Resistance and current problems

[mustang]

Problem 5.
A total charge of 12mC passes through a cross-sectional area of a nichrome wrie is 3.1s.
If the number of charges that pass through the cross-sectional area furing the given time interval doubles, what is the resulting current? In units of A.
Note: What formula(s) should I use?

Problem 7.
How long does it take for 7.5 C of charge to pass through a cross-sectional area of a copper wire if I=19A? In units of s.
Note: I don't know where to start?

Problem 11.
A typical color television draws 2.8 A of current when connected across a potential difference of 121V.
What is the effective resistance of the television set? Answer in omega.
Note: What formula(s) should I use?

 

[Jimmy]

Current can be defined as the charge which passes a given point in one second. Specifically, 1 ampere of current is 1 coulomb of charge passing a given point in one second.

If you have 12 milli-coulombs pass in 3.1 seconds then it's a simple matter to calculate how many coulombs are passing in 1 second. That would be the current in amperes. Current (in amperes) equals Charge (in Coulombs) divided by time (in seconds). I=C/T

If charge doubles, then it's easy to see what happens to current by looking at the formula. Current is directly proportional to charge.

--

Here we are given 19 amperes of current and are asked to find the time it takes 7.5 coulombs of charge to pass a given point. First ask yourself how many coulombs are passing a given point in one second. If we have 19 amperes of current, we have 19 coulombs of charge passing any given point in one second. If it takes 1 second for 19 coulombs of charge to pass, how long will it take for 7.5C of charge to pass?

--

Simple ohms law problem:

Voltage(V)= Current(I) * Resistance(R)

To find resistance, R=E/I.

 

[M98Ranger]

For Problem 5, we can use the formula for current, I = Q/t, where Q is the charge and t is the time. Since the number of charges doubles, the new charge would be 24mC. Plugging this into the formula, we get I = 24mC/3.1s = 7.74 A.

For Problem 7, we can use the same formula, I = Q/t, where Q is 7.5 C and I is 19 A. Rearranging the formula to solve for t, we get t = Q/I = 7.5 C/19 A = 0.39 s.

For Problem 11, we can use Ohm's law, R = V/I, where R is the resistance, V is the potential difference, and I is the current. Plugging in the values, we get R = 121V/2.8A = 43.2 ohms.