Electricity and Magnetism Practice Questions — JAMB UTME — Page 6

Question 101Question →

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 102Question →

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 103Question →

A parallel-plate capacitor with air between its plates has a capacitance of $15\text{ }\mu\text{F}$ . If the plate separation is reduced to one-third of its initial value and a dielectric material of relative permittivity $\varepsilon_r = 4.0$ is completely inserted between the plates, what is the new capacitance of the capacitor?

20\text{ }\mu\text{F}

45\text{ }\mu\text{F}

60\text{ }\mu\text{F}

180\text{ }\mu\text{F}

Show answer & explanation

Answer: $180\text{ }\mu\text{F}$

Answer

The new capacitance of the capacitor is

180\text{ }\mu\text{F}

.

The capacitance of a parallel-plate capacitor is given by

C = \frac{\varepsilon_r \varepsilon_0 A}{d}

. Reducing plate separation to one-third increases capacitance by a factor of 3. Adding a dielectric with relative permittivity

\varepsilon_r = 4.0

increases capacitance by a factor of 4. Combining both effects increases capacitance by a total factor of

3 \times 4 = 12

, yielding

12 \times 15\text{ }\mu\text{F} = 180\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Express the initial capacitance

C_1

using the formula for a parallel-plate air capacitor.

C_1 = \frac{\varepsilon_0 A}{d} = 15\text{ }\mu\text{F}

Air has a relative permittivity of 1.

2

Write the formula for the modified capacitance

C_2

with plate separation

d' = \frac{d}{3}

and dielectric constant

\varepsilon_r = 4.0

.

C_2 = \frac{\varepsilon_r \varepsilon_0 A}{d'} = \frac{4.0 \varepsilon_0 A}{\frac{d}{3}} = 4.0 \times 3 \times \left(\frac{\varepsilon_0 A}{d}\right) = 12 C_1

Reducing distance to

d/3

increases capacitance by a factor of 3, and inserting the dielectric increases capacitance by a factor of 4.

3

Calculate the value of the new capacitance.

C_2 = 12 \times 15\text{ }\mu\text{F} = 180\text{ }\mu\text{F}

Multiplying the combined scaling factor by the initial capacitance gives the final answer.

Key Concept

Parallel Plate Capacitance and Dielectrics

Estimated Time:1m 30s

Question 104Question →

Two identical isolated metal spheres carrying charges of $+8.0 \times 10^{-6}\text{ C}$ and $-2.0 \times 10^{-6}\text{ C}$ are brought into contact and then separated to a distance of $0.30\text{ m}$ in a vacuum. What is the magnitude of the electrostatic force of repulsion, in newtons ( $\text{N}$ ), between the spheres after contact? (Take Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{C}^{-2}$ )

Show answer & explanation

Answer: 0.9

Answer

The magnitude of the electrostatic force of repulsion between the spheres after contact is

0.9\text{ N}

.

When identical conducting spheres touch, their total electric charge is conserved and shared equally. The net charge is

+8.0\,\mu\text{C} + (-2.0\,\mu\text{C}) = +6.0\,\mu\text{C}

, giving each sphere a charge of

+3.0\,\mu\text{C}

. Applying Coulomb's law with a distance of

0.30\text{ m}

yields

0.9\text{ N}

.

Step-by-Step Solution

1

Calculate the net combined charge of the two identical spheres when brought into contact.

Q_{\text{total}} = q_1 + q_2 = (+8.0 \times 10^{-6}\text{ C}) + (-2.0 \times 10^{-6}\text{ C}) = +6.0 \times 10^{-6}\text{ C}

.

According to the principle of conservation of charge, charges add algebraically.

2

Determine the charge on each individual sphere after separation.

q' = \frac{Q_{\text{total}}}{2} = \frac{+6.0 \times 10^{-6}\text{ C}}{2} = +3.0 \times 10^{-6}\text{ C}

.

Identical conducting spheres share total charge equally when in contact.

3

Compute the force of repulsion using Coulomb's law.

