Alternating Currents Questions and Answers AP Inter 2nd Year Physics Chapter 7

Regular practice with AP Inter 2nd Year Physics Study Material Chapter 7 Alternating Currents Questions and Answers helps students stay prepared for examinations.

AP Inter 2nd Year Physics 7th Lesson Alternating Currents Questions and Answers

I. Multiple Choice Questions

Question 1.
The r.m.s. value of an alternating current of peak value 10 A is:
1) 5 A
2) 7.07 A
3) 10 A
4) 14.14 A
Answer:
2) 7.07 A
I_rms = \(\frac{\mathrm{I}_0}{\sqrt{2}}=\frac{10}{1.414}\) = 7.07 A

Question 2.
The frequency of AC mains in India is:
1) 25 Hz
2) 50 Hz
3) 100 Hz
4) 60 Hz
Answer:
2) 50 Hz
The frequency of AC mains in India is 50Hz.

Question 3.
The power factor of a pure inductive circuit is:
1) 1
2) 0
3) -1
4) 0.5
Answer:
2) 0
In a purely inductive circuit, the phase difference Φ between current and voltage is 90°. Power factor cosΦ = cos 90° = 0.

Question 4.
In an AC circuit, the instantaneous current is given by i – 10 sin(100π t) A. The frequency of the alternating current is:
1) 50 Hz
2) 100 Hz
3) 25 Hz
4) 200 Hz
Answer:
1) 50 Hz
Comparing i = 10 sin 100πt with i = i₀ sin ωt, we have ω = 100π
We know ω = 2πf
∴ Frequency f = \(\frac{\omega}{2 \pi}=\frac{100 \pi}{2 \pi}\) = 50Hz

Question 5.
In a series LCR circuit at resonance, the current is:
1) minimum
2) maximum
3) zero
4) infinite
Answer:
2) maximum
At resonance, X_L = X_C then Z = R. So when impedance is at lowest value, then current is maximum.

Question 6.
The phase difference between current and voltage in a purely capacitive AC circuit is:
1) 0°
2) I leads V by 90°
3) I lags behind V by 90°
4) 180°
Answer:
2) I leads V by 90°
For pure capacitor V = v_msinωt, I = I_msin(ωt + 90°) ∴ Current I leads V by 90°

Question 7.
The reactance of a capacitor decreases with:
1) decrease in capacitance
2) increase in frequency
3) both (1) and (2)
4) decrease in frequency
Answer:
2) increase in frequency
Capacitive reactance X_C = \(\frac{1}{2 \pi \mathrm{fC}}\)
⇒ X_C ∝ \(\frac{1}{\mathrm{f}}\)
X_C decreases with increase in f.

Question 8.
In a purely resistive AC circuit, the average power dissipated is given by:
1) VI
2) VI/2
3) VI cosΦ
4) zero
Answer:
1) VI
In a purely resistive circuit, average power dissipated is P = VI cosΦ.
Power factor cosΦ = 1 ∴ P = VI cosΦ = VI(1) = VI

Question 9.
The power factor of a series LCR circuit at resonance is:
1) 0
2) 0.5
3) 1
4) -1
Answer:
3) 1
At resonance, the phase difference between voltage and current is zero.

Question 10.
A transformer is designed to:
1) increase or decrease AC voltage
2) convert AC into DC
3) increase the frequency of AC
4) store electric charge
Answer:
1) increase or decrease AC voltage
Transformed works on the principle of mutual induction. Transformer is used to change the AC voltage levels i.e, step up and step down, keeping the frequency and power are same.

II. Fill in the Blanks

Question 1.
The r.m.s. value of an AC of peak value I₀ is ________
Answer:
I₀ /\(\sqrt{2}\) .
I_rms = \(\frac{\mathrm{I}_0}{\sqrt{2}}\) = 0.707 I₀

Question 2.
The phase difference between current and voltage in a purely resistive AC circuit is _________.
Answer:
0°
In a pure resistor V = v_msinωt, I = I_msin(ωt) ∴ Phase difference is 0°

Question 3.
The power factor of a pure inductive or capacitive circuit is ________.
Answer:
0
In pure inductive Φ = 90° ∴ Power factor = cos Φ = cos 90° = 0.

Question 4.
In an LCR circuit at resonance, the impedance is equal to __________
Answer:
resistance (R)
At resonance, X_L = X_C. Hence, Impedance Z = R

Question 5.
The average power consumed in a purely inductive circuit is _________
Answer:
zero.
At pure inductor stores energy in its magnetic field during one half cycle and returns it all to the source in the next. The net energy and hence average energy consumed over a full cycle is zero.

Question 6.
The expression for the capacitive reactance _________
Answer:
X_C = 1/ωC or 1 / 2πfC.

Question 7.
The expression for the inductive reactance ___________
Answer:
X_L = ωCL or 2 π f L.

Question 8.
In a purely capacitive circuit, the current _________ the voltage by 90°.
Answer:
leads
For pure capacitor V = v_msinωt, I = I_msin(ωt + 90°)
∴ Current I leads V by 90°.

