In which of these situations does more charge flow past a given point on a wire?
a. Current of 4 A in 1 min
b. Current of 2 A in 0.5 min
c. Current of 8 A in 2 min
d. All are the same charge
I think the answer is C? Since the formula for finding current is 1 = C / s = 1 Ampere. So wouldn't I just have to multiply the current and minutes to get the net charge?
How long would it take for a net charge of 2.5 C to pass a location in a wire if it is to carry a steady current of 5 mA? I know in order to convert mA to amps, divide by 1000, so 5/1000=5x10^-3A. Now what do I do? I thought I'd have to switch the original formula around so I divide the net charge by the current to get the time?
current in amps is coulombs of charge per second
I = Q/t
so
Q = I t
4*1 = 4
2*.5 = 1
8*2 = 16 coulombs
Q = i t
so
t = Q/i
t = 2.5 /5*10^-3 = (1/2) * 10^3
= 500 seconds
The answer is D
For the first question, you are correct that the formula for calculating charge is Q = I × t, where Q is the net charge in coulombs, I is the current in amperes, and t is the time in seconds. To determine which situation has more charge flow, you need to calculate the charge for each option.
a. Current of 4 A in 1 min:
Q = 4 A × 60 s = 240 C
b. Current of 2 A in 0.5 min:
Q = 2 A × 30 s = 60 C
c. Current of 8 A in 2 min:
Q = 8 A × 120 s = 960 C
d. All are the same charge (not applicable in this case)
Based on these calculations, option c has the highest charge flow since 960 C of charge passes in that scenario.
Now, for the second question, you are also on the right track. To calculate the time required for a net charge of 2.5 C to pass, you can use the formula t = Q / I.
Given:
Q = 2.5 C
I = 5 × 10^-3 A
Plugging these values into the formula, we get:
t = 2.5 C / (5 × 10^-3 A) = 500 s
Therefore, it would take 500 seconds for a net charge of 2.5 C to pass a location in the wire with a steady current of 5 mA.
Yes, you're on the right track! To find the net charge, you can indeed multiply the current and time. However, in this case, you need to consider the units as well.
For the first question, let's compare the options:
a. Current of 4 A in 1 min
b. Current of 2 A in 0.5 min
c. Current of 8 A in 2 min
To determine which situation would result in more charge flowing past a given point on a wire, you need to calculate the product of the current and time for each situation.
a. Charge = Current (4 A) x Time (1 min) = 4 A x 1 min = 4 C-min
b. Charge = Current (2 A) x Time (0.5 min) = 2 A x 0.5 min = 1 C-min
c. Charge = Current (8 A) x Time (2 min) = 8 A x 2 min = 16 C-min
Comparing the results, it is clear that option c (Current of 8 A in 2 min) results in the highest charge of 16 C-min. So, the answer is c.
For the second question, you're right that you need to convert the current from mA to Amps by dividing by 1000. So, 5 mA = 5 x 10^(-3) A.
To find the time, you need to use the formula:
Time = Net charge / Current
Plugging in the values:
Time = 2.5 C / 5 x 10^(-3) A
To divide by a fraction, you can multiply by its reciprocal. In this case, the reciprocal of 5 x 10^(-3) A is 1 / (5 x 10^(-3)) A.
Time = 2.5 C * (1 / (5 x 10^(-3)) A)
Simplifying this expression:
Time = 2.5 C * (1 / (5 x 10^(-3)) A) = (2.5 C) * (200 A/ 1 C) = 500 s
Therefore, it would take 500 seconds for a net charge of 2.5 C to pass a location in a wire if it carries a steady current of 5 mA.