If you adjusted the frequency to be lower than the resonant frequency, would the current amplitude have been less than, equal to, or greater than the current amplitude you found at resonance? explain.

Current amplitude will be less than the current amplitude at resonance.

Explanation:

Resonance in series RLC circuit, is a condition where the capacitive and inductive reactances are equal, thus making the circuit purely resistive due to which the overall impedance of the circuit at resonace is minimum resulting in the maximum amplitude of current as there is inverse relationship between impedance and current.

When the frequency is less than resonance frequency, the circuit will be capacitive in nature (since, capacitive reactance is inversely proportional to the frequency wheras inductive reactance is directly proportional to it) and the overall impedance of the circuit is more than that at resonance hence the current in the circuit will also be less than that at resonance.

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.

moisture is lackin, and organic acids are scarce.