Solving Electric Transients in RC Circuits: Time Constant & Voltage Calculations

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An RC circuit with a time constant of 40 microseconds will have a capacitor voltage drop to 5 volts after 40 microseconds when initially charged to 80 volts. The voltage decay can be calculated using the formula V = V0e^(-t/RC), which rearranges to t = RC ln(V/V0). For a second scenario with a time constant of 0.0001 seconds, the capacitor discharges from 100 volts to 20 volts in 0.0001 seconds. This also utilizes the same voltage decay formula for calculations. Understanding these calculations is crucial for solving electric transients in RC circuits.
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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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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.
 
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