610 C passes through a flashlight in 0.510 h. What is the average current?
current=charge/timeinseconds=610/(.51*3600)
0.33
To determine the average current, we need to use Ohm's Law, which states that current (I) is equal to the charge (Q) divided by the time (t).
The charge is calculated by multiplying the current (I) by the time (t).
Given:
Charge (Q) = 610 C
Time (t) = 0.510 h = 1836 s
To calculate the average current, divide the charge by the time:
I = Q / t
I = 610 C / 1836 s
Calculating this:
I ≈ 0.332 A
Therefore, the average current passing through the flashlight is approximately 0.332 Amperes.
To determine the average current, you need to recall Ohm's law, which states that the current (I) is equal to the amount of charge (Q) flowing through a conductor divided by the time (t) it takes for that charge to pass through.
The formula for calculating average current is:
I = Q / t
In this case, you are given the amount of charge (Q) passing through the flashlight, which is 610 C (coulombs), and the time (t) it takes for that charge to pass, which is 0.510 h (hours).
Now, you need to convert the time from hours to seconds since the unit of charge is in coulombs and the unit of time is usually in seconds.
1 hour = 3600 seconds
Therefore, you can convert 0.510 hours to seconds as follows:
0.510 h * 3600 s/h = 1836 s
Now, you can substitute the values into the formula:
I = 610 C / 1836 s
Calculating this equation will give you the average current flowing through the flashlight. Let's do the math:
I = 610 C / 1836 s ≈ 0.332 A (Amperes)
So, the average current through the flashlight is approximately 0.332 Amperes.