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Alternating Current Circuits

Question
ICSEENIPH12029988
Wired Faculty App

An alternating current of frequency f is flowing through an ideal choke coil of self inductance L. If V° and I ° be the peak values of the voltage and current, find the average power consumed, if any, by the choke coil.

Solution

For a clock coil R > > WL,
Power factor, cos straight ϕ space equals space straight R over ωL space equals space 0
Average power, Veff . Ieff . cos straight ϕ
Average power = 0
Therefore, 
Pav = 0

Question
ICSEENIPH12030007
Wired Faculty App

When the primary of a transformer is connected to a 120 V AC mains, the current in the primary is 18.5 mA. Find the voltage across the secondary when it delivers 1.5 mA current through it, assuming the transformer to be an ideal one. State any one type of energy loss in a transformer. 

Solution

For an ideal transformer,

ip x Vs = is  x Vs

Here,

 ip = 18.5 mA

Vp =120 V

tS = 1.5 mA

and VS = ?

Using values in equation, 

18.5 x 120 = 1.5 x Vs

straight V subscript straight s space equals space fraction numerator 18.5 space straight x space 120 over denominator 1.5 end fraction space equals space 1480 space volts 
Hysteresis loss is an energy loss in a transformer.

Question
ICSEENIPH12030008
Wired Faculty App

A 30 mF capacitor, 0.2 H inductor and a 50 Ω resistor are connected in series to an A.C.


source whose emf is given by E = 310 Sin (314 t) where E is in volt and t is in second. Calculate:  

(i) the impedance of the circuit and

(ii) peak value of current in the circuit. 

Solution

It is given that

E = 310 Sin (314 t).

Comparing this equation with standard equation

E = E0 Sin wt

We have E0 = 310 V and w = 314

(i) Impedance of the circuit is, 
straight Z space equals space square root of left parenthesis straight X subscript straight L space minus space space straight X subscript straight C right parenthesis squared plus straight R squared end root space space equals space square root of open parentheses ωL space minus space 1 over ωL close parentheses squared plus space straight R squared end root space space space equals square root of open parentheses 314 space straight x space 0.2 space minus fraction numerator 1 over denominator 314 space straight x space 30 space straight x space 10 to the power of negative 6 end exponent end fraction close parentheses squared plus space left parenthesis 50 right parenthesis squared end root space space space equals square root of 4379.96 end root space space space space equals space 66.18 space straight capital omega ii right parenthesis space Peak space value space of space current space is comma space straight i subscript straight o space equals space straight E subscript straight o over straight Z space equals space fraction numerator 310 over denominator 66.18 end fraction space equals space 4.68 space straight A with straight o on top

Question
ICSEENIPH12030034
Wired Faculty App

In a graphical variation of emf induced with time for the output of an AC generator, mark the peak value of emf induced and the time period.

Solution

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