Question

a current of 55 ma flows through a circuit for 2 minutes 40 seconds. find the change in number of e- flowing.

Answer 1

Answer:

$5.5 \times {10}^{19} electrons$

Explanation:

$we \: know \: that \: q = \: it \\ so \\ q = 55 \times {10}^{ - 3} \times (120 + 40) \\ q = 8.8 \: c \\ n \: e \: = 8.8 \\ n = \frac{8.8}{1.6 \times {10}^{ - 19} } \\ n \: = 5.5 \times {10}^{19} electrons$

ANSWERED BY GAUTHMATH