Electricity and Magnetism Practice Questions — JAMB UTME — Page 4

Question 61Question →

A DC supply with an electromotive force (e.m.f.) of $24.0\text{ V}$ and an internal resistance of $2.0\text{ }\Omega$ is connected across three identical $12.0\text{ }\Omega$ resistors connected in parallel. What is the terminal potential difference across the battery in volts?

Show answer & explanation

Answer: 16

Answer

The terminal potential difference across the battery is

16.0\text{ V}

.

The three identical

12.0\text{ }\Omega

resistors in parallel combine to yield an equivalent external resistance of

4.0\text{ }\Omega

. Adding the battery's internal resistance of

2.0\text{ }\Omega

gives a total circuit resistance of

6.0\text{ }\Omega

. The total current drawn from the battery is

I = \frac{24.0\text{ V}}{6.0\text{ }\Omega} = 4.0\text{ A}

. The terminal potential difference is the voltage drop across the external circuit,

V = 4.0\text{ A} \times 4.0\text{ }\Omega = 16.0\text{ V}

.

Step-by-Step Solution

1

Find the equivalent external resistance of the three parallel resistors.

R_p = 4.0\text{ }\Omega

Three identical resistors

R = 12.0\text{ }\Omega

connected in parallel have an equivalent resistance of

R_p = \frac{12.0}{3} = 4.0\text{ }\Omega

.

2

Find the total circuit resistance by adding internal resistance to the parallel combination.

R_{total} = 6.0\text{ }\Omega

Internal resistance

r = 2.0\text{ }\Omega

acts in series with the external parallel combination:

R_{total} = R_p + r = 4.0 + 2.0 = 6.0\text{ }\Omega

.

3

Calculate the total current supplied by the cell.

I = 4.0\text{ A}

Using the circuit formula

I = \frac{E}{R_{total}}

, we divide the e.m.f. of

24.0\text{ V}

by the total resistance of

6.0\text{ }\Omega

.

4

Compute the terminal potential difference across the cell.

V = 16.0\text{ V}

The potential drop across the external parallel network is

V = I R_p = 4.0\text{ A} \times 4.0\text{ }\Omega = 16.0\text{ V}

, which equals

E - Ir = 24.0\text{ V} - (4.0\text{ A} \times 2.0\text{ }\Omega) = 16.0\text{ V}

.

Key Concept

Terminal Potential Difference and Internal Resistance

Estimated Time:1m 30s

Question 62Question →

An electric circuit consists of a voltage source with an electromotive force (e.m.f.) of $6.0\text{ V}$ and an internal resistance of $1.0\text{ }\Omega$ connected across two identical $6.0\text{ }\Omega$ resistors arranged in parallel. If an ammeter of negligible resistance is connected in series with the voltage source, what current will be displayed on the ammeter?

1.5\text{ A}

2.0\text{ A}

0.75\text{ A}

0.46\text{ A}

Show answer & explanation

Answer: $1.5\text{ A}$

Answer

The ammeter will record a total main circuit current of

1.5\text{ A}

.

The correct response demonstrates proper circuit reduction. First, combining the two

6.0\text{ }\Omega

parallel resistors gives an equivalent resistance of

3.0\text{ }\Omega

. Second, adding the

1.0\text{ }\Omega

internal resistance yields a total resistance of

4.0\text{ }\Omega

. Applying Ohm's law (

I = \frac{E}{R+r}

) gives

I = \frac{6.0\text{ V}}{4.0\text{ }\Omega} = 1.5\text{ A}

, which is the total current measured by an ammeter placed in the main circuit.

Step-by-Step Solution

1

Calculate the equivalent external resistance (

R_p

) of the two parallel resistors.

R_p = \frac{6.0 \times 6.0}{6.0 + 6.0} = \frac{36.0}{12.0} = 3.0\text{ }\Omega

Resistors in parallel combine according to

\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}

.

2

Determine the total resistance (

R_{\text{total}}

) of the complete circuit, including internal resistance.

R_{\text{total}} = R_p + r = 3.0\text{ }\Omega + 1.0\text{ }\Omega = 4.0\text{ }\Omega

Internal resistance acts in series with the equivalent external load resistance.

3

Apply Ohm's law for a complete circuit to find the total current (

I

).

I = \frac{E}{R_{\text{total}}} = \frac{6.0\text{ V}}{4.0\text{ }\Omega} = 1.5\text{ A}

The total current supplied by the source equals the electromotive force divided by the total circuit resistance.

