate the total electric charge passing through the conductor.

Q = I \times t = 3.2\,\text{A} \times 300\,\text{s} = 960\,\text{C}

Electric current is defined as the rate of flow of charge (

I = Q/t

).

3

Determine the number of electrons using charge quantization.

n = \frac{Q}{e} = \frac{960\,\text{C}}{1.6 \times 10^{-19}\,\text{C}} = 6.0 \times 10^{21}

Total charge is equal to the number of carrier electrons multiplied by the elementary charge (

Q = n \cdot e

).

Key Concept

Quantization of Electric Charge and Current

Question 12Question →

A uniform cylindrical metallic conductor has an initial resistance of $12.0\,\Omega$ . The conductor is stretched uniformly until its length increases by $50\%$ , while maintaining constant mass and density. If a constant potential difference of $27.0\,\text{V}$ is subsequently applied across the ends of the stretched conductor, what is the electric current passing through it?

1.0\,\text{A}

1.5\,\text{A}

2.25\,\text{A}

0.67\,\text{A}

Show answer & explanation

Answer: $1.0\,\text{A}$

Answer

The electric current passing through the stretched conductor is

1.0\,\text{A}

.

When a metallic conductor of fixed mass and volume is stretched, increasing its length by a factor of

n = 1.5

causes its cross-sectional area to decrease by a factor of

1.5

. Because resistance is directly proportional to length and inversely proportional to cross-sectional area (

R = \rho L / A

), the new resistance becomes

n^2

times the initial resistance (

1.5^2 \times 12.0\,\Omega = 27.0\,\Omega

). By Ohm's law,

I = V / R = 27.0\,\text{V} / 27.0\,\Omega = 1.0\,\text{A}

.

Step-by-Step Solution

1

Determine the new length and cross-sectional area of the stretched conductor

L_2 = 1.5 L_1

and

A_2 = \frac{A_1}{1.5}

Increasing the length by

50\%

means

L_2 = L_1 + 0.5 L_1 = 1.5 L_1

. Since the volume

V = A \cdot L

remains constant during stretching,

A_1 L_1 = A_2 L_2

, which gives

A_2 = A_1 / 1.5

.

2

Calculate the new resistance of the conductor

R_2 = 27.0\,\Omega

Resistance is given by

R = \rho \frac{L}{A}

. Substituting the new length and area gives

R_2 = \rho \frac{1.5 L_1}{A_1 / 1.5} = (1.5)^2 \rho \frac{L_1}{A_1} = 2.25 R_1 = 2.25 \times 12.0\,\Omega = 27.0\,\Omega

.

3

Apply Ohm's law to find the current

I = 1.0\,\text{A}

Using

I = \frac{V}{R_2}

, substitute

V = 27.0\,\text{V}

and

R_2 = 27.0\,\Omega

to get

I = \frac{27.0\,\text{V}}{27.0\,\Omega} = 1.0\,\text{A}

.

Key Concept

Resistance variation with length and cross-sectional area under constant volume constraint (

R \propto L^2

when stretched)

Estimated Time:2m 0s

Question 13Question →

An electric iron draws a steady current of $2.5\,\text{A}$ when connected to a mains supply. What total electric charge passes through the heating element of the appliance in $2.0\,\text{minutes}$ ?

5.0\,\text{C}

48\,\text{C}

300\,\text{C}

600\,\text{C}

Show answer & explanation

Answer: $300\,\text{C}$

Answer

The total electric charge passing through the heating element is

300\,\text{C}

.

Electric charge

Q

is calculated using the formula

Q = I \times t

. Converting

2.0\,\text{minutes}

into seconds gives

120\,\text{s}

. Multiplying the current of

2.5\,\text{A}

by

120\,\text{s}

gives

300\,\text{C}

.

Step-by-Step Solution

1

Convert the given time duration from minutes into seconds.

t = 2.0\,\text{minutes} \times 60\,\text{s/min} = 120\,\text{s}

.

