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Question
The power supply to the primary coil of a transformer is 200 W. Find
(i) Current in primary coil if the e.m.f. supply to it is equal to 220V.
(ii) The number of turns in the primary coil is equal to 80 and that in secondary is 800. What is the transformation ratio?
(iii) Name the type of transformer.
(iv) What will be the output voltage?
(v) What is the current in the secondary coil for an ideal transformer?
(vi) What is the output power?
(vii) Is output and input power equal?
(viii) Compare the current flowing in a secondary coil and in a primary coil.
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Solution
(i) P(Power) = 200W = VI = Ep × Ip
∴ P = 200 = 200 × Ip
∴ Current in the primary coil = `200/220`
Ip = 0.909 A
(ii)
Transformation ratio `"E"_"s"/"E"_"p" = "N"_"s"/"N"_"p" = 800/80 = 10`
(iii) It is a step up transformer.
(iv) Using `"E"_"s"/"E"_"p" = "N"_"s"/"N"_"p" = 10`
` therefore "E"_"s"/200 = 10`
∴ Es = 2200 V
(v) Using Ep × Ip = 2200 × Is
∴ 200 = 2200 × Is
∴ Is = `2/22 = 0.090` A
(vi) Outpput power = Es × Is = `(2200 xx 0.0909)/100 = 200`W
(vii) Yes, It is equal, because it is an ideal transformer, for an ideal transformer
Efficiency = η = `"Outout power"/"Input power" = 1 "or" 100%`
i.e., there is no energy loss due to any cause.
(viii) Is = 0.0909 A and Ip = 0.909 A
Less current flows through the secondary coil than in the primary coil because it is a step-up transformer.
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