Key Concept

Terminal Circuit Current with Internal Resistance and Parallel Resistor Combinations

Question 63Question →

An electric iron rated at $1200\,\text{W}$ is operated for $5\,\text{hours}$ each day for a period of $30\,\text{days}$ . If electrical energy costs $\text{₦}20.00$ per kilowatt-hour ( $\text{kWh}$ ), what is the total cost of electricity consumed by the iron over this period in Naira ( $\text{₦}$ )?

Show answer & explanation

Answer: 3600

Answer

The total cost of electricity consumed by the iron over the 30-day period is 3600 Naira.

To find the cost of electrical energy in commercial units, express power in kilowatts (

1.2\,\text{kW}

) and time in total hours (

150\,\text{h}

). The energy consumed is

1.2 \times 150 = 180\,\text{kWh}

. At a tariff of

\text{₦}20.00

per

\text{kWh}

, the total cost is

180 \times 20 = 3600\,\text{Naira}

.

Step-by-Step Solution

1

Convert the power rating of the appliance from watts to kilowatts

P = \frac{1200\,\text{W}}{1000} = 1.2\,\text{kW}

Commercial energy consumption is calculated in kilowatt-hours (kWh), requiring power in kilowatts.

2

Calculate the total operating time in hours

t = 5\,\text{hours/day} \times 30\,\text{days} = 150\,\text{hours}

The usage duration across the month must be expressed in total hours.

3

Determine electrical energy consumed in kWh

E = P \times t = 1.2\,\text{kW} \times 150\,\text{h} = 180\,\text{kWh}

Energy is the product of power in kilowatts and time in hours.

4

Calculate total cost of energy consumed

\text{Total Cost} = 180\,\text{kWh} \times \text{₦}20.00/\text{kWh} = \text{₦}3600

Total cost is found by multiplying energy in kWh by the unit tariff rate.

Key Concept

Commercial electrical energy unit (kWh) and billing calculation

Question 64Question →

Two electric lamps rated at $60\,\text{W}, 220\,\text{V}$ and $100\,\text{W}, 220\,\text{V}$ respectively are connected in series across a $220\,\text{V}$ supply line. What is the total electric power dissipated by the combination?

37.5\,\text{W}

80.0\,\text{W}

160.0\,\text{W}

40.0\,\text{W}

Show answer & explanation

Answer: $37.5\,\text{W}$

Answer

The total electric power dissipated by the two lamps in series is

37.5\,\text{W}

.

For two appliances designed for the same rated voltage

V

connected in series across that voltage

V

, the equivalent power is given by the formula

P_{\text{total}} = \frac{P_1 P_2}{P_1 + P_2}

. Substituting

P_1 = 60\,\text{W}

and

P_2 = 100\,\text{W}

gives

P_{\text{total}} = \frac{60 \times 100}{60 + 100} = 37.5\,\text{W}

.

Step-by-Step Solution

1

Calculate the resistance of each lamp from its rating.

R_1 = \frac{V^2}{P_1} = \frac{220^2}{60} = 806.67\,\Omega

and

R_2 = \frac{V^2}{P_2} = \frac{220^2}{100} = 484.00\,\Omega

Electrical devices are rated at a specific voltage, allowing their resistance to be determined via

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

.

2

Find the total resistance of the series circuit.

R_{\text{total}} = R_1 + R_2 = 806.67 + 484.00 = 1290.67\,\Omega

Resistors in series add directly.

3

Calculate total power drawn from the

220\,\text{V}

supply.

P_{\text{total}} = \frac{V^2}{R_{\text{total}}} = \frac{220^2}{1290.67} = 37.5\,\text{W}

Total power is total voltage squared divided by equivalent resistance.

Key Concept

Power combination in series circuits

Estimated Time:1m 30s

Question 65Question →

Match each electrical operational scenario on the left with its corresponding effect on electrical energy or power on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Doubling current quadruples heat energy (

H \propto I^2

); halving voltage reduces power to one-quarter (

P \propto V^2

); connecting two identical resistors in parallel doubles total power (

P \propto 1/R_{\text{eq}}

); and operating a 500 W appliance for 2 hours consumes 1.0 kWh.

Each left-side scenario correctly aligns with fundamental laws of electrical energy and power: doubling current quadruples heat energy via

H \propto I^2

; halving voltage reduces power to one-quarter via

P \propto V^2

; parallel resistor connection halves total resistance and doubles power; and operating a 0.5 kW appliance for 2 hours consumes 1.0 kWh.