Standard units require time to be measured in seconds when evaluating charge in coulombs.

2

Apply the electric charge formula

Q = I \times t

.

Q = 2.5\,\text{A} \times 120\,\text{s} = 300\,\text{C}

.

Electric current is defined as the total charge passing a given cross-section per unit time.

Key Concept

Electric Current and Charge Relationship

Question 14Question →

A metallic conductor wire of cross-sectional area $2.5 \times 10^{-6}\,\text{m}^2$ has a resistance of $10.0\,\Omega$ at $0\,^\circ\text{C}$ . The temperature coefficient of resistance of the material is $5.0 \times 10^{-3}\,^\circ\text{C}^{-1}$ . The conductor contains a free-electron density of $5.0 \times 10^{28}\,\text{m}^{-3}$ . When the wire is heated to $100\,^\circ\text{C}$ and connected across a potential difference of $60\,\text{V}$ , what is the drift velocity of the conduction electrons in the wire in millimeters per second ( $\text{mm/s}$ )? (Take elementary charge $e = 1.6 \times 10^{-19}\,\text{C}$ .)

Show answer & explanation

Answer: 0.2

Answer

The drift velocity of the conduction electrons is

0.2\,\text{mm/s}

.

The resistance increases from

10.0\,\Omega

to

15.0\,\Omega

when heated from

0\,^\circ\text{C}

to

100\,^\circ\text{C}

. Applying

60\,\text{V}

results in a current of

4.0\,\text{A}

. Combining this with the cross-sectional area and electron density gives a drift velocity of

2.0 \times 10^{-4}\,\text{m/s}

, which equals

0.2\,\text{mm/s}

.

Step-by-Step Solution

1

Calculate the resistance at the operating temperature (

100\,^\circ\text{C}

)

R_{100} = 15.0\,\Omega

Resistance varies with temperature according to

R_T = R_0(1 + \alpha \Delta T)

.

2

Find the current in the wire using Ohm's Law

I = 4.0\,\text{A}

Current is given by

I = V / R_{100}

.

3

Determine current density

J

J = 1.6 \times 10^6\,\text{A/m}^2

Current density is total current per unit cross-sectional area,

J = I / A

.

4

Calculate the electron drift velocity

v_d

v_d = 2.0 \times 10^{-4}\,\text{m/s} = 0.2\,\text{mm/s}

Drift velocity relates to current density by

v_d = J / (n e)

.

Key Concept

Temperature Dependence of Resistance and Microscopic Model of Electric Current

Estimated Time:2m 0s

Question 15Question →

A resistor of resistance $15\,\Omega$ is connected across a direct-current source. If a steady current of $0.80\,\text{A}$ flows through the resistor, what is the potential difference across its terminals?

12\,\text{V}

18.75\,\text{V}

0.053\,\text{V}

14.2\,\text{V}

Show answer & explanation

Answer: $12\,\text{V}$

Answer

The potential difference across the terminals of the resistor is

12\,\text{V}

.

According to Ohm's law, the potential difference

V

across a resistor is equal to the product of the electric current

I

passing through it and its resistance

R

(

V = I \times R

). Substituting

I = 0.80\,\text{A}

and

R = 15\,\Omega

yields

V = 12\,\text{V}

.

Step-by-Step Solution

1

Identify the given physical quantities

Resistance

R = 15\,\Omega

and electric current

I = 0.80\,\text{A}

.

These are the given parameters needed to find potential difference.

2

Apply Ohm's Law formula for potential difference

V = I \times R

Ohm's law states that potential difference is directly proportional to current for a ohmic resistor of constant resistance.

3

Substitute the values and calculate

V = 0.80\,\text{A} \times 15\,\Omega = 12\,\text{V}

Multiplying current by resistance yields potential difference in volts.

