Plugging in the values of L and C in our example circuit we arrive at a resonant frequency of 159155 Hz. F 0 o 0 2 p 1 2 p L C. X C 12pifC.
Then hit the square button.
First calculate the. F 0 1 2 p 1 L C 1 2 p 1 300 10 3 H 800 10 4 F 103 10 2 H z. We now know that at the resonant frequency fr the admittance of the circuit is at its minimum and is equal to the conductance G given by 1R because in a parallel resonance circuit the imaginary part of admittance ie. Let us consider a series connection of R L and C.