Question 9.
In a transformer, the core is laminated to reduce __________ losses.
Answer:
eddy current
In a transformer, the core is laminated to reduce eddy current losses.

Question 10.
The power factor of an AC circuit is given by _______.
Answer:
cosΦ
The power factor of an AC circuit is given by cosΦ

III. One Word Answer Questions

Question 1.
What is the r.m.s value of <111 AC whose peak value is 141 V?
Answer:
V_rms = \(\frac{V_m}{\sqrt{2}}=\frac{141}{1.414}\) = 99.7V ≃ 100V [∵ peak value V₀ = 141V]

Question 2.
What is the time period of AC mains in India?
Answer:
The time period (T) of AC mains in India is 0.02 seconds.
∴ Time period T = \(\frac{1}{\text { frequency }}=\frac{1}{50}\) = 0.02s = 20 milli seconds [∵ standard AC frequency in India is 50Hz.]

Question 3.
In which type of AC circuit is the phase difference between current and voltage zero?
Answer:
In a purely resistive AC circuit, the phase difference between current and voltage is zero.

Question 4.
What is the condition f resonance in a series LCR circuit?
Answer:
Condition for resonance: X_L = X_C i.e., Inductive reactance = Capacitive reactance.

Question 5.
At resonance, what is the value of the power factor in a series LCR circuit?
Answer:
At resonance, the Power factor is 1. Power factor = \(\frac{R}{Z}=\frac{R}{R}\) = 1 [∵ Z = R]

Question 6.
Give the expression for impedance of a series LCR circuit.
Answer:
Impedance Z = \(\sqrt{R^2+\left(X_L-X_C\right)^2}\)
Here R is resistance, X_L is inductive reactance and X_C is capacitive reactance.

Question 7.
Which device is used to reduce AC without much loss of energy?
Answer:
Step down transformer is the device used to reduce AC without much loss of energy.

Question 8.
Write the expression for power factor in an AC circuit.
Answer:
Power factor cosΦ = \(\frac{\mathrm{R}}{\mathrm{Z}}\) ; R= Resistance, Z= Impedance, Φ is phase angle.

Question 9.
Name the type of current in which electrons flow in one direction only.
Answer:
In direct current(DC), the electrons flow in one direction.

IV. Very Short Answer Questions

Question 1.
A transformer converts 200 V AC into 2000 V AC. Calculate the number of turns in the secondary if the primary has 10 turns.
Answer:
Given V_p = 200 V, V_s= 2000 V, N_p = 10, N_s = ?
Transformer ratio formula: \(\frac{N_s}{N_p}=\frac{V_s}{V_p}\) ⇒ N_s = \(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{~V}_{\mathrm{p}}}\)N_p = \(\frac{2000}{200}\) × 10=100 turns

Question 2.
A pure inductor of 21 mH is connected to a source of 220V, Find the inductive reactance if the frequency of the source is 50Hz.
Answer:
Given Inductance L = 21mH = 21 × 10^-3 H, Voltage V = 220V, Frequency f = 50Hz
∴ Inductive reactance X_L = 2πfL = 2 × 3.14 × 50 × 21 × 10^-3 = 6.597Ω = 6.Ω

Question 3.
What is transformer ratio?
Answer:
Transformer Ratio is the ratio of output emf to input emf of a transformer, (or)
It is the ratio between number of turns in the secondary coil and the number of turns in the primary coil of a transformer.
Formula: Transformer ratio = \(\frac{N_s}{N_p}=\frac{V_s}{V_p}\)

Question 4.
Write the expression for the reactance of (i) an inductor and (ii) a capacitor.
Answer:
(i) Reactance of an inductor is X_L = ωL
Here to is angular frequency and L= Inductance

(ii) Reactance of a capacitor is X_C = \(\frac{1}{\omega C}\)
Here to is angular frequency and C = Capacitance

Question 5.
What is the phase difference between A.C emf and current in the following: pure resistor, pure inductor and pure capacitor.
Answer:

1. In a pure resistor, the phase difference between AC voltage and AC current is zero.
2. In a pure inductor, the current lags behind the voltage by π/2.
3. In a pure capacitor, the current leads ahead the voltage by π/2.

Question 6.
Define power factor. On which factors does the power factor depend?
Answer:
Power Factor is the cosine of the phase angle (Φ) between the voltage and the current.
Formula 1: P = V I cos Φ. Here cos Φ is called power factor.
Formula 2: Power factor cos Φ = \(\frac{\mathrm{R} \text { (Resistance) }}{\mathrm{Z} \text { (Impedance) }}\)
Power factor depends on the phase difference between voltage and current or equivalently on the ratio R/Z.

Question 7.
What is meant by wattless component of current?
Answer:
Wattless current: In a purely inductive or capacitive circuit, the power factor cos<ft=0 . Then no power is dissipated even though current is flowing in the circuit.

In such cases, the current is said to be wattless current.

