Question

A 60\ W60 W light bulb (normally ran at 115\ V115 V) is attached to a transformer. The voltage source in the transformer is 35\ V35 V and N_p = 25N ​p ​​ =25. How many turns does the secondary coil need to be to run this lightbulb at 100\ W100 W?

Answer 1

137 turns

Step-by-step explanation:

The light is rated 60 W and normally operated at 115V,

using electrical power formula;

P (power) = I (currents in Ampere) × V ( potential difference in Volts)

substitute the power and V mentioned above into the formula

60 = I × 115

I = 60 / 115 = 0.522 A

with the Current above we can calculate the voltage expected for 100W in the secondary coil

P = I × V

100 = 0.522 × V

V = 100 ÷ 0.522 = 191.6 Volts is the voltage expected by the secondary coil of the transformer to deliver

using transformer formula with Vs being Voltage in the secondary coil and Vp voltage in primary coil, Ns being number of turns in the secondary coil and Np number of turns in the primary coil.

Vs = 191.6 volts, Vp = 35 volts Np = 25 and Np= ?

Vp/ Vs = Np / Ns

make Ns subject of the formula

Ns = Vs × Np / Vp

Ns = 25 × 191.6 / 35 = 136.9 turns approx 137 turns