Alternating Current (AC) Circuits Practice Questions — JAMB UTME

Question 1Question →

An alternating current (AC) series circuit contains a resistor of resistance $R = 30\ \Omega$ , an inductor of inductive reactance $X_L = 80\ \Omega$ , and a capacitor of capacitive reactance $X_C = 40\ \Omega$ . The circuit is connected across an AC supply with a peak voltage of $V_0 = 100\sqrt{2}\text{ V}$ . What is the average power dissipated in the circuit?

120\text{ W}

240\text{ W}

13.3\text{ W}

400\text{ W}

Show answer & explanation

Answer: $120\text{ W}$

Answer

The average electrical power dissipated in the circuit is

120\text{ W}

.

The average power dissipated in an AC series circuit depends solely on the resistive component. First, converting peak voltage

V_0 = 100\sqrt{2}\text{ V}

yields an RMS voltage of

V_{\text{rms}} = 100\text{ V}

. Computing the circuit impedance yields

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

. This produces an RMS current of

I_{\text{rms}} = \frac{100}{50} = 2\text{ A}

. Finally, substituting into

P_{\text{avg}} = I_{\text{rms}}^2 R

gives

(2)^2 \times 30 = 120\text{ W}

.

Step-by-Step Solution

1

Calculate the root-mean-square (RMS) voltage of the AC supply.

V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100\text{ V}

AC power calculations require RMS voltage values rather than peak voltage values.

2

Calculate the net impedance (

Z

) of the series RLC circuit.

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{30^2 + (80 - 40)^2} = \sqrt{30^2 + 40^2} = \sqrt{2500} = 50\ \Omega

Resistance and reactances combine quadratically due to phase differences between voltage across components.

3

Calculate the RMS current (

I_{\text{rms}}

) flowing through the circuit.

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{100\text{ V}}{50\ \Omega} = 2\text{ A}

Ohm's law for AC circuits states

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}

.

4

Calculate the average power dissipated in the circuit.

P_{\text{avg}} = I_{\text{rms}}^2 R = (2\text{ A})^2 \times 30\ \Omega = 120\text{ W}

Power is dissipated only in resistive elements, as pure inductors and capacitors store and return energy without net loss.

Key Concept

Impedance and Power in Series AC Circuits

Question 2Question →

An alternating voltage source of $90\text{ V}$ (RMS) is connected in series across a circuit containing a resistor of resistance $12\ \Omega$ , an inductor of reactance $20\ \Omega$ , and a capacitor of reactance $11\ \Omega$ . What is the root-mean-square (RMS) current flowing through the circuit?

2.1\text{ A}

4.3\text{ A}

6.0\text{ A}

7.5\text{ A}

Show answer & explanation

Answer: $6.0\text{ A}$

Answer

The RMS current flowing through the circuit is

6.0\text{ A}

.

The total impedance

Z

of a series RLC circuit is given by

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

. Substituting the given values gives

Z = \sqrt{12^2 + (20 - 11)^2} = \sqrt{144 + 81} = 15\ \Omega

. Dividing the RMS voltage (

90\text{ V}

) by this impedance yields an RMS current of

6.0\text{ A}

.

Step-by-Step Solution

1

Calculate the net reactance of the series AC circuit

X = X_L - X_C = 20\ \Omega - 11\ \Omega = 9\ \Omega

Inductive and capacitive reactances oppose each other in phase by

180^\circ

.

2

Determine the total impedance (Z) of the series RLC circuit using phasor addition

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\ \Omega

Resistance and net reactance are

90^\circ

out of phase, so impedance is calculated using the Pythagorean theorem.

3

Calculate the RMS current using Ohm's law for AC circuits

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{90\text{ V}}{15\ \Omega} = 6.0\text{ A}

The RMS current is equal to the RMS voltage divided by total circuit impedance.

Key Concept

Impedance and RMS Current in Series RLC Circuits

Question 3Question →

An AC series circuit consists of a resistor of resistance $R = 6\ \Omega$ connected in series with an inductor of inductive reactance $X_L = 8\ \Omega$ . What is the total impedance of the circuit?

