Question

transformer has 26 turns on the primary coil and 720 turns on the secondary coil. If this transformer is to produce an output of 4100 V with a 16-mA current, what input current and voltage are needed

Answer 1

Corrected Question:

A step-up transformer has 26 turns on the primary coil and 720 turns on the secondary coil. If this transformer is to produce an output of 4100 V with a 16-mA current, what input current and voltage are needed?

Answer:

The input current and voltage needed are 443 mA and 148 V respectively.

Explanation:

In a step-up transformer, the relationship between the number of turns in its primary coil ($N_{p}$), the number of turns in its secondary coil ($N_{s}$), the input voltage ($V_{p}$), the output voltage ($V_{s}$), the input current ($I_{p}$), and the output current ($I_{s}$) is given by;

$\frac{N_{s} }{N_{p} }$ = $\frac{V_{s} }{V_{p} }$ = $\frac{I_{p} }{I_{s} }$

This implies that;

$\frac{N_{s} }{N_{p} }$ = $\frac{V_{s} }{V_{p} }$  ---------------------(i)

$\frac{N_{s} }{N_{p} }$ = $\frac{I_{p} }{I_{s} }$  ---------------------(ii)

From the question;

$N_{p}$ = 26 turns

$N_{s}$ = 720 turns

$V_{s}$ = 4100V

$I_{s}$ = 16mA = 16 x 10⁻³A

(a) Substitute these values into equation (ii) as follows;

$\frac{720}{26}$ = $\frac{I_{p} }{16*10^{-3}}$

Solve for $I_{p}$;

$I_{p}$ = 720 x 16 x 10⁻³ / 26

$I_{p}$ = 443 x 10⁻³

$I_{p}$ = 443 mA

Therefore the input current needed is 443mA

(b) Also, substitute those values into equation (i) as follows;

$\frac{720}{26}$ = $\frac{4100 }{V_{p} }$

Solve for $V_{p}$;

$V_{p}$ = 4100 x 26 / 720

$V_{p}$ = 148 V

Therefore, the input current needed is 148 V