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

[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...

 

[Nate810]

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.

 

[quicknote]

I would be happy to help you figure out these problems. The first step in solving these types of problems is to understand the concept of time constant in an RC circuit. The time constant, denoted by the symbol "τ" (tau), is a measure of how quickly the capacitor in the circuit charges or discharges. It is calculated by multiplying the resistance (R) and the capacitance (C) in the circuit, τ = RC.

Now, let's apply this concept to the first problem. We are given a time constant of 40 microseconds, which means that τ = 40 microseconds. The voltage of the capacitor (V) starts at 80 volts and we want to find out after what time (t) it will reach 5 volts. To solve this, we can use the equation V = V(0) e^(-t/τ). Plugging in the given values, we get 5 = 80 e^(-t/40μs). Solving for t, we get t = 100 microseconds. Therefore, after 100 microseconds, the capacitor voltage will be equal to 5 volts.

Moving on to the second problem, we are given a time constant of 0.0001 seconds, which means that τ = 0.0001 seconds. The capacitor is initially charged to 100 volts, and we want to find out how much time it takes for the voltage to decrease to 20 volts. Using the same equation as before, we get 20 = 100 e^(-t/0.0001s). Solving for t, we get t = 0.0001 seconds, or 100 microseconds. This means that after 100 microseconds, the capacitor voltage will decrease to 20 volts.

I hope this helps you understand how to solve these types of problems in RC circuits. Remember to always use the equation V = V(0) e^(-t/τ) and plug in the given values to find the unknown variable. Good luck!