Question 8.
Why capacitor blocks Direct Current (D.C.)?
Answer:
D.C. has constant magnitude and direction. Once a capacitor is fully charged, it behaves like an open circuit for D.C. When current passes through the capacitor it blocks direct current D.C.

Question 9.
On what principle metal detector works? Explain it.
Answer:
Metal detector works on the principle of electromagnetic induction.
The primary search coil in metal detector generates an alternating magnetic field. When a metal object is nearby, due to the influence of the magnetic field, current will be induced in the metal. The induced current creates a secondary magnetic field which is detected by the metal detector’s receiver coil, and it generates a sound as an alarm.

Question 10.
How the phenomenon of resonance in the series LCR circuit be used in tuning mechanism of a radio and TV set?
Answer:
In the series LCR circuit, energy alternates between the inductor and capacitor, enabling the radio or TV to pick up a precise frequency. Resonance in the circuit enables radio or T.V. tuning by selecting a specific station’s frequency while rejecting others.

V. Short Answer Questions

Question 1.
Obtain an expression for the current through an inductor when an AC emf is applied.
Answer:
Current through an Inductor with AC emf :
Consider a pure inductor with self inductance L connected to an AC of voltage v = v_m sin ωt
Let I be the instantaneous current at any time t passing through the inductor.
Applying the Knchhoffs law on the circuit, we have v – L \(\frac{d I}{d t}\) = 0…………(1)

We know amplitude of current I_m = \(\frac{\mathrm{v}_{\mathrm{m}}}{\omega \mathrm{~L}}\)
∴ (2) ⇒ I = I_m sin (ωt – \(\frac{\pi}{2}\))
Current in an AC inductor circuit is I = I_m sin (ωt – \(\frac{\pi}{2}\))
Comparing this with v = _m sin ωt, we claim that the current lags the voltage by π/2.

Question 2.
Obtain an expression for the current in a capacitor when an AC emf is applied.
Answer:
Current through an Capacitor with AC :
Consider a pure capacitor C connected to an AC voltage v = v_m sin ωt ………..(1)

Let I be the instantaneous current at any time t passing through the capacitor.
The potential difference across the capacitor of capacitance C is v = \(\frac{\mathrm{q}}{\mathrm{C}}\) ……….. (2)
From (1) & (2), \(\frac{\mathrm{q}}{\mathrm{C}}\) = v_m sin ωt (∵ v = v_m sin ωt)
⇒ q = C v_m sin ωt
We know I = \(\frac{d q}{d t}\) ⇒ I = \(\frac{d}{d t}\) (C v_m sin ωt)
⇒ I = ω C v_m cos ωt (∵ \(\frac{d}{d x}\) sinkx = k cos kx)
I = ω C v_m sinωt + \(\frac{\pi}{2}\)) ……… (2) [∵ cos(ωt) = sin(ωt + \(\frac{\pi}{2}\))]
We know amplitude of current I_m = \(\frac{v_m}{X_C}=\frac{v_m}{1 / \omega C}\) = v_mωC
(2) ⇒ I = I_m sin (ωt – \(\frac{\pi}{2}\))
Current in an AC capacitor circuit is I = I_m sin (ωt – \(\frac{\pi}{2}\))
Comparing this with v = v_m sin ωt, we claim that the current leads the voltage by π/2.

Question 3.
Derive the expression for the impedance of a series LCR circuit by using phasor diagrams technique.
Answer:
The Phasor diagram for series LCR circuit is as shown in figure.
Here, voltage V_L is in upward direction and voltage V_C is in downward direction.
So, net voltage up to point A is V_L – V_C assuming V_L > V_C
Maximum voltage is denoted by V_m.

From the phasor diagram, OB = \(\sqrt{(\mathrm{OC})^2+(\mathrm{CB})^2}\)
∴ V_m = \(\sqrt{V_R^2+\left(V_L-V_C\right)^2}\) = \(\sqrt{\left(I_m R\right)^2+\left(I_m X_L-I_m X_C\right)^2}\)
⇒ v_m = I_m\(\sqrt{R^2+\left(X_L-X_C\right)^2}\)
\(\frac{V_m}{I_m}\) = Z = \(\sqrt{R^2+\left(X_L-X_C\right)^2}\) for X_L > X_C
This is the expression for impedance of series LCR circuit.

Question 4.
Define resonance in a series LCR circuit. Derive the condition for resonance and explain what happens to current and impedance at resonance.
Answer:
IN RESONANCE LCR CIRCUIT :
An LCR circuit that allows the maximum current at a specific frequency of AC is known as a series resonance (LCR) circuit. Consider an LCR circuit of given frequency n.
At resonance, capacitive reactance (X_C) is equal to the inductive reactance (X_L).
Thus at resonance X_L = X_C
Impedance Z = \(\sqrt{R^2+\left(X_L-X_C\right)^2}\) = \(\sqrt{R^2+0}\) = R ⇒ Z = R
Again, at resonance X_L = X_C ⇒ ωL = \(\frac{1}{\omega C},\) ⇒ ω² = \(\frac{1}{\mathrm{LC}}\) ⇒ ω = \(\frac{1}{\sqrt{\mathrm{LC}}}\)
We know ω = 2πn ⇒ 2πn = \(\frac{1}{\sqrt{\mathrm{LC}}}\) ⇒ n = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\) Here n is resonant frequency.
Current and Impedance at Resonance:
i) We know I₀ = \(\frac{E_0}{R}\)
∴ at resonance if R is minimum then current in the circuit becomes maximum.

ii) At resonance, X_L = X_C ⇒ Z = R.
Hence if the impedance of the circuit is minimum then resistance also becomes minimum.

