# An unknown capacitor is charged to 6v, its maximum value, then discharged through a 1k ohm resistor. If the capacitor voltage is 3v, 0.3 seconds after starting to discharge, what is the capacitance of the unknown capacitor?

To start, the equation to use can be identified as the capactior discharge equation V(t) = Vmaxe-t/RC. From here, we can see that this equation can be solved to give capacitance, and identify which values are needed to solve this. Firstly, we identifly the variables we are given in the question. We are given Vmax as 6v, as the max value the capacitor is charged to, as well as the value of the resistor which the capacitor is discharged through, 1000 ohms. We are given a value of V(t), 3v, at t=0.3s. So, we have all the known values we need to solve for capacitance. We start by plugging in our known values. This gives us the equation 3 = 6e-0.3/1000C. Now, all we do is rearange this equation to find C, our unknown capacitance. Divide both sides by 6 to give 3/6 = 0.5 = e-0.3/1000C. Now that the power of e is on its own, we can take natural log, ln() of both sides. This gives ln(0.5) = -0.3/1000C. By multiplying though by C, and dividing through by ln(0.5), we get C = -0.3/1000ln(0.5). This equation gives a value of 4.33x10-4 F, or 433uF.

Answered by Alexander R. Physics tutor

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