Key Concept

Ohm's Law (

V = IR

)

Question 16Question →

A resistance thermometer has a resistance of $4.0\,\Omega$ at $0\,^\circ\text{C}$ and $6.0\,\Omega$ at $100\,^\circ\text{C}$ . The thermometer is connected in series with a fixed $10.0\,\Omega$ resistor across a $24.0\,\text{V}$ DC power source having an internal resistance of $1.0\,\Omega$ . If a steady current of $1.2\,\text{A}$ flows through the circuit, what is the temperature of the thermometer's environment?

250\,^\circ\text{C}

300\,^\circ\text{C}

450\,^\circ\text{C}

350\,^\circ\text{C}

Show answer & explanation

Answer: $250\,^\circ\text{C}$

Answer

The temperature of the thermometer environment is

250\,^\circ\text{C}

.

By applying the complete circuit equation

E = I(R_\theta + R_{\text{fixed}} + r)

, the total circuit resistance is found to be

20.0\,\Omega

. Subtracting the fixed resistance of

10.0\,\Omega

and the cell internal resistance of

1.0\,\Omega

yields the thermometer resistance

R_\theta = 9.0\,\Omega

. Substituting this into the thermometric relation

\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100\,^\circ\text{C}

gives

\frac{9.0 - 4.0}{6.0 - 4.0} \times 100 = 250\,^\circ\text{C}

.

Step-by-Step Solution

1

Calculate total circuit resistance using Ohm's Law and internal resistance equation

R_{\text{total}} = \frac{E}{I} = \frac{24.0\,\text{V}}{1.2\,\text{A}} = 20.0\,\Omega

The electromotive force (e.m.f) of the source equals total current multiplied by total resistance including internal resistance.

2

Determine the resistance of the thermometer at the unknown temperature (

R_\theta

)

R_\theta = R_{\text{total}} - R_{\text{fixed}} - r = 20.0\,\Omega - 10.0\,\Omega - 1.0\,\Omega = 9.0\,\Omega

The circuit components are in series, so total resistance is the sum of external resistances and internal resistance.

3

Apply the linear resistance thermometer temperature scale formula

$\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100\,^\circ\text{C} = \frac{9.0 - 4.0}{6.0 - 4.0} \times 100 = \frac{5.0}{2.0} \times 100 = 250\,^\circ\text{C}$

Resistance varies linearly with temperature between the ice point (

0\,^\circ\text{C}

) and steam point (

100\,^\circ\text{C}

).

Key Concept

Integration of Ohm's Law, internal resistance of a cell, and resistance thermometry

Estimated Time:2m 0s

Question 17Question →

A conductor wire with a cross-sectional area of $2.0 \times 10^{-6}\,\text{m}^2$ carries a steady current of $1.6\,\text{A}$ . If the conduction electron density of the material is $5.0 \times 10^{28}\,\text{m}^{-3}$ and the elementary charge is $1.6 \times 10^{-19}\,\text{C}$ , what is the average drift velocity of the electrons in the wire?

1.0 \times 10^{-4}\,\text{m/s}

1.0 \times 10^{-6}\,\text{m/s}

1.6 \times 10^{-23}\,\text{m/s}

4.0 \times 10^{-16}\,\text{m/s}

Show answer & explanation

Answer: $1.0 \times 10^{-4}\,\text{m/s}$

Answer

The average drift velocity of the electrons is

1.0 \times 10^{-4}\,\text{m/s}

.

The correct answer is derived from the fundamental relationship

I = n A e v_d

. Solving for drift velocity gives

v_d = \frac{I}{n A e} = \frac{1.6}{5.0 \times 10^{28} \times 2.0 \times 10^{-6} \times 1.6 \times 10^{-19}} = 1.0 \times 10^{-4}\,\text{m/s}

.

Step-by-Step Solution

1

Identify the drift velocity formula relating current to charge carrier parameters

The electric current is given by

I = n A e v_d

, where

I

is current,

n

is electron density,

A

is cross-sectional area,

e

is elementary charge, and

v_d

is drift velocity.

This formula connects macroscopic electric current to microscopic charge dynamics.

