In A.C. circuit in which inductance and capacitance are joined in series. Current is found to be maximum when the value of inductance is 0.5H and the value of capacitance is 8μF. The angular frequency of applied alternating voltage will be:400 Hz5000 Hz2×105 Hz500 Hz
Question
In A.C. circuit in which inductance and capacitance are joined in series. Current is found to be maximum when the value of inductance is 0.5H and the value of capacitance is 8μF. The angular frequency of applied alternating voltage will be:400 Hz5000 Hz2×105 Hz500 Hz
Solution
In an AC circuit with series inductance and capacitance, the current is maximum when the circuit is at resonance. The condition for resonance is when the inductive reactance equals the capacitive reactance.
The formula for inductive reactance (XL) is XL = 2πfL, where f is the frequency and L is the inductance.
The formula for capacitive reactance (XC) is XC = 1/(2πfC), where C is the capacitance.
At resonance, XL = XC, so we can set the two formulas equal to each other and solve for f:
2πfL = 1/(2πfC)
Solving for f gives us:
f = 1/(2π√(LC))
Substituting the given values of L = 0.5 H and C = 8μF = 8 x 10^-6 F, we get:
f = 1/(2π√((0.5 H)(8 x 10^-6 F)))
Solving this gives us:
f ≈ 5000 Hz
So, the angular frequency of the applied alternating voltage is 5000 Hz.
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