Step-by-Step Solution

1

Apply Joule's Law of Heating (

H = I^2 R t

) to determine the effect of changing current.

Doubling current from

I

to

2I

results in

H' = (2I)^2 R t = 4 I^2 R t = 4H

, so heat quadruples.

Heat generation depends on the square of current when resistance and time are constant.

2

Apply the voltage-power relationship (

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

) for changing voltage.

Halving potential difference to

V/2

yields

P' = \frac{(V/2)^2}{R} = \frac{V^2}{4R} = \frac{P}{4}

, so power drops to one-quarter.

Power is directly proportional to the square of potential difference across a constant resistance.

3

Calculate parallel resistance and total circuit power.

Equivalent resistance of two parallel resistors is

R/2

. Total power is

P_{\text{total}} = \frac{V^2}{R/2} = 2 \frac{V^2}{R}

, which is double the single resistor power.

Parallel combination reduces net resistance by half, thereby doubling total current and power drawn from a constant voltage source.

4

Calculate energy consumption in commercial units (kWh).

E = 0.5\,\text{kW} \times 2\,\text{h} = 1.0\,\text{kWh}

.

Commercial energy is computed by expressing power in kW and time in hours.

Key Concept

Mathematical relationships governing electrical power (

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

), Joule heating (

H = I^2 R t

), and commercial energy units (kWh).

Question 66Question →

A dip needle placed in the magnetic meridian at a location on the Earth's surface makes an angle of $30^\circ$ with the horizontal. If the total intensity of the Earth's magnetic field at this location is $6.0 \times 10^{-5}\text{ T}$ , what is the vertical component of the Earth's magnetic field?

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

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

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

1.2 \times 10^{-4}\text{ T}

Show answer & explanation

Answer: $3.0 \times 10^{-5}\text{ T}$

Answer

The vertical component of the Earth's magnetic field is

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

.

The vertical component of the Earth's magnetic field is given by

B_v = B \sin \theta

. Substituting

B = 6.0 \times 10^{-5}\text{ T}

and

\theta = 30^\circ

yields

B_v = 6.0 \times 10^{-5} \times 0.5 = 3.0 \times 10^{-5}\text{ T}

.

Step-by-Step Solution

1

Identify the given physical quantities from the problem statement.

Total intensity

B = 6.0 \times 10^{-5}\text{ T}

and inclination angle

\theta = 30^\circ

.

These parameters are required to compute the resolved component.

2

Recall the formula for resolving the vertical component of Earth's magnetic field.

B_v = B \sin \theta

The vertical component corresponds to the vertical side of the right triangle formed by total field

B

and angle of dip

\theta

.

3

Substitute the given values into the formula and calculate.

B_v = (6.0 \times 10^{-5}\text{ T}) \times \sin(30^\circ) = (6.0 \times 10^{-5}) \times 0.5 = 3.0 \times 10^{-5}\text{ T}

.

Since

\sin(30^\circ) = 0.5

, multiplying

6.0 \times 10^{-5}

by

0.5

yields

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

.

Key Concept

Resolution of Earth's total magnetic field into horizontal and vertical components

Question 67Question →

An electric heating element rated at $1000\,\text{W}$ and $220\,\text{V}$ is connected to a $110\,\text{V}$ power line. Assuming the electrical resistance of the heating element remains constant, what is the power dissipated by the heater when operating at this reduced voltage?

250\,\text{W}

500\,\text{W}

1000\,\text{W}

2000\,\text{W}

Show answer & explanation

Answer: $250\,\text{W}$

Answer

250\,\text{W}

The electrical power rating of an appliance defines its fixed resistance via

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

. With

R = 48.4\,\Omega

, connecting the appliance to a

110\,\text{V}

supply yields

P = \frac{110^2}{48.4} = 250\,\text{W}

. Alternatively, because

P \propto V^2

for a constant resistance, halving the voltage reduces the power by a factor of

(1/2)^2 = 1/4

, giving

1000\,\text{W} \times \frac{1}{4} = 250\,\text{W}

.

Step-by-Step Solution

1

Calculate the resistance

R

of the heater from its rated values.

R = \frac{V_{\text{rated}}^2}{P_{\text{rated}}} = \frac{220^2}{1000} = \frac{48400}{1000} = 48.4\,\Omega

The resistance of a heating element is determined by its physical design and rated specifications.

2

Calculate the power

P_{\text{new}}

dissipated when connected to the

110\,\text{V}

supply.

