Most of you might have noted that electric power stations generate *alternating* current rather than direct current even though the majority of electrical and electronic appliances require direct current. At the power generation stage, alternating current is preferred since the current can be controlled, without power loss, using reactive elements (inductors and capacitors). Transmission of electric power over long distances without appreciable losses is possible by using transformers. The idea, as most of you should know, is to transmit electrical energy at *low current* and high voltage so that the Joule heating (which is proportional to the square of the current) in the transmission lines is minimized.

You will find some earlier posts on alternating currents on this site. You can access all the posts by trying a search for ‘alternating current’ using the search box provided on this page. Or, you may click on the label ‘alternating current’ below this post.

Let us discuss a few more questions on alternating currents:

**(1) A short circuited coil is placed in a time varying magnetic field. Electric power is dissipated due to the current induced in the coil. If the number of turns were quadrupled and the wire radius halved, the electric power dissipated would be**

**(a) halved **

**(b) same **

**(c) doubled **

**(d) quadrupled**

The above question appeared in IIT 2002 screening test paper.

Electric power dissipation is given by *P* = *V*^{2}/*R*. where *V* is the voltage induced in the coil and R is the resistance of the coil. When the number of turns is quadrupled, the induced voltage *V* is quadrupled so that *V*^{2} becomes 16 times. But the resistance of the coil also becomes 16 times since resistance *R *= *ρL /A* where* ρ* is the resistivity (specific resistance),* L* is the length and *A* is the area of cross section of the wire. The length becomes 4 times when the number of turns is quadrupled and the area of cross section becomes one-fourth when the radius is halved. The resistance therefore becomes 16 times.

The power dissipated is therefore unchanged [Option (b)].

**(2****) An electric heater consumes 1000 watts power when connected across a 100 volt D.C. supply. If this heater is to be used with 200V, 50 Hz A.C.supply, the value of the inductance to be connected in series with it is**

**(a) 5.5 H **

**(b) 0.55 H **

**(c) 0.055 H **

**(d)1.1 H **

**(e)11 H **

The current drawn by the heater is 1000 W/100V = 10 A . When the heater is used with A.C. supply, it will draw 10 A itself. (Note that the current and the voltage values are R.M.S. values when you deal with electric power). If ‘*L*’ is the inductance required, the expression for the current *I* is

** I** =

*V*/√(

*R*

^{2}+

*L*

^{2}

*ω*

^{2})] where

*V*,

*R*and

*ω*are respectively the alternating voltage, the resistance of the heater and the angular frequency of the A.C.

Substituting, 10 = 200/√[10^{2} + *L*^{2} (100π)^{2}], since ω = 2π*f* where* f*’ is the frequency of the A.C.

Squaring and rearranging, *L*^{2} = 300/100π)^{2} from which *L *= √3/10π = 0.055 H.

(3) An alternating **emf V = 6 cos1000t is applied across a series LR circuit of 3 mH inductance and 4 Ω resistance. The amplitude of the current is**

**(a) 0.6 A **

**(b) 1.2 A **

**(c) 1.4 A **

**(d) 1.8 A**

Amplitude (maximum value or peak value) of current (*I*_{max}) is given by

*I*_{max} = *V*_{max}/√(*R*^{2}+*L*^{2} *ω*^{2}) = 6/√[4^{2}+(3×10^{-3})^{2}×1000^{2}] = 1.2 A

[Note that the values of *V*_{max} and *ω* are obtained from the expression for the emf *V* which is in the form, *V* = *V*_{max} cos *ωt*].

Here is a very simple question which you should answer carefully:

**(4) The voltage V applied across an A.C. circuit and the current I flowing in it are given by**

** V = 12 cos ωt volt and I = 20 sin ωt milliampere respectively. **

**The power dissipated in the circuit is**

**(a) 120 watt **

**(b) 120 milliwatt **

**(c) 240 watt **

**(d) 249 milliwatt **

**(e) zero**

In alternating current circuits the power is given by *P* = *V _{r}*

_{ms}

*I*

_{rms}cos

*Φ*where

*Φ*is the phase difference between the applied voltage and the resulting current. Since the voltage is a cosine function and the current is a sine function, the phase difference

*Φ*is π/2. [Note that the voltage can be written as

*V*= 12 sin(ωt + π/2) volt]. Therefore cos

*Φ*(which is called the power factor) is zero. The correct option is (e).

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