F = \frac{k(q')^2}{r^2} = \frac{(9.0 \times 10^9)(3.0 \times 10^{-6})^2}{(0.30)^2} = \frac{9.0 \times 10^9 \times 9.0 \times 10^{-12}}{0.09} = 0.9\text{ N}

.

Coulomb's law defines the electrostatic force between two point charges.

Key Concept

Charge conservation, redistribution by conduction, and Coulomb's law

Question 105Question →

A galvanometer with an internal resistance of $40\text{ }\Omega$ gives a full-scale deflection when a current of $10\text{ mA}$ passes through it. What resistance must be connected in series with the galvanometer to convert it into a voltmeter capable of measuring potential differences up to $10\text{ V}$ ?

960\text{ }\Omega

1000\text{ }\Omega

1040\text{ }\Omega

60\text{ }\Omega

Show answer & explanation

Answer: $960\text{ }\Omega$

Answer

The required multiplier resistance is

960\text{ }\Omega

.

To convert a galvanometer into a voltmeter, a multiplier resistor

R_m

is connected in series. The total resistance of the voltmeter combination is

R_{\text{total}} = R_g + R_m = \frac{V}{I_g} = \frac{10\text{ V}}{0.010\text{ A}} = 1000\text{ }\Omega

. Subtracting the galvanometer's internal resistance (

40\text{ }\Omega

) yields

R_m = 960\text{ }\Omega

.

Step-by-Step Solution

1

Convert the full-scale deflection current to amperes.

I_g = 10\text{ mA} = 10 \times 10^{-3}\text{ A} = 0.010\text{ A}

.

Standard SI units must be used for electrical calculations.

2

Apply the voltmeter multiplier conversion formula.

V = I_g(R_g + R_m) \implies R_m = \frac{V}{I_g} - R_g

.

The multiplier resistor

R_m

is connected in series with the galvanometer resistance

R_g

.

3

Substitute the known values into the equation.

R_m = \frac{10}{0.010} - 40 = 1000 - 40 = 960\text{ }\Omega

.

Subtracting internal resistance gives the external resistance needed for full-scale voltage rating.

Key Concept

Voltmeter Conversion using a Series Multiplier Resistor

Question 106Question →

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 107Question →

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 108Question →

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 109Question →

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 110Question →

An electric lamp rated at $60\,\text{W}$ is kept switched on for $50\,\text{seconds}$ . What is the total electrical energy consumed by the lamp during this time?

3000\,\text{J}

300\,\text{J}

1.2\,\text{J}

110\,\text{J}

Show answer & explanation

Answer: $3000\,\text{J}$

Answer

3000\,\text{J}

Electrical energy is calculated using the formula

E = P \times t

. Substituting the given values

P = 60\,\text{W}

and

t = 50\,\text{s}

gives

E = 60 \times 50 = 3000\,\text{J}

.

Step-by-Step Solution

1

Identify the given physical quantities.

Power

P = 60\,\text{W}

and time

t = 50\,\text{s}

.

Power is given in watts (

\text{J/s}

) and time is given in seconds.

2

Apply the electrical energy formula

E = P \times t

.

E = 60\,\text{W} \times 50\,\text{s} = 3000\,\text{J}

.

Total electrical energy is the product of power dissipation and time duration.

Key Concept

Electrical Energy and Power

Estimated Time:45s

Question 111Question →

An electric cell of electromotive force $E$ and internal resistance $r$ is connected across a parallel combination of two resistors with resistances $6.0\text{ }\Omega$ and $12.0\text{ }\Omega$ . The potential difference across the parallel combination is $4.0\text{ V}$ . When the $12.0\text{ }\Omega$ resistor is removed from the circuit, the current supplied by the cell becomes $0.75\text{ A}$ . What is the internal resistance $r$ of the cell in ohms?

Show answer & explanation

Answer: 2

Answer

The internal resistance of the cell is

2.0\text{ }\Omega

.

Analyzing the circuit under both states yields two simultaneous equations for the e.m.f.

E

in terms of internal resistance

r

:

E = 4.0 + 1.0r

and

E = 4.5 + 0.75r

. Solving these equations gives

r = 2.0\text{ }\Omega

.

