The 300 turn primary of a transformer has resistnace `0.82Omega` and the resistance of its secondary of 1200 turns is `6.2Omega`. Find the voltage across the primary if the power output from the secondary at 1600 V is 32 kW. Calculate the power losses in both coils when the transformer efficiency is `80%`.

Explanation

`N_(p)=300" "R_(p)=0.82Omega`
`N_(s)=1200" "R_(s)=6.2Omega`
`"Output power "=32kW, = 32,000W`
`V_(s)=1600V`
`V_(s)I_(s)=32,000`
`1600I_(s)=32,000`
`I_(s)=(32,000)/(1,600)=20A`
`I_(s)=20A, R_(s)=6.2Omega`
`therefore" Power loss in secondary coil"`
`P=I_(s)^(2)xxR_(s)" (1)"`
`P_(s)=(20)^(2)xx6.2`
`=400xx6.2`
`=2480W`
`=2.48kW`
`therefore " Power loss in secondary coil"`
`P_(s)=2.48kW`
`"Efficiency "eta=(V_(s)I_(s))/(V_(p)I_(p))`
`"But "eta=80" (given)"`
`therefore (80)/(100)=(V_(s)I_(s))/(V_(p)I_(p))`
`=(32,000)/(V_(p)I_(p))`
`therefore V_(p)I_(p)=(32,000xx100)/(80)`
`=400xx100=4xx10^(4)W`
`"For a transformer "(V_(p))/(V_(s))=(N_(p))/(N_(s))`
`therefore (V_(p))/(1600)=(300)/(1200)`
`V_(p)=(300xx1600)/(1200)`
`=400V`
We know `V_(p)I_(p)=4xx10^(4)`
We know `V_(p)I_(p)=4xx10^(4)=100A`
`therefore " Power loss in the primary coil"`
`=I_(p)^(2)R_(p)`
`=(100)^(2)xx0.82`
`8200W`
`P_(p)=8.2 kW`
`therefore" Power loss in the primary coil"`
`P_(p)=8.2 kW`