RC circuit discharge: finding time when voltage drops from 80V to 5V

[cunhasb]

I would really appreciate if anyone could help me figure out these problems...

An RC circuit has a time constant of 40 microseconds. If the capacitor is first charged to a voltage of 80 volts and the RC circuit is then closed upon itself, after what time will the capacitor voltage be equal to 5 volts?... and

An RC circuit has a time constant of 0.0001 second. If the capacitor C is charged to 100 volts and, with the battery removed, the capacitor is then allowed to discharge through the resistor, at what time after being connected to the resistor will the voltage be 20 volts? (Assume that, in discharging, the voltage V at any time is equal to V=V(0) e^(-1/RC), where V(0) is the voltage to which C was charged.)

Thank you so much guys...

 

[member 553083]

1. For the first problem, the answer is 40 microseconds. To calculate this, you need to use the formula V = V0e^(-t/RC). This can be rearranged to solve for t (time): t = RC ln(V/V0). Plugging in the given values yields t = 40 μs. 2. For the second problem, the answer is 0.0001 second. To calculate this, you need to use the formula V = V0e^(-t/RC). This can be rearranged to solve for t (time): t = RC ln(V/V0). Plugging in the given values yields t = 0.0001 s.