# Transient analysis in presence of faults

While reading Shenkman's Transient Analysis of Electric Power Circuits I came across the following problem:

... This circuit represents the equivalent of a d.c. supply network. At the instant of time $$\t=0\$$, the short-circuit fault occurs at node ‘‘$$\a\$$’’ and when the short-circuit current $$\i_{\text{sc}}\$$ through the breaker reaches the value $$\I=500 \text A\$$, the circuit breaker opens practically instantaneously. Find the transient response of current $$\i_2\$$ after the fault. The circuit parameters are $$\R_1 =1 \text{ Ohm}, R=R_2 =9 \text{ Ohm}, L_1 =0.01 \text{H}, L_2 = 0.45 \text{H}\$$ and $$\V_s =1100 \text{V}.\$$ First stage (the period between a short circuit $$\t=0\$$ and opening the circuit breaker, BR, $$\t=t\$$ Since the circuit is divided into two sub circuits: the left one with current $$\i_1\$$ and the right one with current $$\i_2\$$ , we shall obtain two time constants and two natural responses: ... $$\i_{1,n}=A_1e^{−100t}\$$, $$\i_{2,n}=A_2e^{−20t}\$$. The forced responses in these circuits are: $$\ (1) i_{1,f}= V_s/ R_1 =1100/ 1 =1100 \text{A}, (2) i_{2,f} =0.\$$ The initial conditions of the above two currents may be obtained by inspection of the given circuit prior to short-circuiting: $$\ (1) i_1(0−)= V_s/\big(R_1+ (R2||R3)\big) =200 \text{A}, (2) i_2(0−) = i_1 (0 −)/2=100 \text{A}. \$$

Now reading the above analysis for currents at $$\t=0-\$$ and $$\t=0+\$$ I could think of the following circuits, as $$\t \to \infty\$$, the circuit breaker opens at some point disconnecting the middle branch and the inductances are now in series, but the author suggests somehow the entire right branch is opened. I can't understand that.:

• <<< The forced responses in these circuits are: (1)i1,f=Vs/R1=1100/1=1100A, ... >>> ? Commented Feb 8 at 8:57