Electricity and Magnetism Practice Questions — JAMB UTME — Page 3

Question 41Question →

An alternating current (AC) circuit consists of a $40\ \Omega$ resistor, an inductor with a reactance of $70\ \Omega$ , and a capacitor with a reactance of $40\ \Omega$ connected in series across an AC supply. What is the power factor of the circuit?

0.80

0.60

0.27

1.00

Show answer & explanation

Answer: 0.80

Answer

0.80

The total impedance of the series RLC circuit is found using phasor addition:

Z = \sqrt{40^2 + (70 - 40)^2} = 50\ \Omega

. The power factor is the ratio of resistance to impedance:

\cos \theta = \frac{40}{50} = 0.80

.

Step-by-Step Solution

1

Calculate the net reactance (

X_{\text{net}}

) of the circuit

X_{\text{net}} = X_L - X_C = 70\ \Omega - 40\ \Omega = 30\ \Omega

In a series RLC circuit, inductive and capacitive reactances oppose each other in phase.

2

Calculate total impedance (

Z

) using phasor addition

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = 50\ \Omega

Resistance and net reactance act at right angles in the impedance phasor diagram.

3

Calculate the power factor (

\cos \theta

)

\cos \theta = \frac{R}{Z} = \frac{40}{50} = 0.80

The power factor is defined as the cosine of the phase angle, which equals the ratio of resistance to total impedance.

Key Concept

Power factor of a series RLC AC circuit

Estimated Time:1m 30s

Question 42Question →

A series alternating current (AC) circuit contains an inductor of inductance $L = \frac{1}{\pi^2}\ \text{H}$ , a capacitor of capacitance $C = 25\ \mu\text{F}$ , and a resistor of resistance $R = 50\ \Omega$ . What is the resonant frequency of the circuit in hertz ( $\text{Hz}$ )?

Show answer & explanation

Answer: 100

Answer

The resonant frequency of the circuit is

100\ \text{Hz}

.

At resonance, the inductive reactance

X_L = 2\pi f L

equals the capacitive reactance

X_C = \frac{1}{2\pi f C}

. Equating both yields

f_0 = \frac{1}{2\pi\sqrt{LC}}

. Substituting

L = \frac{1}{\pi^2}\ \text{H}

and

C = 25 \times 10^{-6}\ \text{F}

gives

\sqrt{LC} = \frac{5 \times 10^{-3}}{\pi}\ \text{s}

, leading to

f_0 = \frac{1}{2\pi \left(\frac{5 \times 10^{-3}}{\pi}\right)} = 100\ \text{Hz}

.

Step-by-Step Solution

1

Write down the formula for the resonant frequency of a series RLC circuit.

f_0 = \frac{1}{2\pi\sqrt{LC}}

Resonance occurs when the inductive reactance equals the capacitive reactance (

X_L = X_C

).

2

Substitute the values of inductance

L = \frac{1}{\pi^2}\ \text{H}

and capacitance

C = 25 \times 10^{-6}\ \text{F}

into

\sqrt{LC}

.

\sqrt{LC} = \sqrt{\frac{1}{\pi^2} \times 25 \times 10^{-6}} = \frac{5 \times 10^{-3}}{\pi}\ \text{s}

Simplifying the square root removes the fraction containing

\pi

.

3

Calculate the resonant frequency

f_0

.

f_0 = \frac{1}{2\pi \times \frac{5 \times 10^{-3}}{\pi}} = \frac{1}{10^{-2}} = 100\ \text{Hz}

The factor of

\pi

cancels out in the denominator, resulting in a whole number value.

Key Concept

Resonant Frequency in AC Series Circuits

Estimated Time:1m 30s

Question 43Question →

At a point on the Earth's surface, the total magnetic intensity is $4.0 \times 10^{-5}\text{ T}$ and the angle of dip is $60^\circ$ . What is the horizontal component of the Earth's magnetic field at this point?

