Alternating Current Important Questions

Very Short Answer Questions (1 Mark Questions)

Ques. Two identical loops, one that is made up of copper and another made up of aluminium are rotated at the same speed in the same magnetic field, the induced emf will be? The current produced will be more in copper, why?

Ans. The induced EMF will be equal in both the loops. The induced current will be more in the copper wire due to its low resistance.

Ques. Define wattless current.

Ans. When the average power consumed in a circuit is zero it is known as wattless current or Idle current.

Ques. The instantaneous current flowing from an a.c source is l = 6 sin 314 t. What is the rms value of current?

Ans. By comparing the equation that has been given with the standard form

I = I₀ sin ωt

\(I_{0}=6 \mathrm{~A}, I_{r m s}=\frac{I_{0}}{\sqrt{2}}=\frac{6}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{6}{2} \times 1.414=4.242 \mathrm{~A}\)

Ques. A heating element is marked 210 V, 630 W. What will be the value of current drawn by the element when it is connected to a 210 V DC source?

Ans. I = P/V = 630/210 = 3A

Ques. Why is the core of a transformer laminated? (Comptt. Delhi 2013)

Ans. The core of a transformer is laminated to minimize eddy currents in the iron core to reduce energy loss in the form of heat.

Ques. Why do we prefer AC voltage over DC voltage?

Ans. The reasons due to which we prefer AC voltage over DC voltage are as follows: 

• It can be stepped up and stepped down using a transformer
• The carrying loss is very much less in an AC circuit

Ques. Plot a graph showing the variation of capacitive reactance with the change in the frequency of the AC source. (Comptt. All India 2015)

Ans. The graph would be as follows:

In the graph, X_c represents the capacitive-reactance and v is the frequency in the ac circuit.

Ques. Why is there no power consumption in an ideal inductor connected to an ac source?

Ans. There is no power consumption in an ideal inductor because current and voltage beyond an ideal inductor are out of phase by 90°.

Hence P = VRMS IRMS cos 90° = 0

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Short Answer Questions (2 Marks Questions)

Ques. An electric lamp is having a coil of negligible inductance and is connected in series with a capacitor, and an a.c source is glowing with a certain brightness. How will the brightness of a lamp change on reducing the
(i) capacitance, and
(ii) the frequency? Justify your Answer. (Delhi 2009)

Ans. The brightness of lamp ∝ I₀,
Assuming the lamp has zero resistance and zero inductance.

\(\mathrm{I}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{X}_{\mathrm{C}}}=\frac{\mathrm{V}_{0}}{1 / \omega \mathrm{C}}=\mathrm{E}_{0}(2 \pi v) \mathrm{C}\)

Upon reducing C or v, it would decrease
∴ The brightness of the lamp will decrease.

Ques. State the principle of working of a transformer. Can the transformer used to step up or step down a d.c. voltage? Justify your answer. (All India 2009)

Ans. Transformers operate on the principle of mutual induction. Mutual induction Is the process of inducing an EMF in a circuit by the changing current in another circuit.

Transformers cannot be used to step up or down dc voltages as it requires a changing flux to do so. DC voltage cannot produce a change in the magnetic flux.

Ques. An ideal inductor is in turn put across 220 V, 50 Hz, and 220 V, 100 Hz supplies. Tell whether the current flowing through it in the two cases be the same or different.

Ans. The current through the inductor is given by I = V/X_L = V/ 2πfL current is inversely proportional to the frequency of applied AC.

We can see that the frequency is different and therefore the current will also be different.

Ques. Give two advantages and disadvantages of AC over DC.

Ans. Here are the advantages and disadvantages of AC over DC:

Advantages of AC

• The generation and transmission of ac is more economical than dc.
• The alternating voltage may be easily stepped up or down as per requirements by making use of suitable transformers.

Disadvantages of AC

• It is more fatal than dc.
• It cannot be used for electrolysis.

AC vs DC

Ques. Derive an expression for the impedance of an a.c. circuit consisting of an inductor and a resistor. (Delhi 2008)

Ans. As per the phasor diagram, R is obtained

Clearly is the effective resistance of the series LR circuit which opposes or impedes the flow of a.c through it. It is called its impedance and is denoted by Z.

\(Z=\sqrt{R^{2}+X_{L}^{2}}=\sqrt{R^{2}+\omega^{2} L^{2}} \quad\left[\because X_{L}=\omega L\right]\)

Therefore the average power dissipated each cycle in a capacitor is zero

Ques. The circuit arrangement as shown in the diagram shows that when an a.c. passes through coil A, and the current starts flowing in coil B.

(i) State the underlying principle involved.
(ii) Mention the factors on which the current produced in coil B depends. (All India 2008)

Ans. (i) It is based upon the principle of “mutual induction”.

