Question

Suppose the battery on your phone delivers a current of 0.770 mA (that's milliAmperes) over a time of 3.00 hours. What is the total charge that flows out of the battery during this time?

How many electrons flow out of the battery during this time?

Answer 1

a)

q= 0.77*10^-3*3*60*60

= 8.316 C

b) No. of ele. = 8.316/(1.6*10^-19)

= 5.1975*10^19 electrones

Answer 2

Q = i*T

Thus, Q = 0.770 * 10^-3 * 3 * 3600

Q = 8.316 C

Number of electrons = Q/e = 8.316/(1.6*10^-19) = 5.198 * 10^19 electrons

Answer 3

a)

t=3*60*60*=10,800 s

q=it = =0.77*10^-3*10800

q=8.316 C

b)

N=8.316/1.6*10^-19

N=5.19*10^-19 electrons

Answer 4

a)

q=I*time

time = 3*60*60 sec

so

q= 0.77*10^-3*3*60*60

= 8.316 C

b) No. of ele. = q/charge of electrone

= 8.316/(1.6*10^-19)

= 5.1975*10^19 electrones

Answer 5

I = Q/t

0.77 * 10^-3 = Q / (3*3600)

Q = 8.316 C

Q = N * Qe

8.316 = N * 1.602 * 10^-19

N = 5.19 * 10^19 electrons