2.0 \times 10^{-5}\text{ T}

3.5 \times 10^{-5}\text{ T}

4.0 \times 10^{-5}\text{ T}

8.0 \times 10^{-5}\text{ T}

Show answer & explanation

Answer: $2.0 \times 10^{-5}\text{ T}$

Answer

The horizontal component of the Earth's magnetic field is

2.0 \times 10^{-5}\text{ T}

.

The horizontal component

B_h

of the Earth's magnetic field is given by resolving the total magnetic intensity

B

along the horizontal direction using

B_h = B \cos \theta

. Substituting

B = 4.0 \times 10^{-5}\text{ T}

and

\theta = 60^\circ

yields

B_h = 4.0 \times 10^{-5} \times 0.5 = 2.0 \times 10^{-5}\text{ T}

.

Step-by-Step Solution

1

Identify the given values and formula

Total field

B = 4.0 \times 10^{-5}\text{ T}

, Angle of dip

\theta = 60^\circ

. Formula:

B_h = B \cos \theta

The horizontal component of Earth's magnetic field is obtained by resolving the total magnetic vector along the horizontal plane.

2

Substitute the given values into the equation

B_h = 4.0 \times 10^{-5}\text{ T} \times \cos(60^\circ) = 4.0 \times 10^{-5} \times 0.5 = 2.0 \times 10^{-5}\text{ T}

Since

\cos(60^\circ) = 0.5

, evaluating the product gives the exact horizontal component.

Key Concept

Resolution of Earth's Magnetic Field Components

Question 44Question →

A galvanometer with an internal resistance of $5\text{ }\Omega$ produces a full-scale deflection when a current of $10\text{ mA}$ flows through it. Calculate the shunt resistance, in ohms, required to convert this galvanometer into an ammeter capable of measuring currents up to $50\text{ mA}$ .

Show answer & explanation

Answer: 1.25

Answer

The required shunt resistance is

1.25\text{ }\Omega

.

To extend the range of a galvanometer, a shunt resistor

S

is placed in parallel with it. The potential difference across the galvanometer equals the potential difference across the shunt:

I_g R_g = (I - I_g) S

. Substituting the given values

R_g = 5\text{ }\Omega

,

I_g = 10\text{ mA}

, and maximum current

I = 50\text{ mA}

yields

10\text{ mA} \times 5\text{ }\Omega = (50\text{ mA} - 10\text{ mA}) \times S

, solving to

S = \frac{50}{40} = 1.25\text{ }\Omega

.

Step-by-Step Solution

1

Find the current passing through the parallel shunt resistor

I_s = 40\text{ mA}

By Kirchhoff's current law, the total maximum current divides into the galvanometer current and the shunt current (

I = I_g + I_s

).

2

Calculate the required shunt resistance

S

S = 1.25\text{ }\Omega

Because the galvanometer and shunt resistor are connected in parallel, the potential difference across both branches is equal (

I_g R_g = I_s S

).

Key Concept

Conversion of a galvanometer to an ammeter using a low-resistance shunt in parallel

Question 45Question →

Match each electrical quantity or operational scenario on the left with its corresponding mathematical expression on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electrical power in terms of current and resistance corresponds to

I^2 R

. Electrical energy consumed across voltage

V

for time

t

corresponds to

\frac{V^2 t}{R}

. Rate of heat generation in terms of voltage and current corresponds to

V I

. Total electric charge transformed corresponds to

\frac{E}{V}

.

Each item is matched by applying the fundamental relationships of electrical energy (

E = V I t = I^2 R t = \frac{V^2 t}{R} = Q V

) and electrical power (

P = \frac{E}{t} = V I = I^2 R = \frac{V^2}{R}

).

Step-by-Step Solution

1

Analyze the first scenario (power with current and resistance)

Power formula derived from Ohm's law (

V = IR

) into

P = VI

gives

P = (IR)I = I^2 R

.

Relates current and resistance directly to power dissipation.

2

Analyze the second scenario (energy with voltage, resistance, and time)

Energy

E = P t

. Using

P = \frac{V^2}{R}

, energy becomes

E = \frac{V^2 t}{R}

.

