Question

What is the voltage output of a transformer used for
rechargeable flashlight batteries if its primary has 525 turns, its
secondary has8 turns, and the input voltage is 112 V?

V

(b) What input current is required to produce a 4.00 A
output?

mA

(c) What is the power input?

W

Answer #1

Given:

Input voltage (V_{p}) = 112 V

Primary turns (n_{p})= 525

Secondary turns (n_{S}) = 8

Input current (Ip) = 4.00 A

Here, A transformer consists of two coils - the primary coil and the secondary coil both wound on a soft iron core.

Using the formula,

Secondary voltage
(V_{s})/ Primary
voItage (V_{P}) =
Secondary turns (n_{S})/ Primary turns
(n_{P})

Secondary voltage (V_{s}) = (8 * 112 V) / 525

= 1.706 V

Similarly, Secondary Current (I_{s})/ Primary Current
(I_{P}) = Secondary turns (n_{S})/ Primary turns
(n_{P})

Primary Current (I_{P}) = (4 A * 525)/8

=262.5 A

c) Power input (Pp) = Ip * Vp = 262.5 * 112

= 29400 W

= 29.4 kW

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