P_{\text{new}} = \frac{V_{\text{new}}^2}{R} = \frac{110^2}{48.4} = \frac{12100}{48.4} = 250\,\text{W}

Electric power dissipated across a constant resistance varies with the square of the applied voltage.

Key Concept

Relationship between voltage, resistance, and electrical power dissipation

Estimated Time:1m 30s

Question 68Question →

At a magnetic observation station, the horizontal component of the Earth's magnetic field is $40\text{ }\mu\text{T}$ and the vertical component is $30\text{ }\mu\text{T}$ . What is the total magnetic field intensity of the Earth at this station in $\mu\text{T}$ ?

Show answer & explanation

Answer: 50

Answer

The total magnetic field intensity of the Earth at this station is

50\text{ }\mu\text{T}

.

The horizontal component (

B_h

) and vertical component (

B_v

) of the Earth's magnetic field act at right angles to each other. Therefore, the resultant total magnetic field intensity (

B

) is calculated using vector addition:

B = \sqrt{B_h^2 + B_v^2} = \sqrt{40^2 + 30^2} = 50\text{ }\mu\text{T}

.

Step-by-Step Solution

1

Identify the vector relationship between the horizontal and vertical components of the Earth's magnetic field.

B = \sqrt{B_h^2 + B_v^2}

, where

B_h = 40\text{ }\mu\text{T}

and

B_v = 30\text{ }\mu\text{T}

.

The horizontal and vertical components of the Earth's magnetic field are mutually perpendicular vector components.

2

Substitute the values into the formula and solve for total magnetic field intensity

B

.

B = \sqrt{(40)^2 + (30)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50\text{ }\mu\text{T}

.

Applying the Pythagorean theorem yields the magnitude of the resultant magnetic field vector.

Key Concept

Resolution of Earth's magnetic field into horizontal (

B_h

) and vertical (

B_v

) components.

Estimated Time:1m 0s

Question 69Question →

At the Earth's magnetic poles, the horizontal component of the Earth's magnetic field reaches its maximum value while the vertical component is equal to zero.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false. At the magnetic poles, the angle of dip is

90^\circ

, which means the horizontal component of the Earth's magnetic field is zero and the vertical component is at its maximum.

The statement is false because at the Earth's magnetic poles, the angle of inclination (dip) is

90^\circ

. Substituting this angle into the component resolution formulas yields

B_h = B \cos 90^\circ = 0

and

B_v = B \sin 90^\circ = B

. Therefore, the horizontal component is zero and the vertical component is maximum at the poles.

Step-by-Step Solution

1

Determine the angle of dip (

\theta

) at the Earth's magnetic poles.

The angle of dip at the poles is

\theta = 90^\circ

.

The Earth's magnetic flux lines enter or leave the surface vertically at the magnetic poles.

2

Calculate the horizontal component (

B_h

) using the magnetic field resolution formula.

B_h = B \cos(90^\circ) = 0

.

Since

\cos(90^\circ) = 0

, there is no horizontal magnetic field component at the poles.

3

Calculate the vertical component (

B_v

) using the magnetic field resolution formula.

B_v = B \sin(90^\circ) = B

.

Since

\sin(90^\circ) = 1

, the vertical component accounts for the entire total magnetic field intensity.

4

Compare the calculated component values against the given statement.

The statement claims

B_h

is maximum and

B_v = 0

, which contradicts the physical reality where

B_h = 0

and

B_v

is maximum.

Therefore, the statement is evaluated as false.

Key Concept

Earth's Magnetic Field Components at Magnetic Poles

Question 70Question →

According to Lenz's law, what occurs at the near face of a stationary solenoid when the north pole of a bar magnet is moved rapidly towards it?

A magnetic north pole is induced at the near face to oppose the approaching magnet.

A magnetic south pole is induced at the near face to attract the approaching magnet.

A magnetic south pole is induced at the near face to repel the approaching magnet.

No magnetic polarity is produced because induction generates only electric voltage without magnetic fields.

Show answer & explanation

Answer: A magnetic north pole is induced at the near face to oppose the approaching magnet.

Answer

A magnetic north pole is induced at the near face to oppose the approaching magnet.

Lenz's law states that the direction of an induced current is always such that its magnetic effect opposes the motion or change causing it. When a north pole approaches the solenoid face, the induced current flows counter-clockwise (viewed from the magnet) to form an induced magnetic north pole at that face, thereby exerting a repulsive force that opposes the motion.

Step-by-Step Solution

1

Identify the cause of change in magnetic flux.

The approaching north pole increases magnetic flux linking the solenoid.