Step-by-Step Solution

1

Calculate the equivalent resistance of the parallel resistor network.

R_p = 4.0\text{ }\Omega

Using the parallel resistor formula:

R_p = \frac{R_1 R_2}{R_1 + R_2} = \frac{6.0 \times 12.0}{6.0 + 12.0} = 4.0\text{ }\Omega

.

2

Determine the initial total current delivered by the cell.

I_1 = 1.0\text{ A}

From the terminal voltage across the parallel load:

I_1 = \frac{V_1}{R_p} = \frac{4.0\text{ V}}{4.0\text{ }\Omega} = 1.0\text{ A}

.

3

Formulate the e.m.f. equation for the initial circuit state.

E = 4.0 + 1.0r

Applying the equation

E = V + Ir

gives

E = 4.0 + (1.0)r

.

4

Formulate the e.m.f. equation after removing the

12.0\text{ }\Omega

resistor.

E = 4.5 + 0.75r

With external load

R_2 = 6.0\text{ }\Omega

and current

I_2 = 0.75\text{ A}

,

E = I_2(R_2 + r) = 0.75(6.0 + r) = 4.5 + 0.75r

.

5

Solve for the internal resistance

r

by equating the two e.m.f. expressions.

r = 2.0\text{ }\Omega

Equating the two expressions for

E

:

4.0 + 1.0r = 4.5 + 0.75r \implies 0.25r = 0.50 \implies r = 2.0\text{ }\Omega

.

Key Concept

Internal Resistance and Multi-State Circuit Analysis

Estimated Time:2m 0s

Question 112Question →

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 113Question →

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 114Question →

A $2.0\text{ }\mu\text{F}$ capacitor and a $3.0\text{ }\mu\text{F}$ capacitor are connected in series across a $100\text{ V}$ d.c. power supply. What is the magnitude of the electric charge stored on the $2.0\text{ }\mu\text{F}$ capacitor?

500\text{ }\mu\text{C}

120\text{ }\mu\text{C}

200\text{ }\mu\text{C}

80\text{ }\mu\text{C}

Show answer & explanation

Answer: $120\text{ }\mu\text{C}$

Answer

The electric charge stored on the

2.0\text{ }\mu\text{F}

capacitor is

120\text{ }\mu\text{C}

.

The correct answer is

120\text{ }\mu\text{C}

. For capacitors connected in series, the equivalent capacitance

C_{eq}

is determined using

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}

, giving

C_{eq} = \frac{2.0 \times 3.0}{2.0 + 3.0} = 1.2\text{ }\mu\text{F}

. Multiplying by the source voltage of

100\text{ V}

yields a total charge

Q = C_{eq} V = 120\text{ }\mu\text{C}

. Because capacitors in series store equal amounts of charge, the charge on the

2.0\text{ }\mu\text{F}

capacitor is

120\text{ }\mu\text{C}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance

C_{eq}

of the series combination.

C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{2.0 \times 3.0}{2.0 + 3.0} = 1.2\text{ }\mu\text{F}

Capacitors in series combine according to the reciprocal formula

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}

.

2

Determine the total charge

Q

supplied by the

100\text{ V}

source.

Q = C_{eq} V = 1.2\text{ }\mu\text{F} \times 100\text{ V} = 120\text{ }\mu\text{C}

The total charge is the product of the equivalent capacitance and the total voltage.

3

Identify the charge on the individual

2.0\text{ }\mu\text{F}

capacitor.

Q_1 = Q = 120\text{ }\mu\text{C}

Components connected in series carry the exact same electric charge.

Key Concept

Equivalent Capacitance and Charge Distribution in Series Circuits

Estimated Time:1m 30s

Question 115Question →

A potentiometer wire of length $100\text{ cm}$ has a resistance of $10\text{ }\Omega$ . It is connected in series with a driver cell of electromotive force $3.0\text{ V}$ and internal resistance $2.0\text{ }\Omega$ , alongside an external series resistor of $8.0\text{ }\Omega$ . A test cell of unknown electromotive force $E$ gives a balance point at a length of $60\text{ cm}$ from the zero end of the wire. What is the value of $E$ in volts?