Question 5.
What is power factor? Derive its expression in terms of resistance R and impedance Z. Explain its significance in AC circuits.
Answer:
Power factor of an AC circuit is defined as the cosine of the phase angle (Φ) between voltage and current.
Formula: cosΦ = \(\frac{\mathrm{R}}{\mathrm{Z}}\)
Expression: Consider impedance triangle for series LCR circuit as shown in the figure.
Here Z= hypotenuse, base = R, perpendicular = X_L – X_C and Φ is the phase angle.

From the right angled triangle .
Power factor, cos Φ = \(\frac{\mathrm{R}}{\mathrm{Z}}\)
Significance: If power factor cos Φ = 1 the maximum power is dissipated in the circuit.
If power factor cos Φ = 0 then no power is consumed. This leads to the concept of wattless current dissipated even though current is flowing in the circuit

Question 6.
State the principle on which a transformer works. Describe the working of a transformer with necessary theory.
Answer:
Transformer: Transformer is a device which transforms AC electric power from one circuit to another. A transformer consists of two coils called primary coil and secondary coil wound on a continuous iron core.
Principle: Transformer is based on the principle of mutual induction. When current in primary coil changes, an induced current is produced in the secondary coil.
Transformer ratio \(\frac{N_s}{N_p}=\frac{V_s}{V_p}\)
Here N_s = number of turns in secondary coil, N_p = number of turns in primary coil.
V_s = AC voltage in secondary, V_p = ACc voltage applied to primary,
Working: When an alternating voltage V_p is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf ε_s in it. Let Φ be the flux in each turn in the core at time t.
According to Faraday’s laws of induction, the induced emf in the secondary with N_s turns is ε_s = -N_s \(\frac{d \phi}{d t}\)
The alternating flux <|) also induces an emf in the primary, ε_p = – N_p \(\frac{d \phi}{d t}\)
Here, ε_s = V_s and ε_p = V_p
∴ V_s = – N_s \(\frac{d \phi}{d t}\) ……….. (1) and V_p = – N_p\(\frac{d \phi}{d t}\) ……….. (2)
Dividing eq (1) by eq (2), we get \(\frac{V_s}{V_p}=\frac{N_s}{N_p}\). This is the formula for transformer ratio.
It there is no power loss, then input power = output power
⇒ i_p V_p = i_s V_s ⇒ \(\frac{V_s}{V_p}=\frac{N_s}{N_p}\)
If \(\frac{N_s}{N_p}\) > 1, then the transformer is called step-up transformer
If \(\frac{N_s}{N_p}\) < 1, then the transformer is called step-down transformer

VI. Long Answer Questions

Question 1.
Derive an expression for impedance oI’series I.CR circuit by using the technique of phasors and deduce the expression for resonant frequency.
Solution:
The Phasor diagram for series LCR circuit is as shown in figure.
Here, voltage V_L is in upward direction and voltage V_C is in downward direction.
So, net voltage up to point A is V_L – V_C assuming V_L > V_C
Maximum voltage is denoted by V_m.

From the phasor diagram, OB = \(\sqrt{(\mathrm{OC})^2+(\mathrm{CB})^2}\)
∴ V_m = \(\sqrt{V_R^2+\left(V_L-V_C\right)^2}\) = \(\sqrt{\left(I_m R\right)^2+\left(I_m X_L-I_m X_C\right)^2}\)
⇒ v_m = I_m\(\sqrt{R^2+\left(X_L-X_C\right)^2}\)
\(\frac{V_m}{I_m}\) = Z = \(\sqrt{R^2+\left(X_L-X_C\right)^2}\) for X_L > X_C
This is the expression for impedance of series LCR circuit.

IN RESONANCE LCR CIRCUIT :
An LCR circuit that allows the maximum current at a specific frequency of AC is known as a series resonance (LCR) circuit. Consider an LCR circuit of given frequency n.
At resonance, capacitive reactance (X_C) is equal to the inductive reactance (X_L).
Thus at resonance X_L = X_C
Impedance Z = \(\sqrt{R^2+\left(X_L-X_C\right)^2}\) = \(\sqrt{R^2+0}\) = R ⇒ Z = R
Again, at resonance X_L = X_C ⇒ ωL = \(\frac{1}{\omega C},\) ⇒ ω² = \(\frac{1}{\mathrm{LC}}\) ⇒ ω = \(\frac{1}{\sqrt{\mathrm{LC}}}\)
We know ω = 2πn ⇒ 2πn = \(\frac{1}{\sqrt{\mathrm{LC}}}\) ⇒ n = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\) Here n is resonant frequency.
Current and Impedance at Resonance:
i) We know I₀ = \(\frac{E_0}{R}\)
∴ at resonance if R is minimum then current in the circuit becomes maximum.

ii) At resonance, X_L = X_C ⇒ Z = R.
Hence if the impedance of the circuit is minimum then resistance also becomes minimum.

