A 30-W led bulb and a 90-W led bulb are wired in the parallel to a 120-V source. You plug in a third bulb in parallel with the other two and the current increases to a 4.5-A, what is the wattage of the bulb you added? Please explain how you got the answer in steps.

Answer 1

Use the formulas:
Power (watts) = V^2/R
V=voltage in V = 120 V
R=reisistance in Ω
=>
R=V^2/P............(1)

R(30W)=120^2/30=480Ω
R(90W)=120^2/90=160Ω

Now using equation
V=iR
i=current in amps

With three connected in parallel,
i=4.5A
R(equivalent)=V/i=120/4.5=80/3 Ω

Using the parallel resistance formula
R(equivalent)=1/(1/R1+1/R2+1/R3)
or
80/3=1/(1/480+1/160+1/R3)
or
3/80=(1/480+1/160+1/R3)
solve for R3
R3=1/(3/80-1/480-1/160)
=240/7Ω
=34.2857Ω approx.

Now you can substitute R in equation (1) given above to solve for P, the wattage, which turns out to be much greater than the other two.

Post your answer for a check if you wish.