a current is used to heat the tip of a soldering iron. a transformer that plugs

Question

a current is used to heat the tip of a soldering iron. a transformer that plugs into the wall (Vrms=120V, 60hz ac) has a secondary coil (on the iron side) with 100 turns. the transformer supplies an rms voltage of 24V to the iron delivering 50W of power.

a) What is the resistance in the wire that acts at the heating element for the iron?

b) How many turns are in the primary coil (the wall side of the transfomer)?

c) What is the RMS curent in the primary coil?

d) What is the peak current in the primary coil?

Explanation / Answer

a)

Here, as P = 50 W , V = 24 V

Resistance , R = V^2/P

R = 24^2/50

R = 11.52 Ohm

b)

as Vs/Vp = Ns/Np

24/120 = 100/Np

Np = 500 turns

c)

as Power = Irms * Vrms

50 = 120 * Ip

Ip = 0.417 A

current in primary is 0.417 A

d)

Ipeak = Irms * sqrt(2)

Ipeak = 0.417 * sqrt(2)

Ipeak = 0.589 A

the peak current in primary is 0.589 A

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.