Expresses energy consumption using voltage and resistance over a given time duration.

3

Analyze the third scenario (rate of heat generation with voltage and current)

Rate of heat generation is power

P = V I

.

Direct definition of electrical power as energy converted per unit time.

4

Analyze the fourth scenario (charge transferred from energy and voltage)

Since potential difference is energy per unit charge (

V = \frac{E}{Q}

), rearranging gives

Q = \frac{E}{V}

.

Relates fundamental definitions of potential difference, energy, and charge.

Key Concept

Formulas for Electrical Energy, Power, and Charge

Question 46Question →

A point charge of $+5.0 \times 10^{-8}\text{ C}$ is situated in a vacuum. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}$ , what is the magnitude of the electric field intensity, in $\text{N C}^{-1}$ , at a distance of $0.3\text{ m}$ from the charge?

Show answer & explanation

Answer: 5000

Answer

The magnitude of the electric field intensity at a distance of

0.3\text{ m}

is

5000\text{ N C}^{-1}

.

The electric field intensity

E

at a distance

r

from a point charge

q

in free space is given by

E = \frac{k |q|}{r^2}

. Substituting

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

,

q = 5.0 \times 10^{-8}\text{ C}

, and

r = 0.3\text{ m}

into the expression yields

E = \frac{9.0 \times 10^9 \times 5.0 \times 10^{-8}}{(0.3)^2} = \frac{450}{0.09} = 5000\text{ N C}^{-1}

.

Step-by-Step Solution

1

Identify the given values and formula

q = 5.0 \times 10^{-8}\text{ C}

,

r = 0.3\text{ m}

,

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

, and

E = \frac{k |q|}{r^2}

Electric field intensity surrounding a point charge depends on the charge magnitude and inversely on the square of the distance.

2

Calculate the square of the distance

r^2 = (0.3)^2 = 0.09\text{ m}^2

The inverse-square law requires using

r^2

in the denominator.

3

Substitute values and solve for field intensity

E = \frac{9.0 \times 10^9 \times 5.0 \times 10^{-8}}{0.09} = \frac{450}{0.09} = 5000\text{ N C}^{-1}

Multiplying the terms in the numerator yields

450\text{ N}\cdot\text{m}^2\text{C}^{-1}

, and dividing by

0.09\text{ m}^2

gives

5000\text{ N C}^{-1}

.

Key Concept

Electric field intensity due to a isolated point charge

Question 47Question →

Two identical insulated conducting spheres, $X$ and $Y$ , carry initial charges of $+8.0 \times 10^{-9}\text{ C}$ and $-2.0 \times 10^{-9}\text{ C}$ , respectively. They are brought into brief contact and then separated by a distance of $0.3\text{ m}$ in a vacuum. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}$ , what is the magnitude of the electrostatic force between the two spheres after separation?

9.0 \times 10^{-7}\text{ N}

1.6 \times 10^{-6}\text{ N}

2.5 \times 10^{-6}\text{ N}

3.6 \times 10^{-6}\text{ N}

Show answer & explanation

Answer: $9.0 \times 10^{-7}\text{ N}$

Answer

The magnitude of the electrostatic force between the spheres after separation is

9.0 \times 10^{-7}\text{ N}

.

When two identical conducting spheres touch, the total charge conserves and divides equally between them. The net charge is

(+8.0 \times 10^{-9}\text{ C}) + (-2.0 \times 10^{-9}\text{ C}) = +6.0 \times 10^{-9}\text{ C}

, yielding

+3.0 \times 10^{-9}\text{ C}

on each sphere. Substituting these equal charges and distance

0.3\text{ m}

into Coulomb's law gives

F = \frac{9.0 \times 10^9 \times (3.0 \times 10^{-9})^2}{0.09} = 9.0 \times 10^{-7}\text{ N}

.

Step-by-Step Solution

1

Calculate the total net charge after contact and the charge on each identical sphere.

