A phone charger requires 0.5 A at 5V. It is connected to a transformer with 100 % of efficiency whose primary contains 2200 turns and is connected to 220-V household outlet. (a) How many turns should there be in the secondary?
(b) What is the current in the primary?
(c) What would be the output current and output voltage values if number of secondary turns (N2) doubled of its initial value?

Explanation:

a ) for transformer which steps down voltage , if V₁ and V₂ be voltage of primary and secondary coil and n₁ and n₂ be the no of turns of wire in them

V₁ /V₂ = n₁ / n₂

Here V₁ = 220 V , V₂ = 5V , n₁ = 2200 n₂ = ?

220 /5 = 2200 / n₂

n₂ = 2200 x 5 / 220

= 50

b )

for 100 % efficiency

input power = output power

V₁ I₁ = V₂I₂

I₁ and I₂ are current in primary and secondary coil

220 x I₁ = 5 x .5

I₁ = .01136 A .

c )

If n₂ = 100

V₁ /V₂ = n₁ / n₂

220 / V₂ = 2200 / 100

V₂ = 10 V

V₁ I₁ = V₂I₂

220 x .01136 = 10 I₂

I₂ = .25 A.

energy can cross the boundaries of a closed system in the form of heat and mass

explanation: