A capacitor in a single-loop RC circuit is charged to 64% of its final voltage in 1.5 s.
Find the time constant for the circuit.
Find the percentage of the circuit's final voltage after 3.9 s .
Vc/Vi= %
The voltage across the capacitor is
Vi * [1 - e^-(t/RC)]
in this type of circuit. Vi is the power supply voltage. RC is the time constant.
The final voltage is Vi.
When t = RC (the time constant), the voltage is
Vc = Vi*[1 - e^-1] = 0.632 Vi
When t = 1.5 s,
0.64 Vi = Vs * [1 - e^(-t/RC)]
e^(-t/RC)= 0.36
t/RC = 1.02
RC = 1.5/1.02 = 1.47 seconds
That is the time constant.
After 3.9 s
Vc = Vi* [1 - e^(-3.9/1.47)]
= 0.930 Vi
Unless you show some work on future posts here, I will not be assisting you again. Perhaps someone else will.
Vc = Vi* [1 - e^(-3.9/1.47)]
= 0.930 Vi
To find the time constant (τ) of the circuit, we can use the formula:
τ = RC
Where R is the resistance in the circuit and C is the capacitance of the capacitor.
Now, to solve the equation, we need to use the information given in the first part of the question. The capacitor is charged to 64% of its final voltage in 1.5 seconds.
We know that the time constant (τ) is the time it takes for the capacitor to charge to approximately 63.2% of its final voltage. Since the given time (1.5 seconds) corresponds to a charge of 64% instead, we can approximate the time constant as follows:
τ ≈ t / ln(1 / (1 - 0.64))
Where t is the given time, and ln is the natural logarithm.
Substituting the values, we get:
τ ≈ 1.5 / ln(1 / (1 - 0.64))
Calculating the value, we find:
τ ≈ 1.5 / ln(1 / 0.36)
τ ≈ 1.5 / ln(2.78)
Using the natural logarithm of 2.78 (approximately 1.022), we find:
τ ≈ 1.5 / 1.022
τ ≈ 1.47 seconds
Hence, the time constant (τ) for the circuit is approximately 1.47 seconds.
Moving on to the second part of the question, to find the percentage of the circuit's final voltage after 3.9 seconds, we can use the formula:
Vc/Vi = e^(-t / τ)
Where Vc is the voltage across the capacitor at time t, Vi is the initial voltage across the capacitor, t is the given time, and τ is the time constant of the circuit.
To find the percentage of the final voltage, we can multiply the value of Vc/Vi by 100.
Vc/Vi = e^(-3.9 / 1.47)
Calculating the value, we find:
Vc/Vi ≈ e^(-2.653)
Finally, to find the percentage, we multiply the value by 100:
Vc/Vi ≈ 0.071
Therefore, the capacitor's voltage is approximately 7.1% of its final voltage after 3.9 seconds.