Net charge

Q_{net} = (+8.0 \times 10^{-9}\text{ C}) + (-2.0 \times 10^{-9}\text{ C}) = +6.0 \times 10^{-9}\text{ C}

. Since the spheres are identical, the charge on each sphere is

q = \frac{+6.0 \times 10^{-9}\text{ C}}{2} = +3.0 \times 10^{-9}\text{ C}

.

When identical conductors touch, total charge is conserved and redistributes equally between them.

2

Apply Coulomb's Law using the redistributed charge and separation distance.

F = k \frac{q_1 q_2}{r^2} = (9.0 \times 10^9) \frac{(3.0 \times 10^{-9})(3.0 \times 10^{-9})}{(0.3)^2} = \frac{8.1 \times 10^{-7}}{0.09} = 9.0 \times 10^{-7}\text{ N}

.

Coulomb's Law calculates the magnitude of electrostatic force between two point-like charges at a given distance.

Key Concept

Charge conservation, redistribution between identical conductors, and Coulomb's Law

Estimated Time:1m 30s

Question 48Question →

A point charge of $+6.0 \times 10^{-6}\text{ C}$ experiences an attractive electrostatic force of $0.54\text{ N}$ when placed at a distance of $1.0\text{ m}$ from a second point charge in a vacuum. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}$ , what is the magnitude of the second charge in microcoulombs ( $\mu\text{C}$ )?

Show answer & explanation

Answer: 10

Answer

The magnitude of the second charge is 10 µC.

Using Coulomb's law

F = \frac{k |q_1 q_2|}{r^2}

, substituting

F = 0.54\text{ N}

,

q_1 = 6.0 \times 10^{-6}\text{ C}

,

r = 1.0\text{ m}

, and

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

yields

|q_2| = 1.0 \times 10^{-5}\text{ C}

, which equals

10\text{ }\mu\text{C}

.

Step-by-Step Solution

1

State Coulomb's Law formula

F = \frac{k |q_1 q_2|}{r^2}

Coulomb's Law describes the electrostatic force between two point charges.

2

Substitute given parameters into the equation

0.54 = \frac{9.0 \times 10^9 \times 6.0 \times 10^{-6} \times |q_2|}{1.0^2}

Knowns:

F = 0.54\text{ N}

,

q_1 = 6.0 \times 10^{-6}\text{ C}

,

r = 1.0\text{ m}

,

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

.

3

Solve for the unknown charge magnitude

q_2

|q_2| = 1.0 \times 10^{-5}\text{ C}

Rearranging yields

|q_2| = \frac{0.54}{5.4 \times 10^4} = 1.0 \times 10^{-5}\text{ C}

.

4

Convert the value from Coulombs to microcoulombs

10\text{ }\mu\text{C}

1\text{ }\mu\text{C} = 10^{-6}\text{ C}

, so

1.0 \times 10^{-5}\text{ C} = 10\text{ }\mu\text{C}

.

Key Concept

Coulomb's Law

Question 49Question →

Two capacitors with capacitances of $3\text{ }\mu\text{F}$ and $6\text{ }\mu\text{F}$ are connected in parallel. This combination is then connected in series with a $9\text{ }\mu\text{F}$ capacitor across a $12\text{ }\text{V}$ direct current source. What is the total electrical energy stored in the network?

3.24 \times 10^{-4}\text{ }\text{J}

6.48 \times 10^{-4}\text{ }\text{J}

7.92 \times 10^{-4}\text{ }\text{J}

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

Show answer & explanation

Answer: $3.24 \times 10^{-4}\text{ }\text{J}$

Answer

The total electrical energy stored in the network is

3.24 \times 10^{-4}\text{ }\text{J}

.

First, find the equivalent capacitance of the parallel branch (

3\text{ }\mu\text{F} + 6\text{ }\mu\text{F} = 9\text{ }\mu\text{F}

). Next, combine this with the

9\text{ }\mu\text{F}

capacitor in series, yielding

C_{eq} = \frac{9 \times 9}{9 + 9} = 4.5\text{ }\mu\text{F} = 4.5 \times 10^{-6}\text{ }\text{F}

. Finally, substituting into the stored energy formula

E = \frac{1}{2} C_{eq} V^2

gives

E = 0.5 \times 4.5 \times 10^{-6} \times 144 = 3.24 \times 10^{-4}\text{ }\text{J}

.

