Your RC product must remain the same, 0.006 s.
The first capacitor has a C value given by
8 J = (1/2) C V^2 = (9/2) C
C = 16/9 farad
The first resistor has
R = 6*10^-3/(16/9) = 3.38*10^-3 ohms
To multiply the stored power by 10, you would have to increase C by a factor of 10 while keeping V the same. The higher C will require a lower value R, to keep the RC value the same.
I consider the capacitance value unrealistically high and the resistances unrealistically low in this example. Farad-scale capaciators are huge, and would have appreciable internal resistances that might exceed these R values.
An RC-circuit has a time constant of 6x10-3 seconds.
The capacitor stores a maximum energy of 8 J, when charged using a 3-Volt battery.
Using the same battery, we want to increase tenfold the amount of stored energy, while keeping the same time constant (to ensure that the charge and discharge of the capacitor occur in reasonable amounts of time).
To accomplish this, both the capacitor and resistor must be replaced.
Calculate the value of the resistance of the new resistor.
Please Help. I got this wrong
1 answer
1 answer