2
\$\begingroup\$

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}.\$ fig2,18 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.:enter image description here

\$\endgroup\$
1
  • \$\begingroup\$ <<< The forced responses in these circuits are: (1)i1,f=Vs/R1=1100/1=1100A, ... >>> ? \$\endgroup\$
    – Antonio51
    Commented Feb 8 at 8:57

2 Answers 2

2
\$\begingroup\$

Time and currents are in log scale ...

Here are the results of the simulation with microcap v12.

V2 and V3 are "commands" for switching ON/OFF S1 and S2.
V3 short S2, than V2 opens S1 (when current ~ 500 A).

You can see that currents are at initial conditions 200 A and 100 A.
And in fine, current is 110 A. "High" transient at the breaker opening.

enter image description here

enter image description here

\$\endgroup\$
1
\$\begingroup\$

Transient analysis. I looked at the various situations. enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.