Step-by-Step Solution

1

Calculate equivalent capacitance of the parallel section

C_p = C_1 + C_2 = 3\text{ }\mu\text{F} + 6\text{ }\mu\text{F} = 9\text{ }\mu\text{F}

Capacitors in parallel add directly.

2

Calculate total equivalent capacitance of the network

\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{C_3} = \frac{1}{9\text{ }\mu\text{F}} + \frac{1}{9\text{ }\mu\text{F}} = \frac{2}{9\text{ }\mu\text{F}} \implies C_{eq} = 4.5\text{ }\mu\text{F} = 4.5 \times 10^{-6}\text{ }\text{F}

Capacitors in series combine via reciprocals.

3

Calculate total stored electrical energy

E = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} \times (4.5 \times 10^{-6}\text{ }\text{F}) \times (12\text{ }\text{V})^2 = 3.24 \times 10^{-4}\text{ }\text{J}

Energy stored in a capacitor network depends on equivalent capacitance and potential difference across it.

Key Concept

Equivalent capacitance of series-parallel combinations and energy stored in a capacitor

Question 50Question →

A parallel plate capacitor with a capacitance of $12\text{ }\mu\text{F}$ is fully charged by connecting it across a $50\text{ V}$ direct current power source. What is the total electrostatic energy stored in the capacitor in millijoules ( $\text{mJ}$ )?

Show answer & explanation

Answer: 15

Answer

The total electrostatic energy stored in the capacitor is

15\text{ mJ}

.

The energy stored in a capacitor is given by

E = \frac{1}{2}CV^2

. Substituting

C = 12 \times 10^{-6}\text{ F}

and

V = 50\text{ V}

yields

E = \frac{1}{2} \times 12 \times 10^{-6} \times 2500 = 0.015\text{ J} = 15\text{ mJ}

.

Step-by-Step Solution

1

Convert given values to standard SI units.

C = 12 \times 10^{-6}\text{ F}

,

V = 50\text{ V}

Calculating in SI units ensures the resultant energy is in Joules.

2

Apply the formula for energy stored in a charged capacitor.

E = \frac{1}{2} C V^2

Work done during charging is stored as electrostatic potential energy in the electric field between the plates.

3

Substitute the values and convert Joules to millijoules.

E = \frac{1}{2} \times (12 \times 10^{-6}) \times 2500 = 0.015\text{ J} = 15\text{ mJ}

Multiplying Joules by

10^3

converts the result into millijoules.

Key Concept

Energy stored in a capacitor (

E = \frac{1}{2}CV^2

)

Question 51Question →

A capacitor of capacitance $5\text{ }\mu\text{F}$ is charged to a potential difference of $200\text{ V}$ and then disconnected from the power supply. If it is subsequently connected in parallel across an uncharged capacitor of capacitance $15\text{ }\mu\text{F}$ , what is the final common potential difference across the combination in volts?

Show answer & explanation

Answer: 50

Answer

The final common potential difference across the combination is 50 V.

By charge conservation, the total charge

Q = C_1 V_1 = 5\text{ }\mu\text{F} \times 200\text{ V} = 1000\text{ }\mu\text{C}

is shared across the two parallel capacitors. The total equivalent capacitance is

C_{\text{total}} = C_1 + C_2 = 20\text{ }\mu\text{F}

. Therefore, the final potential difference is

V = \frac{Q}{C_{\text{total}}} = \frac{1000\text{ }\mu\text{C}}{20\text{ }\mu\text{F}} = 50\text{ V}

.

Step-by-Step Solution

1

Calculate the initial electric charge (

Q

) stored on the charged capacitor

Q = 5\text{ }\mu\text{F} \times 200\text{ V} = 1000\text{ }\mu\text{C}

Before connection, all charge is stored solely on the

5\text{ }\mu\text{F}

capacitor.

