Question

What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 515 turns, its secondary 8 turns, and the input voltage is 117 V

Answer 1

Answer:

7532V

Explanation:

For a given transformer, the ratio of the number of turns in its primary coil ($N_{p}$) to the number of turns in its secondary coil ($N_{s}$) is equal to the ratio of the input voltage ($V_{p}$) to the output voltage ($V_{s}$) of the transformer. i.e

$\frac{N_p}{N_s}$ = $\frac{V_p}{V_s}$            ----------------(i)

From the question;

$N_{p}$ = number of turns in the primary coil = 8 turns

$N_{s}$ = number of turns in the secondary coil = 515 turns

$V_{p}$ = input voltage = 117V

Substitute these values into equation (i) as follows;

$\frac{8}{515}$ = $\frac{117}{V_s}$

Solve for $V_{s}$;

$V_{s}$ = 117 x 515 / 8

$V_{s}$ = 7532V

Therefore, the output voltage (in V) of the transformer is 7532