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2 years ago

6

PLEASE HELP QUICK!!

1 answer:

ivolga24 [154]2 years ago

5 0

R01= 14.1 Ω

R02= 0.03525Ω

<h3>Calculations and Parameters</h3>

Given:

K= E2/E1 = 120/2400

= 0.5

R1= 0.1 Ω, X1= 0.22Ω

R2= 0.035Ω, X2= 0.012Ω

The equivalence resistance as referred to both primary and secondary,

R01= R1 + R2

= R1 + R2/K2

= 0.1 + (0.035/9(0.05)^2)

= 14.1 Ω

R02= R2 + R1

=R2 + K^2.R1

= 0.035 + (0.05)^2 * 0.1

= 0.03525Ω

Read more about resistance here:

brainly.com/question/17563681

#SPJ1

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2 years ago

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3 0

3 years ago

The current entering the positive terminal of a device is i(t)= 6e^-2t mA and the voltage across the device is v(t)= 10di/dtV.

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( a) Find the charge delivered to the device between t=0 and t=2 s.

( b) Calculate the power absorbed.

( c) Determine the energy absorbed in 3 s.

Solution:

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dQ = i(t) . dt

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$= 6 * (-0.5)*e^-2t = - 3*( 1 / e^4 - 1) = 2.945 C$

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v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt

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6 0

4 years ago

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Answer:

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4 0

3 years ago