The electric oscillatory circuit of the radio receiver contains a 10mH inductor and two parallel charged

The electric oscillatory circuit of the radio receiver contains a 10mH inductor and two parallel charged capacitors 360 pF and 40 pF. what wavelength is the circuit tuned to?

L = 10 mH = 10 * 10 ^ -3 H.
C1 = 360 pF = 360 * 10 ^ -12 F.
C2 = 40 pF = 40 * 10 ^ -12 F.
s = 3 * 10 ^ 8 m / s.
λ -?
The wavelength must match the wavelength of the receiver circuit.
The wavelength of natural oscillations λ is determined by the formula: λ = c * T, where c is the speed of light in vacuum, T is the period of oscillations.
Let us find the period T of oscillations of the oscillating circuit of the receiver by the formula: T = 2 * P √ (L * C), where L is the inductance of the coil, C is the total capacitance of the capacitors.
For parallel connection of capacitors, the total capacitance C is determined by the formula: C = C1 + C2.
T = 2 * 3.14 √ (10 * 10 ^ -3 Hn * 400 * 10 ^ -12 F) = 125.6 * 10 ^ -7 s.
λ = 3 * 10 ^ 8 m / s * 125.6 * 10 ^ -7 s = 3768 m.
Answer: the receiver is tuned to a wavelength λ = 3768 m.