A transformer having efficiency of `90%` is working on `200 V` and `3 kW` power supply. If the current in the secondary coil is `6A`, the voltage across the secondary coil and current in the primary coil respectively are

Options

A. 300 V, 15 A
B. 450 V, 15 A
C. 450 V, 13.5 A
D. 600 V, 15 A

Answer

B

Explanation

Here, Efficiency of the transformer, `eta`=90%
Input power, `P_("in") = 3 kW = 3 xx 10^3 W=3000 W `
Voltage across the primary coil, `V_P` = 200 V
Current in the secondary coil, `I_S = 6A`
As `P_("in") = I_PV_P`
`therefore ` Current in the primary coil ,
`I_P = (P_("in"))/(V_P) = (3000V)/(200V) = 15A`
Efficiency of the transformer,
`eta = (V_SI_S)/(V_P I_P)`
` therefore 90/100 = (6V_s)/(3000) " or " V_S = (90 xx 3000)/(100 xx 6) = 450V`