Problem 22.16


A 7.5 v battery supplies a 3.5 mA current to a circuit for 8.2 Hr .



Part A -

How much charge has been transferred from the negative to the positive terminal?

Express your answer using two significant figures.



Part B -

How much work has been done on the charges that passed through the battery?

Express your answer using two significant figures.

I will be happy to critique your thinking.

Current*time= charge watch units.

Exactly what is an ampere?

To solve this problem, we need to use the basic formula relating charge, current, and time:

Q = I * t

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

Part A:
We are given that the battery supplies a current of 3.5 mA (milliamperes) for 8.2 hours. The first step is to convert the current from milliamperes to amperes:

3.5 mA = 0.0035 A

Now we can use the formula to calculate the charge transferred:

Q = (0.0035 A) * (8.2 hours)

Q ≈ 0.0287 As

To express the answer using two significant figures, we round it to:

Q ≈ 0.029 As

Part B:
To calculate the work done on the charges that passed through the battery, we need to use another formula:

W = V * Q

Where W is the work done, V is the voltage, and Q is the charge.

We are given that the battery has a voltage of 7.5 V, and we just calculated the charge (Q) to be approximately 0.029 As. Now we can substitute these values into the formula:

W = (7.5 V) * (0.029 As)

W ≈ 0.21825 J

To express the answer using two significant figures, we round it to:

W ≈ 0.22 J

So, for Part A, the amount of charge transferred from the negative to the positive terminal is approximately 0.029 As. And for Part B, the amount of work done on the charges that passed through the battery is approximately 0.22 J.