series resonant circuit has R = 2 Ω, L = 1 mH and C = 0.1 μF, the value of quality factor Q is:
Question
series resonant circuit has R = 2 Ω, L = 1 mH and C = 0.1 μF, the value of quality factor Q is:
Solution
The quality factor (Q) of a series resonant circuit is given by the formula:
Q = (1/R) * sqrt(L/C)
where: R = resistance = 2 ohms L = inductance = 1 mH = 1 * 10^-3 H C = capacitance = 0.1 μF = 0.1 * 10^-6 F
Substituting these values into the formula, we get:
Q = (1/2) * sqrt((1 * 10^-3) / (0.1 * 10^-6))
Solving the equation inside the square root first:
= sqrt(10^-3 / 10^-7) = sqrt(10^4) = 100
Then, substitute this back into the equation:
Q = (1/2) * 100 Q = 50
So, the quality factor Q of the series resonant circuit is 50.
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