Question

The "power cube" transformer for a portable CD player has an output of 6.9 V and 600 mA (both rms) when plugged into a 120 V outlet.a. If the primary coil has 400 turns of wire, how many turns are on the secondary coil?b. What is the peak current in the primary coil?

Answer 1

Answer:

23

0.04879 A

Explanation:

T denotes turns

V denotes voltage

p denotes primary

s denotes secondary

For turns

$\frac{T_s}{T_p}=\frac{V_s}{V_p}\\\Rightarrow T_s=\frac{V_s}{V_p}\times T_p\\\Rightarrow T_s=\frac{6.9}{120}\times 400\\\Rightarrow T_s=23$

The number of turns in the secondary coil is 23

For current

$\frac{T_s}{T_p}=\frac{I_p}{I_s}\\\Rightarrow \frac{23}{400}=\frac{I_p}{0.6}\\\Rightarrow I_p=\frac{23}{400}\times 0.6\\\Rightarrow I_p=0.0345\ A$

The peak current in the primary coil would be

$\sqrt2\times 0.0345=0.04879\ A$