2

Rearrange the equation to solve for drift velocity

v_d

v_d = \frac{I}{n A e}

Isolating the unknown variable

v_d

before substituting known values.

3

Substitute the given values into the expression

v_d = \frac{1.6}{(5.0 \times 10^{28}) \times (2.0 \times 10^{-6}) \times (1.6 \times 10^{-19})}

Inserting

I = 1.6\,\text{A}

,

n = 5.0 \times 10^{28}\,\text{m}^{-3}

,

A = 2.0 \times 10^{-6}\,\text{m}^2

, and

e = 1.6 \times 10^{-19}\,\text{C}

.

4

Evaluate the denominator and compute the final value of

v_d

Denominator

= (5.0 \times 2.0 \times 1.6) \times 10^{28 - 6 - 19} = 16.0 \times 10^3 = 1.6 \times 10^4

. Thus,

v_d = \frac{1.6}{1.6 \times 10^4} = 1.0 \times 10^{-4}\,\text{m/s}

.

Simplifying powers of ten gives the final numerical answer.

Key Concept

Relationship between Electric Current and Drift Velocity

Estimated Time:1m 30s

Question 18Question →

Two cylindrical wires, X and Y, are made of the same uniform conducting material. Wire X has length $L$ , diameter $d$ , and an electrical resistance of $12\,\Omega$ . Wire Y has length $2L$ and diameter $2d$ . What is the resistance of wire Y?

3\,\Omega

6\,\Omega

12\,\Omega

24\,\Omega

Show answer & explanation

Answer: $6\,\Omega$

Answer

The resistance of wire Y is

6\,\Omega

.

The resistance of a uniform conductor is given by

R = \frac{4\rho L}{\pi d^2}

. For wire X,

R_X = 12\,\Omega

. For wire Y with length

2L

and diameter

2d

, the new resistance becomes

R_Y = \frac{4\rho(2L)}{\pi (2d)^2} = \frac{8\rho L}{4\pi d^2} = \frac{1}{2}\left(\frac{4\rho L}{\pi d^2}\right) = \frac{1}{2} R_X = 6\,\Omega

.

Step-by-Step Solution

1

Express the resistance of a cylindrical conductor in terms of length

L

and diameter

d

.

R = \rho \frac{L}{A} = \rho \frac{L}{\pi (d/2)^2} = \frac{4\rho L}{\pi d^2}

The cross-sectional area

A

of a circular wire with diameter

d

is given by

A = \frac{\pi d^2}{4}

.

2

Write the resistance formula for wire X using its given value.

R_X = \frac{4\rho L}{\pi d^2} = 12\,\Omega

Wire X has length

L

and diameter

d

.

3

Substitute the parameters of wire Y (

L_Y = 2L

and

d_Y = 2d

) into the resistance formula.

R_Y = \frac{4\rho (2L)}{\pi (2d)^2} = \frac{8\rho L}{4\pi d^2} = \frac{2\rho L}{\pi d^2}

Doubling diameter increases the cross-sectional area by a factor of

2^2 = 4

.

4

Relate the resistance of wire Y to the resistance of wire X and calculate the final numerical value.

R_Y = \frac{1}{2} \left(\frac{4\rho L}{\pi d^2}\right) = \frac{1}{2} R_X = \frac{12\,\Omega}{2} = 6\,\Omega

Since

R_Y = \frac{1}{2} R_X

, halving

12\,\Omega

yields

6\,\Omega

.

Key Concept

Dependence of Electrical Resistance on Conductor Dimensions

Estimated Time:1m 30s

Question 19Question →

A metallic wire has a resistance of $12.0\,\Omega$ at $0\,^\circ\text{C}$ and a temperature coefficient of resistance of $4.0 \times 10^{-3}\,^\circ\text{C}^{-1}$ . The wire is uniformly stretched until its length increases by $25\%$ . Assuming the density and total volume of the wire remain constant during stretching, what is the resistance of the stretched wire at $50\,^\circ\text{C}$ ?

