An ideal (lossless) transformer has 4 times as many turns

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SUMMARY

An ideal (lossless) transformer with a primary winding having four times as many turns as the secondary winding presents specific resistance and current values when connected to a sinusoidal voltage source. Given a peak voltage of 10V and a 10k ohm resistor on the secondary, the input resistance can be calculated using the formula rin = (N1/N2)^2 x RL, leading to a resistance of 160k ohms. The peak current flowing in the secondary winding can be determined by calculating the secondary voltage and using Ohm's law, resulting in a peak current of approximately 1mA.

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Homework Statement


An ideal (lossless) transformer has 4 times as many turns on the primary winding as it has on the secondary winding. The secordary winding is loaded with a 10k ohm resistor and a sinusoidal voltage source having negligible resistance and a peak voltage of 10V is connected to the primary .

1. the value nearest to the resistance presented to the voltage source??
2. the value nearest to the peak current flowing in the secordary winding??


Homework Equations



Input resistance rin = e1/i1 = e2N1/N2 x n1/i2N2 = e2/i2{N1/N2}^2

rin = (N1/N2)^2 x RL

The Attempt at a Solution



I can't find out the solution 1. & 2., please advise!
 
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Hi bckcookie! If the primary voltage is 10Vpk, what is the secondary voltage?

Determine the secondary current with that load of 10kΩ on the secondary.

Determine the primary current.

What is the result of dividing primary voltage by primary current?
 

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