Question

A multipurpose transformer has a secondary coil with several points at which a voltage can be extracted, giving outputs of 5.60, 12.0, and 480 V. (a) The input voltage is 220 V to a primary coil of 230 turns. What are the numbers of turns in the parts of the secondary used to produce the output voltages

Answer 1

Answer:

Explanation:

A multipurpose transformer can act as step up as well as step down transformer according to the desired setting by a user.

When the voltage at the output is greater than the voltage at the input of the transformer then it acts as step-up transformer and vice-versa acting is a step down transformer.

Given that:

input (primary) voltage of the transformer, $V_i=220~V$

no. of turns in the primary coil, $N_i=230$

• When the output voltage is 5.60 V:

$V_o=5.60~V$

$\frac{N_i}{N_o} =\frac{V_i}{V_o}$

$\frac{N_o}{230}=\frac{5.60}{220}$

$N_o=5.85\approx 6$ turns compensating the losses

• When the output voltage is 12.0 V:

$V_o=12.0~V$

$\frac{N_i}{N_o} =\frac{V_i}{V_o}$

$\frac{N_o}{230}=\frac{12.0}{220}$

$N_o=12.45\approx 13$ turns compensating the losses

• When the output voltage is 480 V:

$V_o=480~V$

$\frac{N_i}{N_o} =\frac{V_i}{V_o}$

$\frac{N_o}{230}=\frac{480}{220}$

$N_o=501.8\approx 502$ turns compensating the losses