Show answer & explanation

Answer: 0.9

Answer

The electromotive force of the test cell is

0.9\text{ V}

.

The e.m.f. of the test cell is balanced by the potential difference across a length of

60\text{ cm}

of the potentiometer wire. Accounting for the driver cell's internal resistance (

2.0\text{ }\Omega

), external series resistor (

8.0\text{ }\Omega

), and wire resistance (

10\text{ }\Omega

), the total resistance of the primary circuit is

20.0\text{ }\Omega

. This yields a primary current of

0.15\text{ A}

and a potential drop across the wire of

1.5\text{ V}

. The resulting potential gradient is

0.015\text{ V/cm}

, which when multiplied by the balance length of

60\text{ cm}

gives an e.m.f. of

0.9\text{ V}

.

Step-by-Step Solution

1

Calculate total resistance in the primary driver circuit

R_{total} = 10\text{ }\Omega + 2.0\text{ }\Omega + 8.0\text{ }\Omega = 20.0\text{ }\Omega

The driver cell's internal resistance, potentiometer wire, and external series resistor are connected in series.

2

Calculate the current flowing through the potentiometer wire

I = \frac{3.0\text{ V}}{20.0\text{ }\Omega} = 0.15\text{ A}

Apply Ohm's law to the complete primary circuit.

3

Find the voltage drop across the potentiometer wire

V_{wire} = 0.15\text{ A} \times 10\text{ }\Omega = 1.5\text{ V}

The potential difference across the wire depends on its resistance and the primary current.

4

Determine the potential gradient along the wire

k = \frac{1.5\text{ V}}{100\text{ cm}} = 0.015\text{ V/cm}

Potential gradient is the potential drop per unit length of the wire.

5

Calculate the e.m.f. of the unknown test cell

E = 0.015\text{ V/cm} \times 60\text{ cm} = 0.9\text{ V}

At the balance point, no current flows from the test cell, so its e.m.f. equals the potential drop across the balance length.

Key Concept

Potentiometer principle and potential gradient

Estimated Time:2m 0s

Question 116Question →

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

Question 117Question →

Two capacitors with capacitances $C_1 = 4.0\text{ }\mu\text{F}$ and $C_2 = 12.0\text{ }\mu\text{F}$ are connected in series across a $120.0\text{ V}$ d.c. power supply. After becoming fully charged, the capacitors are disconnected from the supply and reconnected in parallel with plates of like polarity connected together. What is the total electrostatic energy lost in the reconnection process?

5.4 \times 10^{-3}\text{ J}

1.62 \times 10^{-2}\text{ J}

1.76 \times 10^{-2}\text{ J}

2.16 \times 10^{-2}\text{ J}

Show answer & explanation

Answer: $5.4 \times 10^{-3}\text{ J}$

Answer

The total electrostatic energy lost in the reconnection process is

5.4 \times 10^{-3}\text{ J}

.

The correct answer is derived by finding the total initial energy stored in series (

2.16 \times 10^{-2}\text{ J}

), determining the total combined charge (

720\text{ }\mu\text{C}

) and parallel capacitance (

16.0\text{ }\mu\text{F}

) upon reconnection to find the final stored energy (

1.62 \times 10^{-2}\text{ J}

), and calculating the difference of

5.4 \times 10^{-3}\text{ J}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance and charge of the initial series arrangement.

C_s = \frac{C_1 C_2}{C_1 + C_2} = \frac{4.0 \times 12.0}{4.0 + 12.0} = 3.0\text{ }\mu\text{F}

, so charge on each capacitor

Q = C_s V = (3.0 \times 10^{-6}\text{ F})(120\text{ V}) = 360\text{ }\mu\text{C}

.

Capacitors in series store identical charge equal to the product of equivalent series capacitance and total applied voltage.

2

Calculate the initial total electrostatic energy stored.

E_i = \frac{1}{2} C_s V^2 = \frac{1}{2} (3.0 \times 10^{-6}\text{ F})(120\text{ V})^2 = 2.16 \times 10^{-2}\text{ J}

.

