please see attached image Question 1:The circuit below shows a reactor being disconnected from its supply.During the process of opening, the circuit breaker reignites but interrupts the reignition current at thefirst zero of the high frequency current, trapping energy in L2Compute the peak voltage subsequently appearing across the reactor as a conseguence of this trappingof energy.L1 = 15mH-L2 = 352mH-C = 3.2nF-Vs = 138/1.732 kVсвLIVs

Answered: The circuit below shows a reactor being…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Step 1

Calculate the surge impedance:

$\begin{array}{rcl} Z_{0} & = & {[\frac{L_{1} L_{2}}{(L_{1} + L_{2}) C}]}^{\frac{1}{2}} \\ = & {[\frac{(1.5 \times 10^{- 3}) (352 \times 10^{- 3})}{(1.5 \times 10^{- 3} + 352 \times 10^{- 3}) 3.2 \times 10^{- 9}}]}^{\frac{1}{2}} \\ = & {[\frac{5.25 \times 10^{6}}{1.1744}]}^{\frac{1}{2}} \\ = & 2120.36 Ω \end{array}$

The energy in inductor is equal to energy stored in capacitor:

$\begin{array}{rcl} \frac{1}{2} L_{2} {I_{0}}^{2} & = & \frac{1}{2} C V^{2} \\ L_{2} {I_{0}}^{2} & = & C V^{2} \\ V^{2} & = & \frac{L_{2} {I_{0}}^{2}}{C} \\ V & = & I_{0} \sqrt{\frac{L_{2}}{C}} \end{array}$

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