This homework is a careful walkthrough of the steps to determine the brightness

Question

This homework is a careful walkthrough of the steps to determine the brightness of each bulb. Assume the voltage source is a 120 V AC source, (which we will treat as equivalent to a 120 V DC battery). In addition, assume each bulb is identical, and each has a power rating of 60 W when 120 V is applied. Assume "Ohmic" bulbs! Notice that the switch is in parallel with bulb C 120 V B( For each problem, show algebraic steps or your reasoning where appropriate 1. Considering the power rating ("60 W when 120 V is applied"), calculate the resistance of any one bulb. (2 pt)

Explanation / Answer

1. P = V^2 / R

60 = 120^2 / R

R = 240 ohm

2. C and D are in series. so equivalent of these two will be Rc + Rd

so 2R.

now 2R and B are in parallel.

1/R' = 1/2R + 1/R

R' = 0.67R

now A and 0.67R are in series.

Req = R + 0.67R = 1.67R

Req = 1.67 x 240 = 400 ohm ......Ans

I = V/Req = 120/400 = 0.3 A ..Ans

Va = IR = 0.3 x 240 = 72 volt

PA = Va^2 / R = 72^2 / 240 = 21.6 W

2b) bulb gets rated power when potential across is 120 V.

but PD across it 72 volt hence its consuming less power.

3a) when swtitch is closed then C is shortcircuted.

hence no current will flow through C.

now D and B are in parallel connection.

1/R' = 1/R + 1/R

R' = 0.5R

now A and R' are in series.

Req = R + R' = 1.5R = 1.5 x 240 = 360 ohm

I = V / Req = 120/360 = 0.333 A

Va = 0.33 x 240 = 80 volt

P = Va^2 / R = 80^2 / 240 = 26.67 W

3b) now it consumes more power than before.

so it will be brighter.

4) Ib = 0.30 x 2R / (R + 2R) = 0.2 A

Vb = 0.2 x 240 = 48 V

Pb = i^2 R = 0.2^2 x 240 = 9.6 W

Ic = Id = 0.3-0.2 = 0.1 A

Vc= Vd = 0.1 x 240 = 24 V

P = 0.1^2 240 = 2.4 W

5)when closed:

Ib = 0.33/2 = 0.16 A

Vb = 0.16 x 240 = 40 V

Pb = 40^2 / 240 =6.67 W

Ic = 0

Vc = 0

Pc = 0

Id = 0.33/2 = 0.16 A

Vd = 0.16 x 240 = 40 V

Pd = 40^2 / 240 =6.67 W

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