10\ \Omega

14\ \Omega

2\ \Omega

100\ \Omega

Show answer & explanation

Answer: $10\ \Omega$

Answer

The total impedance of the circuit is

10\ \Omega

.

The net impedance

Z

in an AC series circuit containing a resistor and an inductor is given by

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

. Substituting

R = 6\ \Omega

and

X_L = 8\ \Omega

yields

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

.

Step-by-Step Solution

1

Identify the formula for impedance in an RL series circuit

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

In an AC circuit, resistance and inductive reactance add vectorially at a

90^\circ

phase angle.

2

Substitute the given values

R = 6\ \Omega

and

X_L = 8\ \Omega

into the formula

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

Squaring each component gives the terms required under the radical.

3

Calculate the square root

Z = 10\ \Omega

The square root of 100 gives the net opposition to current flow (impedance).

Key Concept

Impedance of an RL Series AC Circuit

Estimated Time:45s

Question 4Question →

A $150\text{ V}$ (RMS) AC generator is connected across a series combination of a $12\ \Omega$ resistor, an inductor with an inductive reactance of $20\ \Omega$ , and a capacitor with a capacitive reactance of $11\ \Omega$ . What is the average electrical power consumed by this circuit?

900\text{ W}

1200\text{ W}

1500\text{ W}

1875\text{ W}

Show answer & explanation

Answer: $1200\text{ W}$

Answer

The average electrical power consumed by the circuit is

1200\text{ W}

.

To find the average real power dissipated in an AC series circuit, first determine the net reactance

X = X_L - X_C = 20\ \Omega - 11\ \Omega = 9\ \Omega

. Next, calculate total impedance using

Z = \sqrt{R^2 + X^2} = \sqrt{12^2 + 9^2} = 15\ \Omega

. The RMS current in the circuit is

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{150\text{ V}}{15\ \Omega} = 10\text{ A}

. Since energy is dissipated only by resistance, the average real power is

P = I_{\text{rms}}^2 R = (10)^2 \times 12 = 1200\text{ W}

.

Step-by-Step Solution

1

Calculate the net reactance of the circuit.

X = X_L - X_C = 20\ \Omega - 11\ \Omega = 9\ \Omega

Inductive and capacitive reactances oppose each other in phase by

180^\circ

.

2

Calculate the total impedance of the series RLC circuit.

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\ \Omega

Resistance and net reactance add vectorially at a right angle.

3

Determine the RMS current flowing through the circuit.

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{150\text{ V}}{15\ \Omega} = 10\text{ A}

Ohm's law for AC circuits states that current is the supply RMS voltage divided by total impedance.

4

Calculate the average real power dissipated by the circuit.

P_{\text{avg}} = I_{\text{rms}}^2 R = (10\text{ A})^2 \times 12\ \Omega = 1200\text{ W}

In an AC circuit, average power is only dissipated by resistive components, as ideal inductors and capacitors consume zero net real power over a complete cycle.

Key Concept

Power Dissipation in AC Series Circuits

Question 5Question →

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 6Question →

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 7Question →

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 8Question →

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 9Question →

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 10Question →

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 11Question →

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 12Question →

An RLC series circuit connected across a $100\text{ V}$ (RMS) AC voltage source operates at resonance, dissipating an average power of $400\text{ W}$ . If the inductive reactance at resonance is $25\ \Omega$ , what is the total impedance of the circuit when the capacitance is adjusted such that the capacitive reactance increases by $60\ \Omega$ ?

65\ \Omega

85\ \Omega

60\ \Omega

110\ \Omega

Show answer & explanation

Answer: $65\ \Omega$

Answer

The total impedance of the circuit after adjusting the capacitance is

65\ \Omega

.

