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

Answer:

D) 9.4 V, 0.38 A

1) Voltage in the secondary coil: 9.4 V

The transformer equation states that:

$\frac{V_p}{N_p}=\frac{V_s}{N_s}$

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

$V_s = N_s \frac{V_p}{N_p}=(110)\frac{120 V}{1400}=9.4 V$

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 $I_s$,

$I_s = \frac{V_p I_p}{V_s}=\frac{(120 V)(3.0\cdot 10^{-2}A)}{9.4 V}=0.38 A$