A fully charged 7.7 uF capacitor is connected in series with a 2.0×10^5 Omega resistor.

+2 votes

A fully charged 7.7 capacitor is connected in series with a 2.0×105 resistor. What percentage of the original charge is left on the capacitor after 1.8 of discharging?

I know the formulas to use are i(t)=I_0*e^(-t/RC) and q(t)=C*emf*e^(-t/RC)

Don't know where to start....

asked Jul 20, 2012 in Physics

2 Answers

+1 vote

The best way to start a physics problem is writing down what you know.  Write down the numbers that the problem gives you, for example you know:

capacitance =7.7

resistance = 2.0×105 ohms

time = 1.8 s

Then look at the formula and look for what you need to solve the question.

For example if the formula require emf, ask yourself how do i find emf with the information given.

answered Jul 21, 2012 by ~Expert~ (1,318 points)
Thank you. But after I did all of that and had all the formulas down that's where I got stuck. Can't find the emf and according to my instructor you don't have to. Which I don't get how that'll work.
+1 vote

As the other answer said, start by writing down what you know.

Now, assuming your formulas are correct:

q(t)=C*emf*e^(-t/RC)

also using ohm's law, which states: I = V/R  (where I=current, V=voltage or emf, R=resistance)

Using ohms law you solve for V.

V = R*I  or emf = R*I

So you should now plug in the formula for I, which you have as:

i(t)=I_0*e^(-t/RC)  and multiply by R to get emf.

With this emf plug back in the first equation for charge.

answered Jul 22, 2012 by ~Expert~ (919 points)

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