Electromagnetic induction occurs whenever magnetic flux linked with a conductor changes.

2

Apply Lenz's law to determine the polarity of the induced field.

The induced current must produce a magnetic field that opposes the increase in flux caused by the approaching north pole.

Lenz's law states that the induced effect always opposes the cause producing it.

3

Deduce the required magnetic pole at the near face.

A north pole must be set up at the near face because like magnetic poles repel each other, opposing the inward motion.

Repulsion between like poles (North against North) exerts a retarding force on the approaching magnet.

Key Concept

Lenz's Law of Electromagnetic Induction

Question 71Question →

Two point charges, $q_1 = +1.6 \times 10^{-8}\text{ C}$ and $q_2 = +6.4 \times 10^{-8}\text{ C}$ , are fixed in a vacuum at a distance of $0.60\text{ m}$ apart. At what distance from $q_1$ along the line joining the two charges is the net electric field intensity equal to zero?

0.20\text{ m}

0.40\text{ m}

0.12\text{ m}

0.30\text{ m}

Show answer & explanation

Answer: $0.20\text{ m}$

Answer

The distance from

q_1

where the net electric field intensity is zero is

0.20\text{ m}

.

The net electric field is zero where the magnitudes of the electric fields produced by both charges are equal (

E_1 = E_2

). Setting up

\frac{k q_1}{x^2} = \frac{k q_2}{(0.60 - x)^2}

with

q_2 = 4 q_1

yields

\frac{1}{x^2} = \frac{4}{(0.60 - x)^2}

. Taking the square root gives

\frac{1}{x} = \frac{2}{0.60 - x}

, which yields

x = 0.20\text{ m}

from

q_1

.

Step-by-Step Solution

1

Set up the condition for zero net electric field intensity.

The electric field magnitudes produced by

q_1

and

q_2

at distance

x

from

q_1

must be equal in magnitude and opposite in direction:

E_1 = E_2

.

Since both charges are positive, the point of zero net electric field must lie on the line segment connecting them.

2

Substitute the electric field formula into the equilibrium equation.

\frac{k q_1}{x^2} = \frac{k q_2}{(d - x)^2}

, where

d = 0.60\text{ m}

.

Electric field intensity due to a point charge is given by

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

.

3

Simplify the equation by canceling common terms and substituting known charge values.

\frac{1.6 \times 10^{-8}}{x^2} = \frac{6.4 \times 10^{-8}}{(0.60 - x)^2} \implies \frac{1}{x^2} = \frac{4}{(0.60 - x)^2}

.

Dividing both sides by

k \times 1.6 \times 10^{-8}

reduces the numerical coefficients to simple integers.

4

Take the square root of both sides and solve for

x

.

\frac{1}{x} = \frac{2}{0.60 - x} \implies 0.60 - x = 2x \implies 3x = 0.60 \implies x = 0.20\text{ m}

.

Taking the square root removes the quadratic terms and gives a linear relation for the distance

x

from

q_1

.

Key Concept

Electric field superposition and point of zero field intensity between like point charges

Question 72Question →

A charged oil droplet of mass $3.2 \times 10^{-15}\text{ kg}$ remains stationary in a vacuum between two horizontal charged plates where there is a uniform vertical electric field of strength $2.0 \times 10^4\text{ N C}^{-1}$ . Taking $g = 10\text{ m s}^{-2}$ and the elementary charge $e = 1.6 \times 10^{-19}\text{ C}$ , determine the number of excess electrons on the droplet.

Show answer & explanation

Answer: 10

Answer

The number of excess electrons on the droplet is 10.

The droplet is in mechanical equilibrium under two equal and opposite forces: the downward gravitational force

W = mg

and the upward electric force

F_e = qE

. Setting

qE = mg

gives

q = \frac{mg}{E} = 1.6 \times 10^{-18}\text{ C}

. By the quantization of charge (

q = Ne

), dividing this charge by the elementary charge

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

gives exactly 10 excess electrons.

Step-by-Step Solution

1

Calculate the gravitational force (weight) acting on the droplet.

W = mg = (3.2 \times 10^{-15}\text{ kg}) \times (10\text{ m s}^{-2}) = 3.2 \times 10^{-14}\text{ N}

.

For stationary equilibrium, weight provides the downward vertical force.

2

Apply the equilibrium condition to find the electric force.

F_e = W = 3.2 \times 10^{-14}\text{ N}

.

The net vertical force must be zero for the droplet to remain suspended.