Question 2.
Explain the phase relationship between voltage and eurrent in purely resistive, purely inductive, and purely capacitive circuits. Draw phasor diagrams for each case and discuss their significance in AC analysis.
Answer:
1) AC through pure resistor:
Let an alternating emf of V = V_m sin ωt …….. (1)
be applied to the pure resistor of resistance R.
Let I be the instantaneous current at any time t passing through the resistor.
Let V be the potential drop across resistance R, then V = IR

∴ V = IR = V_m sin ωt ⇒ I = \(\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{R}}\)sin ωt.
∴ Instantaneous current, I = I₀ sin ωt. ………….. (2) (∵ I₀ = \(\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{R}}\))
Comparing (1) and (2),
Phase difference between current and voltage is zero.
Phase diagram for purely resistive circuit is as shown figure.

Phase difference between current and voltage is zero.

2) AC through pure Inductor :
Consider a pure inductor with self inductance L connected to an AC of voltage v = v_m sin ωt …….. (1)
Let I be the instantaneous current at any time t passing through the inductor.

Applying the Krichhoffs law on the circuit, we have
v – L \(\frac{d I}{d t}\) = 0 ……….. (1)
⇒ L\(\frac{d I}{d t}\) = v ⇒ \(\frac{d I}{d t}\) = \(\frac{\mathrm{v}}{\mathrm{~L}}\) = \(\frac{v_m \sin \omega t}{L}\) [∵ v = v_m sin ωt]
Integrating it with respect to time, we get
\(\int \frac{d I}{d t}\) dt = \(\frac{\mathrm{v}_{\mathrm{m}}}{\mathrm{L}}\)∫sin (ωt)dt ⇒ I = \(\frac{v_m}{L}\left(\frac{-\cos (\omega t)}{\omega}\right)\) (∵ ∫sin kx = \(\frac{-\cos \mathrm{kx}}{\mathrm{k}}\))
⇒ I = \(\frac{\mathrm{v}_{\mathrm{m}}}{\omega \mathrm{~L}}\) sin(ωt – \(\frac{\pi}{2}\)) …………. (2) [∵ -cos(ωt) = sin(ωt – \(\frac{\pi}{2}\))]
We know amplitude of current I_m = \(\frac{\mathrm{v}_{\mathrm{m}}}{\omega \mathrm{~L}}\)
∴ (2) ⇒ I = I_m sin(ωt – \(\frac{\pi}{2}\))
Current in an AC inductor circuit is I = I_m sin(ωt – \(\frac{\pi}{2}\))
Comparing this with v = v_m sinωt, we claim that the current lags the voltage by π/2. The phase difference between the emf and current is π/2 radians and current lags behind the emf by π/2 radians

3) AC through pure Capacitor:
Consider a pure capacitor C connected to an AC voltage v = v_m sinωt ………. (1)
Let I be the instantaneous current at any time t passing through the capacitor.
The potential difference across the capacitor of capacitance C is v = \(\frac{\mathrm{q}}{\mathrm{C}}\) ………….. (2)
From (1) & (2), \(\frac{\mathrm{q}}{\mathrm{C}}\) = v_m sinωt (∵ v = v_m sinωt) ⇒ q = C v_m sinωt
We know I = \(\frac{d q}{d t}\) ⇒ I = \(\frac{d}{d t}\)(C v_m sinωt) ⇒ I = ω C v_m cosωt (∵ \(\frac{d}{d x}\) sin kx = k cos kx)
I = ω C v_m sin(ωt + \(\frac{\pi}{2}\)) ………… (2) [∵ cos(ωt) = sin(ωt + \(\frac{\pi}{2}\)) ]
We know amplitude of current I_m = \(\frac{\mathrm{v}_{\mathrm{m}}}{\omega \mathrm{~L}}=\frac{\mathrm{v}_{\mathrm{m}}}{1 / \omega \mathrm{C}}\) = v_mωC
Current in an AC capacitor circuit is I = I_m sin (ωt + \(\frac{\pi}{2}\))
Comparing this with v = v_m sin ωt, we claim that the current leads the voltage by π/2.
The phase difference between the current and emf is π/2 radians and current through capacitor is ahead of the applied emf by π/2 radians,

Significance: Phase relationships help to define reactance (XL and XC) and total impedance Z. If power factor cos Φ = 1 the maximum power is dissipated in the circuit.
If power factor cos Φ = 0 then no power is consumed. This leads to the concept of wattless current dissipated even though current is flowing in the circuit

