You have a 9V battery connected to a device you wish to power. 5C of charge flows out of the battery over the course of twohours. What is the average current (in Amps) the flowed during that time?

Answer: 0.00625A

Explanation:

Quantity of charge Q flowing in a circuit per time is equal to the product of the current flowing and time take. Mathematically

Q = It ... 1

Where Q is the charge, I is the current and t is the time

For a charge across a capacitor, Q = CV ... 2

C is the capacitance of the capacitor

V is the voltage

Equating 1 and 2, we have

It = CV

I = CV/t

Given C = 5C, V = 9V and t = 2hrs = 120seconds

I = 5*9/7200

I = 45/7200

I = 0.00625A

Average current that flowed during that time is 0.00625A

0.00625 A.

Explanation:

From Faraday's law of electricity,

Quantity of electric charge = Current * time.

Q = It ... Equation 1

Where Q = Quantity of charge, I = current, t = time.

Making I the subject of the equation,

I = VQ/t ... Equation 2

Given: Q = 5 C t = 2 h = (2*3600) s = 7200 s, V = 9 V

I = 5*9/7200

I = 0.00625 A

Hence the average current flowing through = 0.00625 A.