Show answer & explanation

Answer: 22.5

Answer

The resistance of the stretched wire at

50\,^\circ\text{C}

is

22.5\,\Omega

.

Stretching a wire by

25\%

increases its length by a factor of

1.25

and reduces its cross-sectional area by a factor of

1.25

(since volume is conserved). The resistance at

0\,^\circ\text{C}

scales as

(1.25)^2 = 1.5625

, giving

18.75\,\Omega

. Accounting for the temperature increase to

50\,^\circ\text{C}

via

R(T) = R'_0(1 + \alpha T)

yields

18.75 \times (1 + 4.0 \times 10^{-3} \times 50) = 18.75 \times 1.20 = 22.5\,\Omega

.

Step-by-Step Solution

1

Calculate the resistance of the wire at

0\,^\circ\text{C}

after uniform stretching.

R'_0 = 18.75\,\Omega

Uniform stretching by

25\%

increases length to

L' = 1.25 L_0

. Volume conservation (

V = A L

) requires area to decrease to

A' = A_0 / 1.25

. Since

R = \rho L / A

,

R'_0 = R_0 (L'/L_0)^2 = 12.0 \times (1.25)^2 = 18.75\,\Omega

.

2

Apply the temperature coefficient formula to calculate resistance at

50\,^\circ\text{C}

.

R(50) = 22.5\,\Omega

Using

R(T) = R'_0 (1 + \alpha T)

, substitute

R'_0 = 18.75\,\Omega

,

\alpha = 4.0 \times 10^{-3}\,^\circ\text{C}^{-1}

, and

T = 50\,^\circ\text{C}

to find

R(50) = 18.75 \times (1 + 0.20) = 22.5\,\Omega

.

Key Concept

Combined effects of dimensional deformation and temperature on electrical resistance

Question 20Question →

A uniform metallic conductor of length $200\,\text{m}$ and cross-sectional area $2.0 \times 10^{-6}\,\text{m}^2$ has a resistivity of $1.6 \times 10^{-8}\,\Omega\cdot\text{m}$ at an initial temperature of $20\,^\circ\text{C}$ . The temperature coefficient of resistivity for the material is $5.0 \times 10^{-3}\,^\circ\text{C}^{-1}$ . If the operating temperature of the conductor increases to $120\,^\circ\text{C}$ while it is connected across a constant potential difference of $12\,\text{V}$ , what is the magnitude of the electric current flowing through the conductor in amperes?

Show answer & explanation

Answer: 5

Answer

The electric current flowing through the conductor is

5.0\,\text{A}

.

The temperature change of

100\,^\circ\text{C}

increases the resistivity of the material from

1.6 \times 10^{-8}\,\Omega\cdot\text{m}

to

2.4 \times 10^{-8}\,\Omega\cdot\text{m}

via

\rho = \rho_0(1 + \alpha \Delta T)

. Substituting this updated resistivity into

R = \frac{\rho L}{A}

gives a resistance of

2.4\,\Omega

. Applying Ohm's Law

I = \frac{V}{R}

with a potential difference of

12\,\text{V}

yields

5.0\,\text{A}

.

Step-by-Step Solution

1

Calculate the temperature difference

\Delta T = 100\,^\circ\text{C}

The temperature change relative to the reference temperature dictates the change in resistivity.

2

Calculate the resistivity at the final temperature

\rho = 2.4 \times 10^{-8}\,\Omega\cdot\text{m}

Resistivity depends on temperature according to

\rho = \rho_0(1 + \alpha \Delta T)

.

3

Calculate the total electrical resistance of the conductor

R = 2.4\,\Omega

Resistance is related to physical geometry and resistivity by

R = \frac{\rho L}{A}

.

4

Apply Ohm's law to solve for the current

I = 5.0\,\text{A}

Electric current is determined by potential difference divided by resistance (

I = V / R

).

Key Concept

Temperature dependence of resistivity and Ohm's Law

Estimated Time:2m 0s

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