3

Determine the charge

q

using

F_e = qE

.

q = \frac{F_e}{E} = \frac{3.2 \times 10^{-14}\text{ N}}{2.0 \times 10^4\text{ N C}^{-1}} = 1.6 \times 10^{-18}\text{ C}

.

Electric field strength relates force and charge.

4

Calculate the number of elementary charges using charge quantization

q = Ne

.

N = \frac{q}{e} = \frac{1.6 \times 10^{-18}\text{ C}}{1.6 \times 10^{-19}\text{ C}} = 10

.

Electric charge exists in discrete integer multiples of the elementary charge

e

.

Key Concept

Equilibrium between electrostatic and gravitational forces combined with charge quantization.

Question 73Question →

A step-down transformer connected to a $240\text{ V}$ AC mains supply operates a $12\text{ V}, 48\text{ W}$ lamp at its normal brightness rating. If the efficiency of the transformer is $80\%$ , what is the electric current drawn by the primary winding from the mains supply?

0.16\text{ A}

0.20\text{ A}

0.25\text{ A}

4.00\text{ A}

Show answer & explanation

Answer: $0.25\text{ A}$

Answer

0.25\text{ A}

The output power delivered to the lamp is

48\text{ W}

. Accounting for the transformer's

80\%

efficiency, the input power required at the primary winding is

\frac{48\text{ W}}{0.80} = 60\text{ W}

. Since the primary voltage is

240\text{ V}

, the primary current drawn is

I_p = \frac{60\text{ W}}{240\text{ V}} = 0.25\text{ A}

.

Step-by-Step Solution

1

Determine the power output at the secondary winding

P_s = 48\text{ W}

The lamp operates at its normal rating, so secondary power equals the lamp rating.

2

Calculate the required power input to the primary winding using transformer efficiency

P_p = \frac{P_s}{\eta} = \frac{48\text{ W}}{0.80} = 60\text{ W}

Efficiency is defined as

\eta = \frac{P_s}{P_p}

, meaning primary input power must exceed secondary output power due to losses.

3

Compute the primary current drawn from the supply

I_p = \frac{P_p}{V_p} = \frac{60\text{ W}}{240\text{ V}} = 0.25\text{ A}

Power in an AC primary circuit is given by

P_p = V_p I_p

assuming a purely resistive secondary load.

Key Concept

Transformer Efficiency and Power Transfer in Electromagnetic Induction

Question 74Question →

A series alternating current (AC) circuit consists of a resistor of resistance $6\ \Omega$ and an inductor with inductive reactance $8\ \Omega$ . What is the total impedance of the circuit?

10\ \Omega

14\ \Omega

2\ \Omega

48\ \Omega

Show answer & explanation

Answer: $10\ \Omega$

Answer

The total impedance of the circuit is

10\ \Omega

.

In a series RL alternating current circuit, total impedance

Z

combines resistance

R

and inductive reactance

X_L

as perpendicular vector components. Using

Z = \sqrt{R^2 + X_L^2}

, substituting

R = 6\ \Omega

and

X_L = 8\ \Omega

gives

Z = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\ \Omega

.

Step-by-Step Solution

1

Identify the given values for resistance and inductive reactance.

R = 6\ \Omega

and

X_L = 8\ \Omega

.

These are the resistive and reactive opposition components in the series RL circuit.

2

Apply the impedance formula for a series RL circuit.

Z = \sqrt{R^2 + X_L^2}

Voltage across a resistor and an inductor are

90^\circ

out of phase, requiring vector/phasor addition to determine total impedance.

3

Calculate the magnitude of total impedance.

Z = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\ \Omega

.

Evaluating the square root yields the net opposing effect to AC current flow.

Key Concept

Impedance in a Series RL Circuit

Question 75Question →

An alternating current supply is connected in series with a resistor of resistance $12\ \Omega$ and an inductor of inductive reactance $5\ \Omega$ . What is the total impedance of the circuit?

13\ \Omega

17\ \Omega

7\ \Omega

169\ \Omega

Show answer & explanation

Answer: $13\ \Omega$

Answer

13\ \Omega

In a series R-L alternating current circuit, the voltage across the resistor is in phase with the current, while the voltage across the inductor leads the current by

90^\circ

. Consequently, resistance and inductive reactance combine vectorially. The total impedance is

Z = \sqrt{R^2 + X_L^2} = \sqrt{12^2 + 5^2} = \sqrt{169} = 13\ \Omega

.

Step-by-Step Solution

1

Identify the given values

Resistance

R = 12\ \Omega

, inductive reactance

X_L = 5\ \Omega

These are the given parameters of the series R-L circuit.