(ii)The factors are as follows:

• The distance between the coils.
• Orientation of the coils.
• Number of turns in the coil

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Long Answer Questions (3 Marks Questions)

Ques. Prove and show that an ideal inductor does not dissipate power in an a.c. circuit. (Delhi 2016)
Ans. When a.c. current is applied to an ideal inductor, and the current lags behind the voltage in phase by π/2 radian. we can now write the instantaneous values of voltage and current as follows :

V = V₀ sin ωt and I = I₀ sin (ωt – π/2)

I = I₀ sin -(π/2 – ωt) or I = -I₀ cos ωt

Work done in small time dt is

dW = Pdt = – V₀I₀ sin ωt cos ωt dt      [∵ 2 sin ωt cos ωt = sin 2 ωt]

= \(- \frac{V_0 I_0}{2}\) sin 2 ωt dt

The average power dissipated each cycle in the inductor is

\(\begin{aligned} P_{a v} &=\frac{W}{T}=\frac{1}{T} \int_{0}^{\mathrm{T}} d W=\frac{-V_{0} I_{0}}{2 T} \int_{0}^{T} \sin 2 \omega t d t \\ &=\frac{V_{0} \mathrm{I}_{0}}{2 \mathrm{~T}}\left[\frac{\cos 2 \omega t}{2 \omega}\right]_{0}^{\mathrm{T}}=\frac{\mathrm{V}_{0} \mathrm{I}_{0}}{4 \mathrm{~T} \omega}\left[\cos \frac{4 \pi \mathrm{T}}{\mathrm{T}}-1\right] \\ &=\frac{\mathrm{V}_{0} \mathrm{I}_{0}}{4 \mathrm{~T} \omega}[1-1]=0 \end{aligned}\)

Thus, the average power dissipated during each cycle in an inductor is zero.

Ques. An alternating voltage given by V = 140 sin 314 t is connected across a pure resistor of 50 ohm. Find
(i) the frequency of the source.
(ii) the rms current through the resistor. (All India 2012)

Ans. (i) V₀ = 140 V, ω = 314

2πv = 314 Therefore, V = 314/2π = 50 Hz

(ii) I_rms = V_rms /R (where V_rms = V/√2 )

= (V₀/√2)/R

=V₀/2 R

= 140/√2 × 50

= 140/1.414 × 50

Ques. The figure shows a series LCR circuit which is connected to a variable frequency with a 200 V source with L = 50 mH, C = 80 µF and R = 40 Ω. Determine
(i) The frequency of the source which derives the circuit in resonance.
(ii) the quality factor (Q) of the circuit.

Ans. (i) ω₀ = 1/√LC

= 1/√(50 × 10^-3) (80 × 10^-6)

= 1/√4000 × 10^-9

= 1/√4 × 10^-6

= 10³/2= 1000/2 = 500 rad S^-1

f = ω₀/2r = 500/2r = 80 hertz

(ii) Q = ω₀L/R = (500 × (50 × 10^-3))/40 = 0.625

Ques. Two heating elements which are of resistances R₁ and R₂ when operated at a constant supply of voltage, V, they consume powers P₁ and P₂ respectively. Deduct the expressions for the power of their combination when they are, in turn, connected in
(i) series and
(ii) parallel across the same voltage supply. (All India 2008)

Ans. When two resistances, R₁ and R₂ are operated at a constant voltage supply V, their consumed power will be P₁ and P₂ respectively and is given by,

\(P_1 = \frac{V^2}{R_1}, P_2 = \frac{V^2}{R_2}\)

When they are connected in series, The power will be

\(\frac{1}{\mathrm{P}}=\frac{1}{\mathrm{P}_{1}}+\frac{1}{\mathrm{P}_{2}}=\frac{\mathrm{R}_{1}}{\mathrm{~V}^{2}}+\frac{\mathrm{R}_{2}}{\mathrm{~V}^{2}}=\frac{\mathrm{R}_{1}+\mathrm{R}_{2}}{\mathrm{~V}^{2}}=\frac{\mathrm{V}^{2}}{\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)}\)

And the power in the parallel will be

\(=P_{1}+P_{2}=\frac{V^{2}}{R_{1}}+\frac{V^{2}}{R_{2}}=\frac{\left(R_{1}+R_{2}\right) V^{2}}{R_{1} R_{2}}\)

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Very Long Answer Questions (5 Marks Questions)

Ques. State the main operating principle of a transformer.
How is large scale transmission of electric energy over long distances done with the use of transformers? (All India 2012)

Ans. A transformer is an electrical device used for converting an alternating current at a low voltage into a high voltage or vice versa. If the transformer increases the input voltage then it is called as step-up transformer.

Principle: it operates upon the principle of mutual induction i.e., “The EMF induction on one coil when changing current is applied on another coil.”

