Electric Transients: Solving RC Circuits Problems

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The discussion focuses on solving problems related to RC circuits and their time constants. The first problem involves a circuit with a time constant of 40 microseconds, where a capacitor charged to 80 volts discharges to 5 volts, calculated to take approximately 110.903 microseconds. The second problem deals with a circuit with a time constant of 0.0001 seconds, where a capacitor charged to 100 volts discharges to 20 volts, taking about 160.94 microseconds. The correct formula for voltage as a function of time, V(t) = V(0)e^(-t/RC), is emphasized for solving these problems. The calculations provided confirm the understanding of the concepts involved in RC circuit behavior.
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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...
 
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Can you write out the formula for voltage or charge as a function of time, given the time constant?

Then we can point you in the right direction if necessary.
 
cunhasb said:
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...
Your formula is independent ot the time. The correct formula would be:
V(t) = V(0)e^{-\frac{t}{RC}}
Where RC is the time constant. If you replace V(0) with the initial value of the voltage and V(t) with the final value, you can use logarithms to solve for the time t.
 
Thank you so much guys...

well... I think I got it... correct me if I'm wrong...

1) Rc= 40 micro V(0) = 80 T=? V = 5V

V=v(0) e^(-t/RC)
5=80*e^(-t/RC)
5/80=e^(-t/RC)
RC*ln0.0625=-t
t=110.903 microseconds...

2) Rc= 0.0001 V(0) = 100 T=? V = 20V

V=v(0) e^(-t/RC)
20=100*e^(-t/RC)
20/100=e^(-t/RC)
RC*ln0.2=-t
t=160.94 microseconds...
 
That's it.
 
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