Question

A transformer has 1400 turns on the primary and 110 on the secondary.?If the primary is connected to a 120V outlet and draws 3.0x10^-2 Amps, what are the voltage and current of the secondary? A) 1500 V, 2.4x10^-3 A B) 1500 V, 0.38 A C) 9.4 V, 2.4x10^-3 A D) 9.4 V, 0.38 A

Answer 1

D) 9.4 V, 0.38 A

1) Voltage in the secondary coil: 9.4 V

The transformer equation states that:

`(V_p)/(N_p)=(V_s)/(N_s)`

where

Vp = 120 V is the voltage in the primary coil

Np = 1400 is the number of turns in the primary coil

Vs = ? is the voltage in the secondary coil

Ns = 110 is the number of turns in the secondary coil

Solving the formula for Vs, we find

`V_s = N_s (V_p)/(N_p)=(110)(120 V)/(1400)=9.4 V`

2) Current in the secondary coil: 0.38 A

A transformer is considered to be 100% efficient: it means that there is no loss of power, so the power in input is equal to the power in output

`P_i = P_o\\V_p I_p = V_s I_s`

where

Vp = 120 V is the voltage in the primary coil

`I_p = 3.0\cdot 10^(-2) A` is the current in the primary coil

Vs = 9.4 V is the voltage in the secondary coil

`I_s` is the current in the secondary coil

Solving the equation for
`I_s`,

`I_s = (V_p I_p)/(V_s)=((120 V)(3.0\cdot 10^(-2)A))/(9.4 V)=0.38 A`