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 the first zero of the high frequency current, trapping energy in L2 Compute the peak voltage subsequently appearing across the reactor as a conseguence of this trapping of energy. L1 = 15mH L2 = 352mH 2nF Vs = 138/1.732 kV св L, Vs

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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 the
first zero of the high frequency current, trapping energy in L2
Compute the peak voltage subsequently appearing across the reactor as a conseguence of this trapping
of energy.
L1 = 15mH
-
L2 = 352mH
-
C = 3.2nF
-
Vs = 138/1.732 kV
св
LI
Vs
Transcribed Image Text: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 the first zero of the high frequency current, trapping energy in L2 Compute the peak voltage subsequently appearing across the reactor as a conseguence of this trapping of energy. L1 = 15mH - L2 = 352mH - C = 3.2nF - Vs = 138/1.732 kV св LI Vs
Expert Solution
Step 1

Calculate the surge impedance:

Z0=L1L2L1+L2C12=1.5×10-3352×10-31.5×10-3+352×10-33.2×10-912=5.25×1061.174412=2120.36 Ω

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

12L2I02=12CV2L2I02=CV2V2=L2I02CV=I0L2C

 

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