Exercise Problems

Question 1.
The instantaneous emf of an A.C. source is given by ε = 300 sin 314t, what is the rms value of the emf.
Solution:
Given instantaneous emf ε = 300 sin 314t. Comparing this with ε = E₀ sinωt we get peak emf E₀ = 300V
RMS value of emf is E_rms = \(\frac{E_0}{\sqrt{2}}=\frac{300}{\sqrt{2}}=\frac{300}{1.414}\) = 212.13V

Question 2.
A Capacitor has a capacitance of \(\frac{1}{\pi}\)μF. Find its reactance for a frequency of 50 Hz.
Solution:
Given Frequency t = 50Hz , Capacitance C = 1 × 10^-6F
Reactance X_C = \(\frac{1}{\omega C}=\frac{1}{2 \pi f C}=\frac{1}{2 \pi \times 50 \times \frac{1}{\pi} \times 10^{-6}}=\frac{1}{100 \times 10^{-6}}=\frac{1}{10^{-4}}\) = 10 kΩ

Question 3.
An Inductor 200 mH, capacitor 500 μF. resistor 100 Ω arc connected in series with a 100 V, AC source. Calculate the current amplitude at resonance and resonance frequency.
Solution:
Given Inductance L = 200mH = 0.2H, Capacitance C = 500 μF= 500 × 10^-6 F
Resistance R = 100Ω, RMS voltage V_RMS = 100V
(i) Current amplitude I_m = \(\frac{V_m}{R}=\frac{\sqrt{2} V_m}{R}=\frac{1.414 \times 100}{100}\) = 1.414 A
(ii) Resonance frequency f = \(\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \times 3.14 \sqrt{0.2 \times 500 \times 10^{-6}}}=\frac{100}{6.28}\) = 15.915 Hz

Question 4.
Determine the impedance of a series LCR circuit, if the reactance of ‘C and ‘L’ arc 250 Ω and 220 Ω. respectively and R is 40 Ω.
Solution:
Given Resistance R = 40Ω, Inductive reactance X_L = 250Ω, Capacitive reactance X_C = 220 Ω
Impedance Z = \(\sqrt{R^2+\left(X_L-X_C\right)^2}\) = \(\sqrt{40^2+(250-220)^2}\)
= \(\sqrt{1600+900}\) = \(\sqrt{2500}\) = 50Ω

Question 5.
An LCR series circuit with L = 100 mH, C = 100 μF, R = 120 Ω is connected to an A.C, source of emf ε = 30 sin (100 t) volt. Find the impedance. Peak current.
Solution:
Given Inductance L = 100 mH = 100 × 10^-3 H, Capacitance = 100 μF = 100 × 10^-6 F
Resistance R = 120Ω, emf E = 30 sin 100t ⇒ E₀ = 30V, ω = 100 rad/s
Inductance reactance X_L = ω_L = 100 × 100 × 10^-3 = 10⁴ × 10^-3 = 10
Capacitive reactance X_C = \(\frac{1}{\omega C}=\frac{1}{100 \times 100 \times 10^{-6}}=\frac{1}{10^{-2}}\) = 100Ω
∴ Impedance Z = \(\sqrt{R^2+\left(X_L-X_C\right)^2}=\sqrt{120^2+(10-100)^2}=\sqrt{120^2+90^2}\) = 150Ω
∴ Peak value of current I₀ = \(\frac{E_0}{Z}=\frac{30}{150}\) = 0.2A

Question 6.
A Sinusoidal Voltage of Peak Value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 μF.
Find (a) the impedance of the circuit (b) the phase difference between the voltage across the source and the current (C) The power dissipated in the circuit.
Solution:
Given f = 50 Hz
∴ Angular frequency ω = 2πf = 2π × 50= 314.16 rad/sec
Inductive reactance X_L = ωL = 314.16 × 25.48 × 10^-3 = 8Ω [∵ L = 25.48mH]
Capacitive reactance C_C = \(\frac{1}{\omega C}=\frac{1}{314.16\left[396 \times 10^{-6}\right]}\) = 4Ω [∵ C = 796 μF]
(a) Impedance Z = \(\sqrt{R^2+\left(X_L-X_C\right)^2}=\sqrt{3^2+[8-4]^2}=\sqrt{9+16}\) = 5Ω
(b) tan Φ = \(\frac{X_L-X_C}{R}=\frac{8-4}{3}=\frac{4}{3}\) ⇒ tan Φ = \(\frac{4}{3}\) ⇒ Φ = tan^-1\(\left(\frac{4}{3}\right)\)
∴ The phase difference Φ = Tan^-1\(\left(\frac{4}{3}\right)\) = Tan^-1(1.33) = 53.1°
RMS voltage V_rms = \(\frac{\mathrm{V}_0}{\sqrt{2}}=\frac{283}{1.414}\) = 200V [∵ V₀ = 283V]
RMS Current I_rms = \(\)\(\) = 40A [∵ Z = 5 from (1)]
∴ Power dissipated P = I_rms² R = (40)² × 3 = 1600 × 3 = 4800ω

