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The number of turns in the primary and the secondary coil of an ideal transformer are 100 and 5000 respectively. If 3.3 kW power is supplied to the transformer at 220 V, find (I) current in the - Physics

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Question

The number of turns in the primary and the secondary coil of an ideal transformer are 100 and 5000 respectively. If 3.3 kW power is supplied to the transformer at 220 V, find

  1. current in the primary coil, and
  2. output voltage.
Numerical
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Solution

Given: Np = 100,

Ns = 5000,

P = 3.3 kW = 3300 W,

Vp = 220 V

I. Current in the Primary Coil (Ip):

P = VpIp

⇒ Ip = `P/V_p`

= `3300/220`

= `330/22`

= 15 A

II. Output Voltage (Vs):

Using the transformer formula:

`V_s/V_p = N_s/N_p`

Vs = `V_p xx N_s/N_p`

= `220 xx 5000/100`

= 220 × 50

= 11000 V

= 11 kV

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