Initial stored energy is determined by the series combination connected across the supply voltage.

3

Determine total charge and equivalent capacitance after parallel reconnection.

Q_p = Q_1 + Q_2 = 360\text{ }\mu\text{C} + 360\text{ }\mu\text{C} = 720\text{ }\mu\text{C}

and

C_p = C_1 + C_2 = 4.0\text{ }\mu\text{F} + 12.0\text{ }\mu\text{F} = 16.0\text{ }\mu\text{F}

.

Connecting like-polarity plates aggregates the individual charges and sums the capacitances in parallel.

4

Calculate the final potential difference and final stored energy.

V_p = \frac{Q_p}{C_p} = \frac{720\text{ }\mu\text{C}}{16.0\text{ }\mu\text{F}} = 45\text{ V}

, so

E_f = \frac{1}{2} C_p V_p^2 = \frac{1}{2} (16.0 \times 10^{-6}\text{ F})(45\text{ V})^2 = 1.62 \times 10^{-2}\text{ J}

.

Charge redistributes until both capacitors reach a common potential difference

V_p

.

5

Calculate the energy lost during reconnection.

\Delta E = E_i - E_f = 2.16 \times 10^{-2}\text{ J} - 1.62 \times 10^{-2}\text{ J} = 5.4 \times 10^{-3}\text{ J}

.

The energy dissipated as heat and spark during charge redistribution is the difference between initial and final total energies.

Key Concept

Energy dissipation during charge sharing between reconnected capacitors

Question 118Question →

A uniform conductor of length $4.0\,\text{m}$ and cross-sectional area $2.0 \times 10^{-7}\,\text{m}^2$ has an electrical resistance of $0.50\,\Omega$ . What is the electrical resistivity of the material of the conductor in units of $10^{-8}\,\Omega\cdot\text{m}$ ?

Show answer & explanation

Answer: 2.5

Answer

The electrical resistivity of the material is

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

, which gives a numerical value of

2.5

in units of

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

.

The electrical resistance

R

of a conductor is given by

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

, where

\rho

is the resistivity,

L

is the length, and

A

is the cross-sectional area. Rearranging to solve for resistivity yields

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

. Substituting

R = 0.50\,\Omega

,

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

, and

L = 4.0\,\text{m}

gives

\rho = \frac{0.50 \times 2.0 \times 10^{-7}}{4.0} = 2.5 \times 10^{-8}\,\Omega\cdot\text{m}

. Expressed in units of

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

, the answer is

2.5

.

Step-by-Step Solution

1

Identify the relevant formula linking resistance, resistivity, length, and area.

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

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

2

Rearrange the equation to solve for resistivity

\rho

.

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

Isolating

\rho

allows direct evaluation using the given quantitative values.

3

Substitute the known numerical values into the equation.

\rho = \frac{0.50 \times (2.0 \times 10^{-7})}{4.0} = 2.5 \times 10^{-8}\,\Omega\cdot\text{m}

Performing the algebraic calculation gives the resistivity in SI units.

Key Concept

Electrical Resistivity and Conductor Dimensions

Question 119Question →

A $6.0\text{ }\mu\text{F}$ capacitor is charged to a potential difference of $100\text{ V}$ using a direct-current source and then disconnected. It is subsequently connected in parallel across an uncharged $4.0\text{ }\mu\text{F}$ capacitor. What is the total electrostatic potential energy lost in the system during the redistribution of charge, in millijoules (mJ)?

Show answer & explanation

Answer: 12

Answer

The total electrostatic potential energy lost in the system during the redistribution of charge is 12 mJ.

The initial energy stored in the charged capacitor is

U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2}(6.0 \times 10^{-6}\text{ F})(100\text{ V})^2 = 30\text{ mJ}

. When connected in parallel to the uncharged capacitor, the total charge

Q = 600\text{ }\mu\text{C}

is conserved across an equivalent capacitance of

C_{eq} = 6.0\text{ }\mu\text{F} + 4.0\text{ }\mu\text{F} = 10.0\text{ }\mu\text{F}

. The common potential becomes

V_f = \frac{Q}{C_{eq}} = 60\text{ V}

, leading to a final stored energy

U_f = \frac{1}{2} C_{eq} V_f^2 = 18\text{ mJ}

. The energy lost is

\Delta U = U_i - U_f = 30\text{ mJ} - 18\text{ mJ} = 12\text{ mJ}

.

