The voltage source is 120 V and each light bulb has a power rating of 100 W 1. C

Question

The voltage source is 120 V and each light bulb has a power rating of 100 W

1. Considering the power rating, calculate the resistance of each bulb

2. With the switch opened, show that the total resistance, the total current, the voltage drop across bulb A, and the average power dissipated in bulb A are as follows.  "Show that" means explain your reasoning, work it out, verify so that the grader   understands that you understand how you could have come up with all four numbers even if we had not given them below

Rtot = 240 ohm

Itot = .5 A

delta V of A = 72 V

P of A alone = 36 W

3. Next with the switch CLOSED, calculate the total resistance, total current, and voltage drop across bulb A and the average power dissapted in bulb A

4. With the switch open calcuate the volatage drop in bulb B C and D. Calculate average power dissapated in bulbs B, C, and D

5. With the switch closed determine the voltage drop in bulbs B, C, and D. Calculate the average power dissapated in bulbs B, C, and D

Explanation / Answer

since watt of each bulb is same that is 100 W therefore, Resistance of each bulb is same that is let us say R ohm

therefore

R net = ( ( 2R ) || R + R ) = (5/3)R ohm

i (A) = 72/R A.

V (A) = i (A) * R(A) = 72 Volt

tsimilarly

i (B) = 48 /R A

V (B) = 48 V

i (c) = i (D) = 24/R A

V (C) = V(D) = 24 V

hence R is 144 ohm

now calculating R net

R net = (5/3)*144 = 240 ohm

hence

I net = I A = 120 / 240 = 0.5 A

P (A) = V * I = 0.5*72 = 36 W

on closing the switch

total resistance = R || R + R = 216 ohm

voltage drop accross R A = ( 120 / R net ) R A = 80 V

Power dessipated accross R A = 44.4444 W

similarly calculate other things .

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