Question 7.
A coil of inductive reactance 31 Ω has a resistance of 8 Ω. It is placed in series with a condenser of capacitive reactance 25 Ω. The combination is connected to an AC source 110 V. Calculate the power factor of the circuit [Given Cos (36.86) = 0.80, Tan(36.86) = 0.75]
Solution:
Given Inductive reactance X_L = 31Ω, Resistance R = 8Ω, Capacitive reactance X_C = 25Ω, Voltage V = 110V,
Impedance Z = \(\sqrt{R^2+\left(X_L-X_C\right)^2}=\sqrt{8^2+(31-25)^2}=\sqrt{8^2+6^2}\) = 10Ω
Current I = \(\frac{V}{Z}=\frac{110}{10}\) = 11A
∴ Power factor \(\frac{R}{Z}=\frac{8}{10}\) = 0.8

Question 8.
A step-up transformer operates on a 230 V line and a load current of 2 A. The ratio of the primary and secondary windings is I : 25. What is the current in the primary?
Solution:
Given Primary voltage V_P = 230V, Secondary current I_S = 2A, Turns ratio N_P : N_S = 1:25
Transformer ratio \(\frac{N_P}{N_S}=\frac{I_S}{I_P}\) ⇒ \(\frac{1}{25}=\frac{2}{I_P}\)
∴ Primary current I_P = 50A

Question 9.
How much current is drawn by the primary of a transformer which steps down 220 V to 22 V to operate a device with an impedance of 220 Ω.
Solution:
Given Primary voltage V_P = 220V, Secondary voltage V_S = 22V,
Secondary impedance Z_S = 220Ω
Secondary current I_s = \(\frac{V_S}{Z_S}=\frac{22}{220}\) = 0.1A
∴ Primary current I_P = I_S × \(\frac{V_S}{V_P}\) = 0.11\(\left(\frac{22}{220}\right)=\) = 0.01A

Question 10.
In an AC circuit, a condenser, a resistor and a pure inductor are connected in series across an alternator (AC generator). If the voltages across them are 20 V, 35 V, 20 V respectively, find the voltage supplied by the alternator.
Solution:
Given V_C = 20 V, V_R = 35 V, V_L = 20 V, V = ?
Formula: V² = V_R² + (V_L – V_C)² = 35² + (20 – 20)² = 35 V

Question 11.
The primary of a transformer with primary to secondary turns ratio of 1: 2, is connected to an alternator of voltage 200 V. A current of 4A is flowing through the primary coil. Assuming the transformer has no loss. Find the secondary voltage and current are respectively.
Solution:
Given N_P : N_S = 1 : 2 ⇒ \(\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{~N}_{\mathrm{p}}}\) = 2, V_P = 200 V, I_P = 4 A, V_S = ? I_S = ?
Formula for Transformer ratio: \(\)
Now \(\frac{V_s}{V_p}=\frac{N_s}{N_p}\) ⇒ \(\frac{V_s}{200}=\frac{2}{1}\) ⇒ V_S = 400 V
Also \(\frac{I_p}{I_s}=\frac{N_s}{N_p}\) ⇒ \(\frac{4}{\mathrm{I}_{\mathrm{s}}}=\frac{2}{1}\) ⇒ I_S = 2 A

Objective Questions

Question 1.
In an A.C. circuit, I_rms and I₀ are related as
1) I_rms = πI₀
2) I_rms = \(\sqrt{2}\)I₀
3) I_rms = I₀/π
4) I_rms = I₀/\(\sqrt{2}\)
Answer:
4) I_rms = I₀/\(\sqrt{2}\)

Question 2.
A 40 μF’ capacitor is connected to a 200 V, 50 Hz ac supply. The r.m.s value of the current in the circuit is, nearly
1)1.7 A
2) 2.05 A
3) 2.5 A
4) 25.1 A
Answer:
3) 2.5 A

Question 3.
In an ac circuit an alternating voltage 200\(\sqrt{2}\) sin 100t volts is connected to a capacitor of capacity 1 μF. The r.m.s. value of the current in the circuit is 1
1) 10mA
2) 100mA
3) 200 mA
4) 20 mA
Answer:
4) 20 mA

Question 4.
A capacitor of capacity C has reactance X. If capacitance and frequency become double then reactance will be
1) 4X
2) X/2
3) X/4
4) 2X
Answer:
3) X/4

Question 5.
An ac. voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are both equal to 3Ω, the phase difference between the applied voltage and the current in the circuit is
1) π/6
2) π/4
3) π/2
4) zero
Answer:
2) π/4

Question 6.
A coil has resistance 30 ohm and inductive reactance 20 ohm at 50 Hz frequency. If an ac source, of 200 volt, 100 Hz, is connected across the coil, the current in the coil will be
1) 2.0 A
2) 4.0 A
3) 8.0 A
4) \(\frac{20}{\sqrt{13}}\)A
Answer:
2) 4.0 A

Question 7.
What is the value of inductance L for which the current is maximum in a series LCR circuit with C = 10 μF and ω = 1000 s^-1?
1) 1 mH
2) cannot be calculated unless R is known
3) 10 mH
4) 100 mH
Answer:
4) 100 mH

