Introduction to Alternating Current

1 Mark Questions

1. Why is the use of AC voltage preferred over DC voltage? Give two reasons. [All India 2014]

Ans. The use of AC voltage is preferred over DC voltage because of

1. the loss of energy in transmitting the AC voltage over long distances with the help of step- up transformers is negligible as compared to DC voltage.
2. AC voltage can be stepped up and stepped down as per the requirement by using a transformer.
3. The current flowing through a pure inductance 2mH is, I = (15 cos 300t) A. What is the (i) rms and average value of current for a complete cycle? [Foreign 2011]

<math>\mathbf{Q}</math> To calculate different values for AC with the help of given equation, compare the given equation with standard equation of AC.

Current flowing through the inductor,

<math>I = 15 \cos 300t</math>

Comparing with <math>I = I_0 \sin \omega t</math>

Here, peak value of current,

<math>I_0 = 15 \text{ A}</math>

1. For complete cycle, rms value of current

<math>I_{\rm rms} = \frac{I_0}{\sqrt{2}} = \frac{15}{\sqrt{2}} \, A</math>

1. For complete cycle, average value of current is zero

i.e.

<math>I_{av} = 0</math> <math display="block">(1/2\times 2=1)</math>

1. Define the term wattless current. [Delhi 2011]

Wattless Current The current in an AC circuit when average power consumption in AC circuit is zero, is referred as wattless current.

If <math>\phi</math> is the phase difference between voltage and current then power associated with I sin φ component of current is termed as wattless current. (1)

1. Define the term rms value of the current. How is it related to the peak value? [All India 2010c]

Ans. It is defined as the value of Alternating Current (AC) over a complete cycle which would generate same amount of heat in a given resistor that is generated by steady current in the same resistor and in the same time during a complete cycle .lt is also called virtual value or effective value of AC.

Let the peak value of the current be <math>l_0</math> ∴. <math display="block">I_{\rm rms} = \frac{I_0}{\sqrt{2}}</math>

<math>I_{\rm rms} = \frac{I_0}{\sqrt{2}}</math>

where, <math>I_0</math> = peak value of AC.

1. How much average power over a complete cycle, does an AC source supply to a capacitor? [Delhi 2009c]