At resonance, the net reactance is zero and the circuit behaves purely resistively. Using

P = \frac{V_{\text{rms}}^2}{R}

, the resistance is

R = \frac{100^2}{400} = 25\ \Omega

. Since

X_L = X_C = 25\ \Omega

at resonance, increasing capacitive reactance by

60\ \Omega

gives a new capacitive reactance of

85\ \Omega

. The net reactance magnitude is

|25\ \Omega - 85\ \Omega| = 60\ \Omega

. Combining resistance and net reactance in quadrature yields an impedance of

Z = \sqrt{25^2 + 60^2} = 65\ \Omega

.

Step-by-Step Solution

1

Determine the resistance of the circuit at resonance using the power dissipation formula.

At resonance, impedance equals resistance (

Z = R

) and phase angle is zero, so

P = \frac{V_{\text{rms}}^2}{R} \implies R = \frac{(100)^2}{400} = 25\ \Omega

.

At resonance, inductive reactance and capacitive reactance cancel each other out completely.

2

Identify the initial and modified reactances.

At resonance,

X_L = X_C = 25\ \Omega

. After adjustment, the new capacitive reactance is

X_C' = 25\ \Omega + 60\ \Omega = 85\ \Omega

.

Capacitive reactance was increased by

60\ \Omega

from its resonant value.

3

Calculate the net reactance of the modified circuit.

X_{\text{net}} = |X_L - X_C'| = |25\ \Omega - 85\ \Omega| = 60\ \Omega

.

Net reactance is the magnitude of the difference between inductive and capacitive reactances.

4

Calculate the new total impedance using phasor addition.

Z = \sqrt{R^2 + (X_L - X_C')^2} = \sqrt{25^2 + 60^2} = \sqrt{625 + 3600} = \sqrt{4225} = 65\ \Omega

.

Resistance and net reactance are

90^\circ

out of phase, requiring vector summation (Pythagorean theorem).

Key Concept

Resonance and Impedance in AC Circuits

Question 13Question →

An alternating current (AC) circuit contains an inductor of inductance $L = \frac{0.2}{\pi^2}\ \text{H}$ , a resistor of resistance $R = 40\ \Omega$ , and a variable capacitor $C$ connected in series across a $50\ \text{Hz}$ voltage supply. What capacitance $C$ , in microfarads ( $\mu\text{F}$ ), is required for the circuit to operate at electrical resonance?

Show answer & explanation

Answer: 500

Answer

The capacitance required to achieve electrical resonance is 500 μF.

At electrical resonance in a series RLC circuit, the inductive reactance (

X_L

) equals the capacitive reactance (

X_C

). Setting

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

yields

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

. Substituting

f_0 = 50\ \text{Hz}

and

L = \frac{0.2}{\pi^2}\ \text{H}

into this equation gives

C = 5 \times 10^{-4}\ \text{F}

, which equals

500\ \mu\text{F}

.

Step-by-Step Solution

1

Recall the resonant frequency formula for a series RLC circuit.

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

At resonance, inductive reactance equals capacitive reactance (

X_L = X_C

).

2

Substitute the given numerical parameters into the equation.

50 = \frac{1}{2\pi \sqrt{\frac{0.2}{\pi^2} \cdot C}}

Given frequency

f_0 = 50\ \text{Hz}

and inductance

L = \frac{0.2}{\pi^2}\ \text{H}

.

3

Isolate the square root term and simplify.

\sqrt{0.2 C} = 0.01

Simplifying

2\pi \cdot \frac{1}{\pi} = 2

and rearranging

2 \sqrt{0.2 C} = \frac{1}{50} = 0.02

.

4

Square both sides and solve for

C

in farads.

C = 5 \times 10^{-4}\ \text{F}

0.2 C = (0.01)^2 = 10^{-4}

, so

C = \frac{10^{-4}}{0.2} = 5 \times 10^{-4}\ \text{F}

.

5

Convert capacitance from farads to microfarads.

C = 500\ \mu\text{F}

Multiply farads by

10^6

to express the result in microfarads.

