Madelyn and Scarlett decide to work a problem from the book.An RLC circuit has been set up as illustrated in the simulation.Initially, the switch S is at a, allowing the capacitor to charge. With the capacitor already fully charged, the switch is moved to b at time t = 0. As a result, the capacitor is being allowed to discharge through the resistor and the inductor.Find the frequency f of the oscillations in current that result, using L = 1.95 H, R = 1.2 Ω, and C = 0.30 F. Recall that 2𝜋f = 𝜔. HzFind the time needed for the amplitude to decrease to 25 percent of its maximum value.
Question
Madelyn and Scarlett decide to work a problem from the book.An RLC circuit has been set up as illustrated in the simulation.Initially, the switch S is at a, allowing the capacitor to charge. With the capacitor already fully charged, the switch is moved to b at time t = 0. As a result, the capacitor is being allowed to discharge through the resistor and the inductor.Find the frequency f of the oscillations in current that result, using L = 1.95 H, R = 1.2 Ω, and C = 0.30 F. Recall that 2𝜋f = 𝜔. HzFind the time needed for the amplitude to decrease to 25 percent of its maximum value.
Solution
To solve this problem, we need to use the formulas for the frequency of oscillation and the time constant in an RLC circuit.
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First, we find the frequency of the oscillations. The formula for the frequency f in an RLC circuit is given by:
f = 1 / (2π√(LC))
Substituting the given values of L = 1.95 H and C = 0.30 F into the formula, we get:
f = 1 / (2π√(1.95 * 0.30)) = 0.117 Hz
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Next, we find the time needed for the amplitude to decrease to 25 percent of its maximum value. The formula for the time constant τ in an RLC circuit is given by:
τ = 2L / R
Substituting the given values of L = 1.95 H and R = 1.2 Ω into the formula, we get:
τ = 2 * 1.95 / 1.2 = 3.25 s
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The time needed for the amplitude to decrease to 25 percent of its maximum value is given by:
t = τ * ln(1 / 0.25)
Substituting the calculated value of τ = 3.25 s into the formula, we get:
t = 3.25 * ln(1 / 0.25) = 3.25 * ln(4) = 4.52 s
So, the frequency of the oscillations is 0.117 Hz and the time needed for the amplitude to decrease to 25 percent of its maximum value is 4.52 s.
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