Question

. A 9V battery supplies power to a 2W lightbulb in a flashlight and lasts for 50 hours. (a) How much current runs through the simple flashlight circuit? (b) How many electrons pass through the lightbulb in the time it stays lit? (c) What rating would this battery be rated in A·h? (d) If the battery cost $2.00, how much does the energy cost per kW·h?

Answer 1

Solution:

A 9V battery supplies power to a 2W lightbulb in a flashlight and lasts for 50 hours.

(a) Current runs through the simple flashlight circuit is,

P = V*I

Or, 2W = 9V*I

or,I = 0.22Amphere.[Answer]

(b) We know that,

I = n*(q/t) [ where n= no of electrons]

or,0.22= n*(1.6*10^-19)/(50*3600)

n= 2.475*10^23 [Answer]

(c) Rating of the battery in A-h is 0.22*50 = 11A-h

(d) Battery Cost is $2.00 = 151.709 INR

Heat Lost , H = P*t = (2*10^-3kW)*50h = 0.1 kW-h

Energy Cost per kW-h is (151.709)/0.1 = 1517.09 INR/kW-h

answered by: anonymous