2

Apply the total impedance formula for a series R-L AC circuit

Z = \sqrt{R^2 + X_L^2}

In an AC circuit, resistance and reactance are perpendicular vectors (

90^\circ

out of phase).

3

Substitute the given values and calculate the result

Z = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\ \Omega

Evaluating the square root yields the total opposition to alternating current.

Key Concept

Impedance of Series AC Circuits

Estimated Time:45s

Question 76Question →

A series alternating current circuit comprises a resistor with resistance $30\ \Omega$ , an inductor with inductive reactance $80\ \Omega$ , and a capacitor with capacitive reactance $40\ \Omega$ . What is the total impedance of the circuit?

50\ \Omega

150\ \Omega

70\ \Omega

40\ \Omega

Show answer & explanation

Answer: $50\ \Omega$

Answer

The impedance of the circuit is

50\ \Omega

.

The impedance

Z

of a series RLC circuit is calculated using the formula

Z = \sqrt{R^2 + (X_L - X_C)^2}

. Substituting the given values

R = 30\ \Omega

,

X_L = 80\ \Omega

, and

X_C = 40\ \Omega

yields

Z = \sqrt{30^2 + (80 - 40)^2} = \sqrt{900 + 1600} = 50\ \Omega

.

Step-by-Step Solution

1

Calculate the net reactance (

X_{net}

)

X_{net} = X_L - X_C = 80\ \Omega - 40\ \Omega = 40\ \Omega

Inductive and capacitive reactances are

180^\circ

out of phase in a series AC circuit.

2

Apply the series impedance formula

Z = \sqrt{R^2 + X_{net}^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \Omega

Resistance and net reactance act at

90^\circ

phase to each other, requiring the Pythagorean relation.

Key Concept

Impedance of a Series RLC Circuit

Question 77Question →

A capacitor of capacitance $50\ \mu\text{F}$ is connected across an alternating current (AC) source operating at a frequency of $\frac{100}{\pi}\ \text{Hz}$ . What is the capacitive reactance of the capacitor?

Show answer & explanation

Answer: 100

Answer

The capacitive reactance of the capacitor is

100\ \Omega

.

Capacitive reactance

X_C

is given by the formula

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

. Substituting

C = 50 \times 10^{-6}\ \text{F}

and

f = \frac{100}{\pi}\ \text{Hz}

into the formula yields

X_C = \frac{1}{2\pi (100/\pi) (50 \times 10^{-6})} = \frac{1}{10^{-2}} = 100\ \Omega

.

Step-by-Step Solution

1

Convert capacitance to farads and state all given values

C = 50 \times 10^{-6}\ \text{F}

and

f = \frac{100}{\pi}\ \text{Hz}

Calculations require standard SI base units.

2

Apply the formula for capacitive reactance

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

Capacitive reactance measures the opposition offered by a capacitor to alternating current.

3

Substitute the values and calculate the result

X_C = \frac{1}{2\pi \cdot \frac{100}{\pi} \cdot (50 \times 10^{-6})} = \frac{1}{10,000 \times 10^{-6}} = 100\ \Omega

The factor

\pi

cancels out directly, making the arithmetic simple.

Key Concept

Capacitive Reactance in AC Circuits

Question 78Question →

An alternating current (AC) series circuit contains a resistor, an inductor, and a capacitor. The root-mean-square (RMS) potential differences measured across the resistor, inductor, and capacitor are $80\text{ V}$ , $110\text{ V}$ , and $50\text{ V}$ , respectively. What is the total supply voltage across the circuit in volts?

Show answer & explanation

Answer: 100

Answer

The total supply voltage across the series AC circuit is

100\text{ V}

.

In a series alternating current circuit, the voltages across the resistor, inductor, and capacitor are not in phase. The resistor voltage is in phase with the current, whereas inductor voltage leads by

90^\circ

and capacitor voltage lags by

90^\circ

. The total voltage is calculated using vector addition:

V = \sqrt{V_R^2 + (V_L - V_C)^2}

. Substituting the given values gives

V = \sqrt{80^2 + (110 - 50)^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = 100\text{ V}

.

Step-by-Step Solution

1

Identify the RMS potential differences across each component.

V_R = 80\text{ V}

,

V_L = 110\text{ V}

, and

V_C = 50\text{ V}

.

In a series AC circuit, voltages across reactive components are out of phase with the resistor voltage.

2

Calculate the net reactive voltage difference between the inductor and capacitor.

