Calculating Charge and Current in a Resistor Circuit

[whitehorsey]

1. At time t = 0, a 23 ohms resistor is linked to a 4.4V battery. How much charge has gone through the resistor after 5 seconds?

2. Q = CΔV
ΔV = IR

3. ΔV = IR
I = ΔV/R
= 4.4/24 A.

I'm not sure what to do next.

 

[rude man]

1. At time t = 0, a 23 ohms resistor is linked to a 4.4V battery. How much charge has gone through the resistor after 5 seconds?

2. Q = CΔV
ΔV = IR

3. ΔV = IR
I = ΔV/R
= 4.4/24 A.

I'm not sure what to do next.

Q = ∫I dt

 

[Curious3141]

1. At time t = 0, a 23 ohms resistor is linked to a 4.4V battery. How much charge has gone through the resistor after 5 seconds?

2. Q = CΔV
ΔV = IR

3. ΔV = IR
I = ΔV/R
= 4.4/24 A.

I'm not sure what to do next.

How did the 23Ω become a 24 when you wrote it down in the formula?

Anyway, what is the relationship between a constant current, charge and time?

 

[vanhees71]

Note that the current is not constant in this case!

 

[Curious3141]

Note that the current is not constant in this case!

EDIT: At first, I thought I'd made a ghastly mistake because I'd forgotten a capacitor or something. But this is just a simple circuit with a battery and a resistor. Why can't the current be assumed to be constant?

 

[technician]

Note that the current is not constant in this case!

There is only a resistor and a battery, how is the current not constant?
There is no indication of capacitance in the question details.

 

[siddharth23]

The current is constaant. Find the current and find that charge using the time given. Straight foward!

 

[vanhees71]

Ok, then I didn't understand the question. I thought there is a resistor and a capacitor. So the Q=CU equation doesn't make any sense.

 

[Chestermiller]

1. At time t = 0, a 23 ohms resistor is linked to a 4.4V battery. How much charge has gone through the resistor after 5 seconds?

2. Q = CΔV
ΔV = IR

3. ΔV = IR
I = ΔV/R
= 4.4/24 A.

I'm not sure what to do next.

1 A = ?? coulombs/sec

 

[whitehorsey]

I figured it out. Thanks guys!