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capacitors

A 4 * 10^-6 F capacitor is charged by a 50V battery and then discharged through a 2 megaohm resistor. In 8 seconds the p.d across the capacitor decreases to a value in V of about
A. 63
B. 50
C. 32
D 18
E 12

the only thing i can manage to calculate is charge which is equal to 2 * 10^-4 C.

Reply 1

Morbo

To solve this problem you need to know how the voltage across the capacitor changes with time as you discharge it - the discharge curve.

It turns out, as a textbook will show you, that this is given by the exponential:

V(t) = V(0)[1 - exp(-t/RC)]

So all you have to do is substitute in the values that you are given:

V(0) = 50 V
t = 8 s
R = 2 megaohms
C = 4 microfarads

I get answer C: 32V

Reply 2

mhines87

Not quite Worzo. Just remember that a capacitor will charge or discharge to approx. 63% of the supply voltage in one time constant (CR seconds).

It should be obvious to you as soon as you look at it that the question has been cleverly engineered such that one time constant is 8 seconds. You can therefore say that the voltage across the cap after 8 seconds will have reduced by:

50*0.63 = 31.5V

Therefore the instantaneous voltage across the cap will be: 50-31.5 = 18.5V... which is closest to the 18V option.

This method is faster since you can do it in your head in a matter of seconds.