V_L - V_C = 110\text{ V} - 50\text{ V} = 60\text{ V}

.

Inductive voltage leads current by

90^\circ

while capacitive voltage lags current by

90^\circ

, making them

180^\circ

out of phase with each other.

3

Determine total supply voltage using vector (phasor) addition.

V = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100\text{ V}

.

The resistive voltage and net reactive voltage are perpendicular (

90^\circ

phase angle difference).

Key Concept

Phasor Addition of Voltages in a Series AC Circuit

Question 79Question →

An insulated neutral conductor gains $5.0 \times 10^{13}$ electrons during a electrostatic charging process. Given that the magnitude of the elementary charge is $e = 1.6 \times 10^{-19}\text{ C}$ , what is the magnitude of the net charge acquired by the conductor in microcoulombs ( $\mu\text{C}$ )?

Show answer & explanation

Answer: 8

Answer

The magnitude of the net electric charge acquired by the conductor is

8.0\ \mu\text{C}

.

According to the principle of charge quantization, the total magnitude of charge

Q

acquired by gaining

n

electrons is given by

Q = n e

. Substituting

n = 5.0 \times 10^{13}

and

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

gives

Q = 8.0 \times 10^{-6}\text{ C}

. Expressed in microcoulombs,

8.0 \times 10^{-6}\text{ C} = 8.0\ \mu\text{C}

.

Step-by-Step Solution

1

Apply the principle of quantization of electric charge formula

Formula

Q = n e

established

Electric charge is quantized and exists in integer multiples of the elementary charge.

2

Multiply the number of electrons by the elementary charge value

Q = 8.0 \times 10^{-6}\text{ C}

Calculates total electrostatic charge in base SI units.

3

Convert the value from Coulombs to microcoulombs

8.0\ \mu\text{C}

The unit

1\ \mu\text{C}

equals

10^{-6}\text{ C}

.

Key Concept

Quantization of Electric Charge

Question 80Question →

A small charged sphere of mass $2.0 \times 10^{-4}\text{ kg}$ carrying a positive charge of $+4.0 \times 10^{-8}\text{ C}$ is suspended by a light insulating string between two vertical parallel plates. When a uniform horizontal electric field of magnitude $E$ is applied between the plates, the string deflects and comes to equilibrium at an angle of $45^\circ$ to the vertical. Taking the acceleration due to gravity $g = 10\text{ m}\cdot\text{s}^{-2}$ , calculate the magnitude of the electric field intensity $E$ in $\text{N}\cdot\text{C}^{-1}$ .

Show answer & explanation

Answer: 50000

Answer

The magnitude of the electric field intensity is

50000\text{ N}\cdot\text{C}^{-1}

(or

5.0 \times 10^4\text{ N}\cdot\text{C}^{-1}

).

In electrostatic equilibrium, the sphere experiences three forces: weight (

mg

) vertically downward, electrostatic force (

qE

) horizontally, and tension (

T

) along the thread at

45^\circ

to the vertical. Balancing components gives

T \sin 45^\circ = qE

and

T \cos 45^\circ = mg

. Dividing these yields

\tan 45^\circ = \frac{qE}{mg} = 1

, which gives

qE = mg

. Substituting the given values gives

E = \frac{2.0 \times 10^{-3}}{4.0 \times 10^{-8}} = 50000\text{ N}\cdot\text{C}^{-1}

.

Step-by-Step Solution

1

Calculate the weight of the charged sphere.

W = mg = (2.0 \times 10^{-4}\text{ kg})(10\text{ m}\cdot\text{s}^{-2}) = 2.0 \times 10^{-3}\text{ N}

.

The weight provides the downward vertical force in equilibrium.

2

Relate the electrostatic force to the weight using the angle of deflection.

\tan(45^\circ) = \frac{F_e}{W} \implies 1 = \frac{F_e}{2.0 \times 10^{-3}\text{ N}} \implies F_e = 2.0 \times 10^{-3}\text{ N}

.

In electrostatic equilibrium, the ratio of the horizontal force to the vertical force equals the tangent of the angle with the vertical.

3

Calculate the electric field strength

E

using

F_e = qE

.

E = \frac{F_e}{q} = \frac{2.0 \times 10^{-3}\text{ N}}{4.0 \times 10^{-8}\text{ C}} = 50000\text{ N}\cdot\text{C}^{-1}

.

The electric field intensity is the electric force per unit charge.

Key Concept

Equilibrium of a charged body in a uniform electric field

Estimated Time:2m 0s

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