2

Calculate the total equivalent capacitance (

C_{\text{total}}

) of the parallel network

C_{\text{total}} = 5\text{ }\mu\text{F} + 15\text{ }\mu\text{F} = 20\text{ }\mu\text{F}

Capacitors in parallel add directly (

C_{\text{total}} = C_1 + C_2

).

3

Apply the law of conservation of charge to find the final common voltage (

V

)

V = \frac{Q}{C_{\text{total}}} = \frac{1000\text{ }\mu\text{C}}{20\text{ }\mu\text{F}} = 50\text{ V}

The total charge remains conserved and redistributes across the total combined capacitance.

Key Concept

Charge Redistribution and Conservation in Parallel Capacitors

Question 52Question →

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 53Question →

A galvanometer has an internal resistance of $20\text{ }\Omega$ and produces a full-scale deflection for a current of $15\text{ mA}$ . What resistance of the multiplier, in ohms ( $\Omega$ ), is required to convert this galvanometer into a voltmeter capable of measuring a maximum potential difference of $30\text{ V}$ ?

Show answer & explanation

Answer: 1980

Answer

The required multiplier resistance is

1980\text{ }\Omega

.

To convert a galvanometer into a voltmeter, a multiplier resistor

R_m

is connected in series. The maximum potential difference

V

across the combination is given by

V = I_g(R_g + R_m)

. Substituting

V = 30\text{ V}

,

I_g = 0.015\text{ A}

, and

R_g = 20\text{ }\Omega

yields

R_m = 1980\text{ }\Omega

.

Step-by-Step Solution

1

Convert current from milliamperes to amperes.

I_g = 15\text{ mA} = 0.015\text{ A}

.

Standard electrical formulas require current in amperes.

2

Apply the relationship between voltage range, galvanometer current, internal resistance, and multiplier resistance.

V = I_g(R_g + R_m)

, where

V = 30\text{ V}

,

I_g = 0.015\text{ A}

, and

R_g = 20\text{ }\Omega

.

A multiplier resistor is connected in series with the galvanometer so that the total potential difference is distributed across both components.

3

Rearrange the equation and compute

R_m

.

R_m = \frac{30}{0.015} - 20 = 2000 - 20 = 1980\text{ }\Omega

.

Subtracting the internal resistance of the galvanometer from the total required resistance yields the necessary multiplier resistance.

Key Concept

Galvanometer conversion to a voltmeter using a series multiplier resistor

Estimated Time:1m 30s

Question 54Question →

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 55Question →

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 56Question →

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 57Question →

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 58Question →

A cell supplies a current of $0.6\text{ A}$ when connected across a $2.0\text{ }\Omega$ resistor. When the resistor is replaced with a $5.0\text{ }\Omega$ resistor, the current supplied by the cell drops to $0.3\text{ A}$ . What is the internal resistance of the cell?

1.0\text{ }\Omega

0.0\text{ }\Omega

3.0\text{ }\Omega

0.5\text{ }\Omega

Show answer & explanation

Answer: $1.0\text{ }\Omega$

Answer

The internal resistance of the cell is

1.0\text{ }\Omega

.

The electromotive force (e.m.f.)

E

of a cell is given by

E = I(R + r)

, where

I

is the current,

R

is external resistance, and

r

is internal resistance. Setting up equations for both cases:

E = 0.6(2.0 + r)

and

E = 0.3(5.0 + r)

. Equating them gives

1.2 + 0.6r = 1.5 + 0.3r

, which simplifies to

0.3r = 0.3

, yielding

r = 1.0\text{ }\Omega

.

Step-by-Step Solution

1

Formulate the electromotive force equation for the first circuit setup

E = I_1(R_1 + r) = 0.6(2.0 + r) = 1.2 + 0.6r

The total voltage supplied by the cell equals the total current multiplied by the sum of external and internal resistance.

2

Formulate the electromotive force equation for the second circuit setup

E = I_2(R_2 + r) = 0.3(5.0 + r) = 1.5 + 0.3r

The cell's electromotive force

E

and internal resistance

r

remain unchanged despite changing the external resistor.

