22 June 2013

Current Electricity Numerical solved Physics class 10



Q 1.Calculate number of electrons in one coulomb of charge.

Ans : As we know Q = ne
So, in 1 electron there is 1.6 x 1019C of charge
In 1C there will be1/1.6x1019 electrons=6.25x10 18 electrons

Q.2: . A household uses the following electric appliances:
 (i) Refrigerator of rating 400 W for ten hours each day.
(ii) Two electric fans of rating 80 W each for twelve hours each day.
(iii) Six electric tubes of rating 18 W each for 6 hours each day.
Calculate the electricity bill of the household for the month of June if the cost per unit of electric energy is Rs. 3.00.

Ans:
 A) Electric Energy consumed per day = 400 x 10 + 2 x 80 x 12 + 6 x 18 x 6 = 6568 wh
Total Energy per month = (6568x30)/100=197.040 kWh
Total Cost = 197.040 x 3= Rs 591

Q. (i) A 100 W electric bulb is connected to 220 V mains power supply. Calculate the strength of the electric current passing through the bulb.

(ii) If the same bulb is taken to U.S.A where the main power supply is 110 V, how much electric current will pass through the bulb when connected to mains?

Ans)
(a) Both the expressions are correct. In the first case, I remain constant whereas the second expression is true when V remains constant

i)P=VI
 I=P/V=100/220=10/22 A

R=V/I=220X 22/10=484

ii)I=V/R=110V/484W=(110/484)A

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