Answer:
The reflected resistance in the primary winding is 6250 Ω
Explanation:
Given;
number of turns in the primary winding,
= 50 turns
number of turns in the secondary winding,
= 10 turns
the secondary load resistance,
= 250 Ω
Determine the turns ratio;
![K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7BN_P%7D%7BN_S%7D%20%5C%5C%5C%5CK%20%3D%20%5Cfrac%7B50%7D%7B10%7D%20%5C%5C%5C%5CK%20%3D%205)
Now, determine the reflected resistance in the primary winding;
![\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms](https://tex.z-dn.net/?f=%5Cfrac%7BR_P%7D%7BR_S%7D%20%3D%20K%5E2%5C%5C%5C%5CR_P%20%3D%20R_SK%5E2%5C%5C%5C%5CR_P%20%3D%20250%285%29%5E2%5C%5C%5C%5CR_P%20%3D%206250%20%5C%20Ohms)
Therefore, the reflected resistance in the primary winding is 6250 Ω
Less than or equal to the magnitude of the vector
P.E = mgh
This is the formula for potential energy.
This is where m is mass, g is the acceleration due to gravity, and h is height.
All you have to do is multiply all these numbers together.
Answer:
The wavelength = 0.3333 meters at 900 MHz, therefore, = /4 = 0.08333 meters.
Work package. Hope this helps!