Answer:
The amount of energy they carry is related to their frequency and their amplitude. The higher the frequency, the more energy, and the higher the amplitude, the more energy.
To solve this problem we need to use the induced voltage ratio law with respect to the number of turns in a solenoid. So
![\frac{\epsilon_2}{\epsilon_1} = -\frac{N_2}{N_1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cepsilon_2%7D%7B%5Cepsilon_1%7D%20%3D%20-%5Cfrac%7BN_2%7D%7BN_1%7D)
For the given values we have to
![N_1 = 400](https://tex.z-dn.net/?f=N_1%20%3D%20400)
![N_2 = 100](https://tex.z-dn.net/?f=N_2%20%3D%20100)
![\epsilon_2 = 120V](https://tex.z-dn.net/?f=%5Cepsilon_2%20%3D%20120V)
Replacing we have that,
![\frac{\epsilon_2}{120} = -\frac{100}{400}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cepsilon_2%7D%7B120%7D%20%3D%20-%5Cfrac%7B100%7D%7B400%7D)
![\epsilon = 30V](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%2030V)
Therefore the RMS value for secondary is 30V.
The current can be calculated at the same way, but here are inversely proportional then,
![\frac{I_2}{I_1} = -\frac{N_1}{N_2}](https://tex.z-dn.net/?f=%5Cfrac%7BI_2%7D%7BI_1%7D%20%3D%20-%5Cfrac%7BN_1%7D%7BN_2%7D)
Replacing we have
![\frac{I_2}{10mA} = -\frac{400}{100}](https://tex.z-dn.net/?f=%5Cfrac%7BI_2%7D%7B10mA%7D%20%3D%20-%5Cfrac%7B400%7D%7B100%7D)
![I_2 = 40mA](https://tex.z-dn.net/?f=I_2%20%3D%2040mA)
Therefore the rms value of current for secondary is 40mA
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