A 6.0 V battery is being used to charge a 5.0x10^-6 F capacitor through a resistor R. The capacitor attains a potential difference of 3.6 V in 1 second after the charging begins. The value of the resistance R is?
Been stuck on this for a while and if anyone can point me into the right direction that will be great.
V=6(1-e^t/rc)
3.6/6=1-e^1/rc
take ln of both sides
ln(.6)=0+-1/rc
r=-ln(.6)/5E-6
Oh ok, thank you.
To find the value of resistance R, we can use the formula for the charging of a capacitor through a resistor:
Vc = Vf(1 - e^(-t/RC))
where:
- Vc is the potential difference across the capacitor at time t,
- Vf is the final potential difference across the capacitor (6.0 V in this case),
- t is the time (1 second in this case),
- R is the resistance,
- C is the capacitance (5.0x10^-6 F in this case),
- e is the exponential function (approximately 2.71828).
We have the values of Vc, Vf, and t. We need to solve for R.
Substituting the given values into the formula, we have:
3.6 V = 6.0 V (1 - e^(-1/(R * 5.0x10^-6 F)))
To isolate R, we can rearrange the equation:
e^(-1/(R * 5.0x10^-6 F)) = 1 - (3.6 V / 6.0 V)
e^(-1/(R * 5.0x10^-6 F)) = 0.4
To remove the exponential term, we can take the natural logarithm (ln) of both sides:
ln(e^(-1/(R * 5.0x10^-6 F))) = ln(0.4)
Simplifying:
-1/(R * 5.0x10^-6 F) = ln(0.4)
To isolate R, we can rearrange the equation:
R = -1 / (5.0x10^-6 F * ln(0.4))
Now we can calculate the value of R using a calculator:
R ≈ -1 / (5.0x10^-6 F * ln(0.4))
Thus, you'll find the value of resistance R by evaluating the expression using a calculator.