For a three-way bulb ( 50 W, 90 W, 140 W) that is designed to operate at 115 V,

Question

For a three-way bulb (50 W, 90 W, 140 W) that is designed to operate at 115 V, find the resistance of each of the two filaments. Assume that the wattage ratings are not limited by significant figures, and ignore any heating effects on the resistances.

R50 W filament = ?
?





A 65.0-? resistor is connected in parallel with a 122.0-? resistor. This parallel group is connected in series with a 25.0-? resistor. The total combination is connected across a 15.0-V battery. (a) Find the current in the 122.0-? resistor.
A

(b) Find the power dissipated in the 122.0-? resistor.
W

Explanation / Answer

V^2/R = power

R= V^2/power = 115*115/50 =264.5 omega

for R90 filament R= 115*115/90=146.944

(1/65)+(1/122)= 1/42.4

42.4+25= 67.4 omega (net resistance)

I through the circuit = 15/67.4=0.222

I= I through 122+I through 65

I in 122/I in 65 = 65/122

I in 65 = 1.87*I in 122

I = I in 122+ I in 65

I= I in 122+ 1.87*I in 122

I = 2.87*I in 122

I in 122= I/2.87

I in 122= 0.222/2.87= 0.077 A

power dissipated = (I in 122)^2*R = 0.077^2*122 = 0.729 W

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