# Mathcad - 01-03 and 04 - 2 ⋅ = W e 1 2 C ⋅ V pk 2 ⋅ =...

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The inductor stores 50 MJ! Much higher energy densities can be achieved with magnetic fields than with electric fields W m 5 10 7 × J = W m 1 2 1.0 H 10000 A ( ) 2 = W m 1 2 L sc I 2 = The energy stored in the magnetic field is given by: Solution: L sc 1.0 H = I 10000 A = Given: Problem 1-4 The capacitor stores less than 3 kJ W e 2.642 10 3 × J = W e 1 2 41.6 μ F 11.27 kV
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Unformatted text preview: ( ) 2 ⋅ = W e 1 2 C ⋅ V pk 2 ⋅ = The energy stored in the electric field is given by: V pk 11.27kV = V pk 2 V rms ⋅ = The voltage is the RMS value, so the peak voltage is the RMS times the square root of 2. Solution: C 41.6 μ F ⋅ = V rms 7.97 kV ⋅ = Given: Problem 1-3...
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## This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue.

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