3

Equate both expressions for electromotive force and solve for internal resistance

1.2 + 0.6r = 1.5 + 0.3r \implies 0.3r = 0.3 \implies r = 1.0\text{ }\Omega

Equating the two expressions allows direct solution for the single unknown variable

r

.

Key Concept

Electromotive Force and Internal Resistance of Cells

Question 59Question →

In a metre bridge experiment, a standard resistor of $4.0\text{ }\Omega$ is connected in the left gap and an unknown resistor $R$ is connected in the right gap. When balanced, the balance point is found at a distance of $40.0\text{ cm}$ from the left end of the $100.0\text{ cm}$ bridge wire. What is the value of the unknown resistance $R$ in ohms?

Show answer & explanation

Answer: 6

Answer

The value of the unknown resistance

R

is

6.0\text{ }\Omega

.

The metre bridge operates on the Wheatstone bridge principle. At balance, the ratio of the resistance in the left gap to the length of the left segment equals the ratio of the resistance in the right gap to the length of the right segment. Substituting

4.0\text{ }\Omega

for the left gap and

40.0\text{ cm}

and

60.0\text{ cm}

for the two wire lengths yields

R = 6.0\text{ }\Omega

.

Step-by-Step Solution

1

Determine the length of the wire segment corresponding to the right gap.

l_2 = 100.0\text{ cm} - 40.0\text{ cm} = 60.0\text{ cm}

.

The total length of a standard metre bridge wire is

100.0\text{ cm}

.

2

Apply the Wheatstone bridge principle for the metre bridge balance condition.

\frac{R_{\text{left}}}{l_1} = \frac{R}{l_2} \implies R = R_{\text{left}} \times \frac{l_2}{l_1}

.

At balance, the potential drop per unit length across the two wire segments is proportional to their respective lengths.

3

Calculate the magnitude of the unknown resistor

R

.

R = 4.0 \times \frac{60.0}{40.0} = 6.0\text{ }\Omega

.

Multiplying and simplifying yields the exact resistance value.

Key Concept

Metre Bridge Principle (Wheatstone Bridge)

Question 60Question →

A battery with an electromotive force (e.m.f.) of $9.0\text{ V}$ and an internal resistance of $0.5\text{ }\Omega$ is connected to an external circuit containing two resistors of $4.0\text{ }\Omega$ and $1.5\text{ }\Omega$ connected in series. What is the terminal potential difference across the battery?

8.25\text{ V}

9.00\text{ V}

0.75\text{ V}

6.00\text{ V}

Show answer & explanation

Answer: $8.25\text{ V}$

Answer

The terminal potential difference across the battery is

8.25\text{ V}

.

The correct answer is obtained by finding the total circuit resistance (

6.0\text{ }\Omega

), determining the current (

1.5\text{ A}

), and subtracting the internal voltage drop (

0.75\text{ V}

) from the electromotive force (

9.0\text{ V}

) to give

8.25\text{ V}

.

Step-by-Step Solution

1

Calculate the total resistance of the circuit.

R_{\text{total}} = R_1 + R_2 + r = 4.0\text{ }\Omega + 1.5\text{ }\Omega + 0.5\text{ }\Omega = 6.0\text{ }\Omega

The external resistors and the battery's internal resistance are connected in series.

2

Calculate the total current flowing through the circuit using Ohm's Law.

I = \frac{E}{R_{\text{total}}} = \frac{9.0\text{ V}}{6.0\text{ }\Omega} = 1.5\text{ A}

The total current depends on the electromotive force and total circuit resistance.

3

Calculate the terminal potential difference across the battery.

V = E - I r = 9.0\text{ V} - (1.5\text{ A} \times 0.5\text{ }\Omega) = 9.0\text{ V} - 0.75\text{ V} = 8.25\text{ V}

Terminal voltage is equal to the e.m.f. minus the potential drop across the internal resistance.

Key Concept

Terminal potential difference and internal resistance in DC electric circuits

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