Key Concept

Resonant Frequency in Series AC Circuits

Question 14Question →

An RLC series circuit operating at resonance contains an inductor of inductance $0.1\text{ H}$ and a capacitor of capacitance $10\ \mu\text{F}$ . What is the resonant angular frequency of the circuit in radians per second ( $\text{rad/s}$ )?

Show answer & explanation

Answer: 1000

Answer

The resonant angular frequency of the circuit is

1000\text{ rad/s}

.

The resonant angular frequency

\omega_0

of an AC circuit is determined by the formula

\omega_0 = \frac{1}{\sqrt{LC}}

. Substituting the given values

L = 0.1\text{ H}

and

C = 10^{-5}\text{ F}

gives

\omega_0 = \frac{1}{\sqrt{10^{-6}}} = 1000\text{ rad/s}

.

Step-by-Step Solution

1

Convert given parameters to standard SI units

L = 0.1\text{ H}

and

C = 10 \times 10^{-6}\text{ F} = 10^{-5}\text{ F}

.

Calculations must be performed in base SI units for dimensional consistency.

2

Calculate the resonant angular frequency using

\omega_0 = \frac{1}{\sqrt{LC}}

\omega_0 = \frac{1}{\sqrt{0.1 \times 10^{-5}}} = \frac{1}{\sqrt{10^{-6}}} = 1000\text{ rad/s}

.

At resonance, inductive reactance equals capacitive reactance (

X_L = X_C

), which yields

\omega_0 = \frac{1}{\sqrt{LC}}

.

Key Concept

Resonant Angular Frequency

Question 15Question →

An alternating voltage source of root-mean-square (RMS) voltage $200\ \text{V}$ is connected across a series combination of a resistor with resistance $60\ \Omega$ and a capacitor with capacitive reactance $80\ \Omega$ . Calculate the root-mean-square current in the circuit.

Show answer & explanation

Answer: 2

Answer

The root-mean-square current flowing through the circuit is 2 A.

The total impedance of the series RC circuit is obtained via quadrature sum

Z = \sqrt{R^2 + X_C^2} = \sqrt{60^2 + 80^2} = 100\ \Omega

. Dividing the RMS supply voltage (

200\ \text{V}

) by this impedance gives an RMS current of

2\ \text{A}

.

Step-by-Step Solution

1

Calculate the total impedance of the series RC circuit.

Z = \sqrt{R^2 + X_C^2} = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = 100\ \Omega

In an AC circuit with resistance and capacitive reactance in series, the total Opposition (impedance) is found by phasor addition.

2

Apply Ohm's law for alternating current circuits to find RMS current.

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{200\ \text{V}}{100\ \Omega} = 2\ \text{A}

The RMS current is equal to the RMS voltage divided by the total circuit impedance.

Key Concept

Impedance and RMS Current in AC Series Circuits

Question 16Question →

An alternating current (AC) circuit consists of a resistor of resistance $R = 30\ \Omega$ connected in series with a pure inductor across an AC supply of root-mean-square (RMS) voltage $100\ \text{V}$ . If the average power dissipated in the circuit is $120\ \text{W}$ , what is the inductive reactance of the inductor?

20\ \Omega

40\ \Omega

50\ \Omega

70\ \Omega

Show answer & explanation

Answer: $40\ \Omega$

Answer

The inductive reactance of the inductor is

40\ \Omega

.

In an AC circuit containing a resistor and a pure inductor, power is dissipated solely by the resistance. Using

P = I_{\text{rms}}^2 R

, the current is

I_{\text{rms}} = \sqrt{120 / 30} = 2\ \text{A}

. The total impedance

Z

is

V_{\text{rms}} / I_{\text{rms}} = 100 / 2 = 50\ \Omega

. Applying the phasor formula for impedance

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

, solving for

X_L

yields

X_L = \sqrt{50^2 - 30^2} = 40\ \Omega

.

