Answer:
C) 20.23 μF
Step-by-step explanation:
R = resistance of the resistor = 15 ohm
L = inductance of the inductor = 11.5 mH = 0.0115 H
f = resonance frequency achieved = frequency of applied voltage = 330 Hz
C = Capacitance of the capacitor
For resonance to be possible
![2\pi fL = (1)/(2\pi fC)](https://img.qammunity.org/2020/formulas/physics/college/3ix2difl0u1foc9ifqamv75q0p7wporec1.png)
![2(3.14) (330) (0.0115) = (1)/(2 (3.14) (330) C)](https://img.qammunity.org/2020/formulas/physics/college/whfkrts0zon15gqsv8re9aen067xciyd87.png)
C = 20.23 x 10⁻⁶ F
C = 20.23 μF