A60\ 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?

137 turns

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

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.