Step-by-Step Solution

1

Calculate the RMS current in the circuit using the average power formula

I_{\text{rms}} = 2\ \text{A}

In an AC circuit with a resistor and a pure inductor, average power is dissipated only by the resistor:

P = I_{\text{rms}}^2 R \implies 120 = I_{\text{rms}}^2 \times 30 \implies I_{\text{rms}}^2 = 4 \implies I_{\text{rms}} = 2\ \text{A}

.

2

Determine the total impedance of the circuit

Z = 50\ \Omega

The total impedance is the ratio of RMS voltage to RMS current:

Z = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{100\ \text{V}}{2\ \text{A}} = 50\ \Omega

.

3

Calculate the inductive reactance using the impedance relationship for a series RL circuit

X_L = 40\ \Omega

Impedance in a series RL circuit is given by

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

. Substituting the known values gives

50 = \sqrt{30^2 + X_L^2} \implies 2500 = 900 + X_L^2 \implies X_L^2 = 1600 \implies X_L = 40\ \Omega

.

Key Concept

Power Dissipation and Impedance in Series RL AC Circuits

Question 17Question →

An alternating current (AC) circuit operating at a frequency of $50\ \text{Hz}$ contains a resistor of resistance $R = 30\ \Omega$ , an inductor of inductance $L = \frac{0.9}{\pi}\ \text{H}$ , and a capacitor of capacitance $C = \frac{200}{\pi}\ \mu\text{F}$ connected in series. What is the total impedance of the circuit?

50\ \Omega

70\ \Omega

170\ \Omega

40\ \Omega

Show answer & explanation

Answer: $50\ \Omega$

Answer

The total impedance of the AC circuit is

50\ \Omega

.

The inductive reactance is

X_L = 2\pi (50)\left(\frac{0.9}{\pi}\right) = 90\ \Omega

and the capacitive reactance is

X_C = \frac{1}{2\pi (50)\left(\frac{200 \times 10^{-6}}{\pi}\right)} = 50\ \Omega

. Since resistance

R

and net reactance

(X_L - X_C = 40\ \Omega)

are perpendicular vectors in a phasor diagram, the total impedance is calculated using the Pythagorean relation:

Z = \sqrt{30^2 + 40^2} = 50\ \Omega

.

Step-by-Step Solution

1

Calculate the inductive reactance (

X_L

)

X_L = 2\pi f L = 2\pi \times 50 \times \frac{0.9}{\pi} = 90\ \Omega

Inductive reactance depends on supply frequency and inductance.

2

Calculate the capacitive reactance (

X_C

)

X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times \frac{200 \times 10^{-6}}{\pi}} = \frac{1}{0.02} = 50\ \Omega

Capacitive reactance is inversely proportional to supply frequency and capacitance.

3

Determine the net reactance (

X

)

X = X_L - X_C = 90\ \Omega - 50\ \Omega = 40\ \Omega

Inductive and capacitive reactances are

180^\circ

out of phase.

4

Calculate total impedance (

Z

) using phasor addition

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \Omega

Resistance and net reactance are

90^\circ

out of phase, requiring right-triangle vector summation.

Key Concept

Total Impedance in a Series RLC AC Circuit

Estimated Time:2m 0s

Question 18Question →

An alternating voltage source described by $V(t) = 210\sqrt{2}\sin(100\pi t)\ \text{V}$ is connected in series with a $40\ \Omega$ resistor, an inductor of inductive reactance $100\ \Omega$ , and a capacitor of capacitive reactance $70\ \Omega$ . What is the root-mean-square (RMS) current flowing through the circuit?

4.2\ \text{A}

1.0\ \text{A}

3.0\ \text{A}

7.0\ \text{A}

Show answer & explanation

Answer: $4.2\ \text{A}$

Answer

The root-mean-square (RMS) current flowing through the circuit is

4.2\ \text{A}

.

The peak voltage is

V_0 = 210\sqrt{2}\ \text{V}

, giving an RMS voltage of

V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = 210\ \text{V}

. The impedance of the series RLC circuit is calculated by

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (100 - 70)^2} = 50\ \Omega

. Therefore, the RMS current is

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{210}{50} = 4.2\ \text{A}

.