Working: As the ac flows through the primary, it produces an alternating magnetic flux in the core which also passes through the secondary. This changing flux will set up an induced emf in the secondary, and a self-induced emf in the primary. If there is no leakage of magnetic flux, then the flux linked with each and every turn of the primary will be equal to that linked with each turn of the secondary.

\(\mathrm{V}_{\mathrm{P}}=-\mathrm{N}_{\mathrm{P}} \frac{d \phi}{d t} \quad \text { and } \quad \mathrm{V}_{\mathrm{S}}=-\mathrm{N}_{\mathrm{S}} \frac{d \phi}{d t}\)

where [N_p and N_s are the number of turns in the primary and secondary respectively, and V_p and V_s are their respective voltages]

\(\therefore \quad \frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{N}_{\mathrm{S}}}{\mathrm{N}_{\mathrm{P}}} \ldots (i)\)

This ratio N_s:N_p is called the turns ratio.

Assuming the transformer as an ideal one, so that there are no energy losses, then

Input power = output power

V_pl_p = V_sI_s

where [I_P and I_S are the currents present in the primary and secondary respectively]

\(\therefore \quad \frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{I}_{\mathrm{S}}}{\mathrm{I}_{\mathrm{P}}} \ldots (ii)\)

From (i) and (ii), we get: \(\frac{\mathrm{I}_{\mathrm{P}}}{\mathrm{I}_{\mathrm{S}}}=\frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{N}_{\mathrm{S}}}{\mathrm{N}_{\mathrm{P}}}\)

In the case of a step-up transformer, N_s > N_p i.e., the turns ratio is greater than 1 and therefore V_s > V_p.
The voltage which we get as output is greater than that of the input voltage.

Main Assumptions:

• The primary resistance and the current are small.
• The same flux links both with the primary and secondary windings as the flux leakage from the due core is negligible.
• The terminals of the secondary are open or the current that has been taken from it is small.

For transmissions in long distances, the voltage output of the generator is stepped-up (but that current is reduced and consequently, IR loss is also reduced). It is transmitted over long distances and is stepped- down at distributing substations at the consumer's end.

Ques. Explain the different energy losses in a transformer. Does the Step-up transformer violate the law of conservation of energy, if not, give reason. (All India 2009)

Ans. Two sources for energy loss in a transformer are: 

(1) The primary and secondary are made up of copper wires and some of the energy will be lost due to the heating of these wires.

This power loss (= I²R) can be decreased by using thick copper wires of low resistance.

(2) Eddy current loss: Eddy currents are induced in the iron ore due to the alternating magnetic flux which leads to some energy loss in the form of heat. We can reduce these types of loss by laminating the iron core.

No, a step-up transformer does not violate the law of conservation of energy because whatever we gain through the voltage ratio is lost through the current ratio and vice-versa. This kind of transformer steps up the voltage while it steps down the current.

Ques. Draw a sketch of the basic elements of an a.c. generator. State its principle and briefly explain its working.

Ans. (a) Principle of A.C. generator: Generators work on the principle of electromagnetic induction. When a closed coil is rotated inside a uniform magnetic field with its axis perpendicular to the magnetic field, the magnetic flux which is linked with the coil changes and an emf is induced and hence a current is set up in it.

(b) Let N = number of turns present in the coil
A = Area of each turn’s face
B = magnitude of the magnetic field
θ = angle which is normal to the coil makes with field B at any instant
ω = the angular velocity of the coil rotation.

The magnetic flux which is linked with the coil at any instant f will be,
ϕ = NAB cos θ = NAB cos ωt

By Faraday’s flux rule, the induced emf is given by,

\( \begin{aligned} \mathrm{E} &=-\frac{d \phi}{d t}=\frac{-d}{d t} \mathrm{NAB}(\cos \omega t) \\ \mathrm{E} &=\mathrm{NAB}(\sin \omega t) . \omega \\ \Rightarrow \mathrm{E} &=\mathrm{E}_{0} \sin \omega t \quad \ldots \text { where } [\mathrm{E}_{0}=\mathrm{NAB} \omega] \end{aligned}\)

When a load of resistance R is connected across the terminals, a current I flows in the external circuit.

\(\begin{aligned} &I=\frac{E}{R}=\frac{E_{0} \sin \omega t}{R}=I_{0} \sin \omega t \\ &\ldots \text { where }\left[I_{0}=\frac{E_{0}}{R}\right] \end{aligned}\)

(c) v = 0.5 Hz; N = 100, A = 0.1 m², B = 0.01 T

\(\begin{aligned} &e_{\max }=\mathrm{NAB}(2 \pi v) \\ &e_{\max }=100 \times 0.01 \times 0.1 \times(2 \pi \times 0.5) \\ &e_{\max }=0.314 \text { volt } \end{aligned}\)

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