Question 8.
In a circuit L, C and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by 45°. The value of C is
1) \(\frac{1}{\pi f(2 \pi f L-R)}\)
2) \(\frac{1}{2 \pi f(2 \pi f L-R)}\)
3) \(\frac{1}{\pi f(2 \pi f L+R)}\)
4) \(\frac{1}{2 \pi f(2 \pi f L+R)}\)
Answer:
4) \(\frac{1}{2 \pi f(2 \pi f L+R)}\)

Question 9.
The value of quality factor is
1) \(\frac{\omega L}{R}\)
2) \(\frac{1}{\omega R C}\)
3) \(\sqrt{\mathrm{LC}}\)
4) L/R
Answer:
1) \(\frac{\omega L}{R}\)

Question 10.
An inductor 20 mH, a capacitor 100 pF and a resistor 50 Ω are connected in series across a source of emf, V = 10 sin 314t. The power loss in the circuit is
1) 0.79 W
2) 0.43 W
3) 2.74 W
4) 1.13 W
Answer:
1) 0.79 W

Question 11.
An inductor 20 mH, a capacitor 50 μF and a resistor 40 Ω are connected in series across a source of emf V = 10 sin 340t. The power loss in A.C. circuit is
1) 0.76 W
2) 0.89 W
3) 0.51 W
4) 0.67 W
Answer:
3) 0.51 W

Question 12.
The instantaneous values of alternating current and voltages in a circuit are given as i = \(\frac{1}{\sqrt{2}}\) sin(100πt)ampere e = \(\frac{1}{\sqrt{2}}\) sin(100 πt + \(\frac{\pi}{3}\)) volt
The average power in watts consumed in the circuit is
1) 1/4
2) \(\sqrt{3}\)/4
3) 1/2
4) 1/8
Answer:
4) 1/8

Question 13.
In an a.c. circuit the e.m.f. (e) and the current (i) at any instant are given respectively by
e = E₀ sin ωt, i = I₀ sin (ωt – Φ)
The average power in the circuit over one cycle of a.c. is
1) \(\frac{E_0 I_0}{2}\)cosΦ
2) E₀I₀
3) \(\frac{E_0 I_0}{2}\)
4) \(\frac{E_0 I_0}{2}\)sin Φ
Answer:
1) \(\frac{E_0 I_0}{2}\)cosΦ

Question 14.
In an a.c. circuit with phase voltage V and current I, the power dissipated is
1) V.I
2) depends on phase angle between V and I
3) 1/2 × V.I
4) \(\frac{1}{\sqrt{2}}\) × V . I
Answer:
2) depends on phase angle between V and I

Question 15.
In an A.C. circuit, the current flowing is I = 5 sin (100t – π/2) ampere and the potential difference is V = 200 sin (100t) volts. The power consumption is equal to
1) 20 W
2) 0 W
3) 1000 W
4) 40 W
Answer:
2) 0 W

Question 16.
A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produces oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be
1) f/2
2) f/4
3) 8f
4) f/2\({\sqrt{2}}\)
Answer:
4) f/2\({\sqrt{2}}\)

Question 17.
A step down transformer connected to an ac mains supply of 220 V is made to operate at 11 V, 44 W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?
1)4 A
2) 0.2 A
3) 0.4 A
4) 2 A
Answer:
2) 0.2 A

Question 18.
A transformer is used to light a 100 W and 110 V lamp from a 220V mains. If the main current is 0.5 amp, the efficiency of the transformer is approximately
1) 50%
2) 90%
3) 10%
4) 30%
Answer:
2) 90%

Question 19.
The core of a transformer is laminated because
1) ratio of voltage in primary and secondary may be increased
2) energy losses due to eddy currents may be minimised
3) the weight of the transformer may be reduced
4) rusting of the core may be prevented.
Answer:
2) energy losses due to eddy currents may be minimised

Question 20.
A step-up transformer operates on a 230 V line and supplies a load of 2 ampere. The ratio of the primary and secondary windings is 1 : 25. The current in the primary is
1) 15 A
2) 50 A
3) 25 A
4) 12.5 A
Answer:
2) 50 A

Question 21.
The primary winding of a transformer has 500 turns whereas its secondary has 5000 turns. The primary is connected to an A.C. supply of 20 V, 50 II/. The secondary will have an output of
1) 2 V, 50 Hz
2) 2 V, 5 Hz
3) 200 V, 50 Hz
4) 200 V, 500 Hz.
Answer:
3) 200 V, 50 Hz

Question 22.
A coil of 40 henry inductance is connected in series with a resistance of 8 ohm and the combination is joined to the terminals of a 2 volt battery. The time constant of the circuit is
1) 5 seconds
2) 1/5 seconds
3) 40 seconds
4) 20 seconds
Answer:
1) 5 seconds

Question 23.
The time constant of C-R circuit is
1) 1/CR
2) CIR
3) CR
4) R/C
Answer:
3) CR