Calculate the resistance of an 80-W light bulb and of a 60-W Eight bulb, each fo

Question

Calculate the resistance of an 80-W light bulb and of a 60-W Eight bulb, each for standard use across a 120-V line. Then suppose the two bulbs are connected in series across that line with the resistances unchanged, and calculate the total-power. Calculate the current in each resistor: R_1= 4 Ohm, R_2 = 12 Ohm, and R_3 = 2 Ohm; If R_2 is decreased, the current I_2 will (increase/decrease/remain constant) the current I_2 will (increase/decrease/remain constant) the current I_3 will (Increase/decrease/remain constant).

Explanation / Answer

12. We know,

Power, P = V2/R

So R1 = 120*120/80 = 180 ohm

and R2 = 120*120/60 = 240 ohm

Now, I1 = V/R1 = 120/180 = 0.667 A

and I2= V/R2 = 120/240 = 0.50 A

When they are placed in series the total resistance is about (180+240) = 420 ohms, so the bulbs have a current,

I = 120/420 = 0.286 A

So, P1' = I2R1 = 0.286*0.286*180 = 14.72 W

P2' = I2R2 = 0.286*0.286*240 = 19.63 W

Total power = 14.72+19.63 = 34.35 W

13.

Current I3 = V/R3 = 15/2 = 7.5 A

Current I2 = 15/12 = 1.25 A

Current I3 = 7.5-1.25 = 6.25 A

If R2 is decreased, current I2 will increase, I1 will decrease and I3 will remain unchanged.

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