Step-by-Step Solution

1

Calculate the initial stored charge

Q

and initial energy

U_i

in the charged

6.0\text{ }\mu\text{F}

capacitor.

Q = 6.0 \times 10^{-4}\text{ C} = 600\text{ }\mu\text{C}

and

U_i = 3.0 \times 10^{-2}\text{ J} = 30\text{ mJ}

.

Before connection, all charge and energy reside solely on the first capacitor.

2

Find the equivalent capacitance

C_{eq}

when the two capacitors are connected in parallel.

C_{eq} = 6.0\text{ }\mu\text{F} + 4.0\text{ }\mu\text{F} = 10.0\text{ }\mu\text{F}

.

Capacitances add directly when connected in parallel.

3

Determine the common final potential difference

V_f

across the combination.

V_f = \frac{Q}{C_{eq}} = \frac{600\text{ }\mu\text{C}}{10.0\text{ }\mu\text{F}} = 60\text{ V}

.

Total electric charge is conserved during redistribution between connected capacitors.

4

Calculate the final total energy

U_f

stored in the combined system.

U_f = \frac{1}{2} C_{eq} V_f^2 = \frac{1}{2} (10.0 \times 10^{-6}\text{ F})(60\text{ V})^2 = 18\text{ mJ}

.

Both capacitors now store energy under the new common potential difference.

5

Compute the total energy lost

\Delta U = U_i - U_f

.

\Delta U = 30\text{ mJ} - 18\text{ mJ} = 12\text{ mJ}

.

The difference in energy is dissipated as heat in connecting wires and spark/radiation.

Key Concept

Charge Conservation and Energy Dissipation during Charge Sharing in Capacitors

Estimated Time:2m 0s

Question 120Question →

A galvanometer has an internal resistance of $19.0\text{ }\Omega$ and produces a full-scale deflection for a current of $50\text{ mA}$ . What value of shunt resistance, in ohms ( $\Omega$ ), must be connected in parallel with the galvanometer to convert it into an ammeter capable of measuring currents up to $1.0\text{ A}$ ?

Show answer & explanation

Answer: 1

Answer

The required shunt resistance is

1.0\text{ }\Omega

.

To convert a sensitive galvanometer into an ammeter, a low-resistance resistor called a shunt (

R_s

) is connected in parallel with the galvanometer. This provides an alternative path for the bulk of the total current. Since the potential difference across parallel branches is equal,

I_s R_s = I_g R_g

. Substituting

I_g = 0.05\text{ A}

,

R_g = 19.0\text{ }\Omega

, and

I_s = 1.0\text{ A} - 0.05\text{ A} = 0.95\text{ A}

yields

R_s = \frac{0.05 \times 19.0}{0.95} = 1.0\text{ }\Omega

.

Step-by-Step Solution

1

Convert the galvanometer full-scale deflection current

I_g

to amperes.

I_g = 50\text{ mA} = 0.05\text{ A}

Standard SI units must be used for electrical calculations.

2

Calculate the current

I_s

that must bypass the galvanometer through the shunt resistor.

I_s = I - I_g = 1.0\text{ A} - 0.05\text{ A} = 0.95\text{ A}

By Kirchhoff's current law, the total maximum current splits into galvanometer current and shunt current.

3

Calculate the required shunt resistance

R_s

using the parallel voltage relation.

R_s = \frac{I_g R_g}{I_s} = \frac{0.05\text{ A} \times 19.0\text{ }\Omega}{0.95\text{ A}} = 1.0\text{ }\Omega

Because the galvanometer and shunt resistor are connected in parallel, they share the exact same potential difference.

Key Concept

Galvanometer Conversion to Ammeter using a Shunt Resistor

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