1) A 100.0(ohm) and a 150.0(ohm) resistor, both rated at 2.00(ohm) , are connected in series across a variable potential difference. What is the greatest this potential difference can be without overheating either resistor?
V=
2) What is the rate of heat generated in a 100.0(ohm) resistor under these conditions?
P=
3) What is the rate of heat generated in a 150.0(ohm) resistor under these conditions?
P=
4) Two light bulbs have resistances of 400(ohm) and 800(ohm) .The two light bulbs are connected in series across a 120 V line.
Find the power dissipated in each bulb?
Find the total power dissipated in both bulbs?
Physics Questions
2008-03-17 10:11 am
回答 (2)
2008-03-17 7:37 pm
✔ 最佳答案
(1) Since they are connected in series, current is the common quantity that, applying P = I2R, we can find out for each resistor, its maximum affordable current value is:100-Ω resistor:
2 = 100I2
I = √0.02 A
150-Ω resistor:
2 = 150I2
I = √0.013 A
So without overheating either resistor, we should take the lower one, i.e. √0.013 A
Then, total p.d. across the resistors is:
√0.013 x (150 + 100) = 28.87 V
(2) Rate of heat generation in the 100-Ω resistor = 0.013 x 100 = 1.3 W
(3) Rate of heat generation in the 150-Ω resistor = 0.013 x 150 = 2 W
(4) Total resistance = 1200 Ω
Current drawn from the line = 120/1200 = 0.1 A
Power dissipated in the 400-Ω bulb = 0.12 x 400 = 4 W
Power dissipated in the 800-Ω bulb = 0.12 x 800 = 8 W
Total power dissipation = 12 W
參考: My physics knowledge
2008-03-17 10:42 am
收錄日期: 2021-04-13 15:19:27
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