Step-by-Step Solution

1

Determine the RMS voltage from the voltage equation

V_{\text{rms}} = 210\ \text{V}

The standard equation for AC voltage is

V(t) = V_0 \sin(\omega t)

, where

V_0 = 210\sqrt{2}\ \text{V}

. The RMS voltage is

V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = 210\ \text{V}

.

2

Calculate the total impedance of the series RLC circuit

Z = 50\ \Omega

Impedance is determined using phasor addition:

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (100 - 70)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50\ \Omega

.

3

Calculate the RMS current using Ohm's law for AC circuits

I_{\text{rms}} = 4.2\ \text{A}

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{210\ \text{V}}{50\ \Omega} = 4.2\ \text{A}

.

Key Concept

Impedance and RMS current calculation in series RLC alternating current circuits

Estimated Time:2m 0s

Question 19Question →

A series alternating current (AC) circuit consists of a resistor of resistance $R = 40\ \Omega$ , an inductor with inductive reactance $X_L = 70\ \Omega$ , and a capacitor with capacitive reactance $X_C = 40\ \Omega$ , connected to an AC voltage supply of $100\ \text{V}$ . What is the power factor of the circuit?

Show answer & explanation

Answer: 0.8

Answer

The power factor of the circuit is 0.8.

The total impedance

Z

of the series circuit is found using

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

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

\cos\phi = \frac{R}{Z} = \frac{40}{50} = 0.8

.

Step-by-Step Solution

1

Calculate the net reactance of the circuit.

X = X_L - X_C = 70\ \Omega - 40\ \Omega = 30\ \Omega

In a series RLC circuit, the net reactance is the arithmetic difference between the inductive reactance and the capacitive reactance.

2

Calculate the total impedance of the circuit.

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + 30^2} = \sqrt{2500} = 50\ \Omega

Impedance represents the combined opposition to current flow from resistance and net reactance in quadrature.

3

Determine the power factor of the circuit.

\cos\phi = \frac{R}{Z} = \frac{40}{50} = 0.8

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

Key Concept

Power Factor in AC Circuits

Estimated Time:1m 30s

Question 20Question →

An alternating current (AC) source with an RMS voltage of $200\ \text{V}$ is connected in series with a $16\ \Omega$ resistor, an inductor of inductive reactance $X_L = 18\ \Omega$ , and a capacitor of capacitive reactance $X_C = 30\ \Omega$ . What is the RMS current flowing in the circuit?

10.0\ \text{A}

3.13\ \text{A}

12.5\ \text{A}

16.67\ \text{A}

Show answer & explanation

Answer: $10.0\ \text{A}$

Answer

The RMS current flowing in the circuit is

10.0\ \text{A}

.

To find the RMS current in a series RLC AC circuit, first calculate the net reactance:

|X_L - X_C| = |18\ \Omega - 30\ \Omega| = 12\ \Omega

. Next, determine total impedance using phasor addition:

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{16^2 + 12^2} = 20\ \Omega

. Finally, apply Ohm's law for AC:

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{200}{20} = 10.0\ \text{A}

.

Step-by-Step Solution

1

Calculate the net reactance of the series circuit

|X_L - X_C| = |18\ \Omega - 30\ \Omega| = 12\ \Omega

Inductive and capacitive reactances are

180^\circ

out of phase in a series AC circuit.

2

Calculate the total impedance Z of the circuit

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{16^2 + (-12)^2} = \sqrt{256 + 144} = \sqrt{400} = 20\ \Omega

Impedance is the phasor sum of resistance and net reactance.

3

Calculate the RMS current using Ohm's law for AC circuits

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{200\ \text{V}}{20\ \Omega} = 10.0\ \text{A}

The RMS current is the total RMS voltage divided by the circuit impedance.

Key Concept

Impedance and RMS Current in Series RLC AC Circuits

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