EXERCISE SHORT QUESTIONS
CHAPTER # 16: ALTERNATING CURRENT
Q # 1. A sinusoidal current has rms value of 10 A. What is the maximum or peak value?
Ans.
Given
The RMS value of current is 10 A.
\[Peak\ Value\ (maximum\ value) = I_{0} = \ ?\]
Using formula:
\[I_{rms} = \frac{I_{0}}{\sqrt{2}}\]
\[\Longrightarrow {I_{0} = \sqrt{2\ }\ \ I}_{rms}\]
\[\Longrightarrow I_{0} = \sqrt{2}(10\ A)\]
\[\Longrightarrow I_{0} = 14.14\ A\]
Thus, the maximum value of the current is \(14.14\ A\).
Q # 2. Name the devices that will (i) Permit flow of direct current but oppose the flow of alternating current, (ii) Permit flow of alternating current but not the direct current.
Ans.
An inductor (choke) is a device which permits flow of direct current but opposes the flow of alternating current.
A capacitor is a device which permits flow of alternating current but not the direct current.
Q # 3.How many times per second will an incandescent lamp reaches maximum brilliance when connected to a 50 Hz source?
Ans. The brilliance of the lamp will become maximum twice in one AC cycle because the current also becomes maximum two times in a cycle (i.e., for +ve half cycle and --ve half cycle).
As the frequency ``f'' of AC cycle is 50 Hz.
\[So\ maximum\ brilliance\ shown\ by\ lamp\ per\ second\ = \ Twice\ the\ frequency\ of\ AC\ source\ \]
\[So\ maximum\ brilliance\ shown\ by\ lamp\ per\ second\ = \ 2f = 2 \times 50 = 100\ times\]
Hence, the brilliance will be maximum 100 time in one second.
Q # 4. A circuit contains an iron-cored inductor, a switch and a DC sources arranged in series. The switch is closed and after an interval reopened. Explain why a spark jumps across the switch contacts?
Ans. When a switch of circuit containing iron cored inductor is closed, current increases from zero to maximum value. This changing current produce change of magnetic flux and hence emf is produced.
After an interval, when switch is reopened, the current changes from maximum to zero. Again emf is developed across the coil. This is back emf. This produces spark across the switch contacts.
Q # 5. How does doubling the frequency affect the reactance of (a) an inductor (b) capacitor?
Ans.
Formula for Reactance of Inductor: \(X_{L} = \omega L\)
Doubling frequency: \(X_{L}^{'} = 2\omega L = 2X_{L}\)
Result: Inductive Reactance will become double
Formula for Reactance of Inductor: \(X_{C} = \frac{1}{\omega C}\)
Doubling frequency: \(X_{C}^{'} = \frac{1}{2\omega C} = \frac{1}{2}X_{C}\)
Result: Capacitive Reactance will becomes half
Hence by doubling the frequency, the inductive reactance will become double, while capacitive reaction remains half.
Q # 6. In a RL circuit, will the current lag or lead the voltage? Illustrate your answer by a vector diagram.
Ans.
Figure shows an RL series circuit excited by an AC source. The potential difference across resistor `IR' would be in phase with current I.
Taking the current as the reference, the potential difference across the resistor is represented by the line along the current line because the potential difference is in phase with current.
The potential difference across the inductor IR. As the current lags the voltage by 90 degrees. so the line representing vector is drawn at right angle to the current line.
Figure shows that the current and the applied voltage are not in phase. The phase θ by which the current leads the voltage is given by the expression:
Q # 7. A choke coil placed in series with an electric lamp in an AC circuit causes the lamp to become dim. Why is it so? A variable capacitor added in series in this circuit may be adjusted until the lamp glows with normal brilliance. Explain, how this is possible?}
\[I = \frac{V}{R}\]
(a) When a choke coil
is connected in series with an electric lamp
If, now, a choke coil of inductive reactance \(X_{L}\) is placed in series with the electric lamp, the new impedance of the circuit will be:
\[Z_{1} = \sqrt{R^{2} + X_{L}^{2}}\]
Therefore, the current flowing through the circuit in this case will be:
\[{I_{1} = \frac{V}{Z}}_{1} = \frac{V}{\sqrt{R^{2} + X_{L}^{2}}}_{}\]
From the comparison of both currents, we see that \(I_{1}\)is smaller than \(I\) and that is why the electric lamp is dimmed on placing a choke coil in the circuit.
(b) A Variable
capacitor added in series with an electric lamp
When a variable capacitor also is in series with the circuit, its capacitive reactance \(X_{C}\) opposes \(X_{L}\) and thus the impedance of the circuit is
\[Z_{2} = \sqrt{R^{2} + \left( X_{L} - X_{C} \right)^{2}}\]
Therefore, the current flowing through the circuit in this case will be:
\[{I_{2} = \frac{V}{Z}}_{2} = \frac{V}{\sqrt{R^{2} + \left( X_{L} - X_{C} \right)^{2}}}_{}\]
If the \(X_{L} = X_{C}\), then
\({I_{2} = \frac{V}{Z}}_{2} = \frac{V}{\sqrt{R^{2} + (0)^{2}}}_{} = \frac{V}{R}_{} = I\)
Hence, the current \(I_{2}\) becomes equal to the current \(I\)for \(X_{L} = X_{C}\), as if there is no reactance in the circuit and hence the lamp glow with normal brilliance.
Q # 8. Explain the condition under which electromagnetic waves are produced from a source.
Ans. When alternating voltage is applied across the ends of a metallic antenna, and oscillating electric field comes into existence which accelerates the electrons again and again as the polarities of the antenna changes after half a cycle.
The accelerated electrons radiate energy carried by changing electric field. A changing electric field creates a magnetic field and a changing magnetic field creates electric field. Thus each field will generate the other and the whole package of electric and magnetic fields will move along propelling itself through space.
Q # 9. How the reception of a particular radio station is selected on your radio set?
Ans. A particular radio station can be selected on a radio set by tuning it. When the frequency of the LC-oscillator in the radio set is equal to the frequency of the radio wave from a particular radio station, a resonance is produced. The current of this signal becomes maximum and can detected and amplified.
Q # 10. What is meant by A.M and F.M?
Ans.
Amplitude Modulation: In this type of modulation, the amplitude of the carrier wave is increased or diminished as the amplitude of the superposing modulating signal increases or decreases.
Frequency Modulation: In this type of modulation, the frequency of the carrier wave is increased or diminished as the amplitude of the superposing modulating signal increases or decreases. But the carrier wave amplitude remains constant.
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