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Calculate current drawn by primary coil of a transformer, Which steps down 200 V to 20 V to operate a device of 20 ohm resistance. Assume efficiency of transformer 80 %.

A

`0.125 A`

B

`0.225 A`

C

`0.325 A`

D

`0.425 A`

Text Solution

Verified by Experts

The correct Answer is:
A
Given : `V_(p) = 200 V, R= 20 Omega, V_(s) = 20 V, eta = 80%`
Current through the secondary coil is
`I_(s)=(V_(s))/(R )=(20V)/(20 Omega)=1 A`
Efficiency of transformer `eta = ("Output power")/("Input power")=(V_(s)I_(s))/(V_(p)I_(p))`
or `I_(p)=(V_(s)I_(s))/(V_(p)eta)=(20xx1xx100)/(200xx80)=0.125 A`
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