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schematic

simulate this circuit – Schematic created using CircuitLab

For t<0 the switch is opened.

Att=0 the switch is closed. The problem asks for the behaviour of i(t) after closing the switch. I tried to solve it with KVL but result is different from the book. The current i(t) for t<0 is 1.73A. So I converted the above circuit to the Laplace domain.

schematic

simulate this circuit

Then I applied KVL and I found a current and

enter image description here

Using the inverse Laplace the result is different from the one in the book wich is found using differential equations in the time domain. Can someone try to solve the circuit? Thanks

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  • \$\begingroup\$ What solution does "the book" provide? What result do you get, expressed as a function of time? \$\endgroup\$ – Elliot Alderson Aug 27 '18 at 0:37
  • \$\begingroup\$ How do you get the 17mV initial condition; it looks wrong? Probably better to do the analysis with the current source rather than convert to a voltage source. \$\endgroup\$ – Chu Aug 27 '18 at 8:49
  • \$\begingroup\$ @ElliotAlderson the result of the book is $$ I(s) = -63.6e^{-638.2t}+7.72e^{-861.8t}+17.27+19 $$ \$\endgroup\$ – Federico Zucchi Aug 27 '18 at 9:15
  • \$\begingroup\$ @ElliotAlderson mine is $$I(s) =4247.46e^{-638.2t}+4245.73e^{-861.8t}$$ \$\endgroup\$ – Federico Zucchi Aug 27 '18 at 9:25
  • \$\begingroup\$ @Chu I found it considering the inductor as a wire in stationary conditions and it is the same of my book. Which value do you think is correct? \$\endgroup\$ – Federico Zucchi Aug 27 '18 at 9:28
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The voltage, \$\small V(s)\$ at the top node, for each branch, is given by:

$$\small V(s)= \frac{19}{s}\small -I_1(s)$$

$$\small V(s)=I(s)\left(\frac{s}{100}\small+10\right)-0.017$$

$$\small V(s)=\left((I_1(s)-I(s)\right)\frac{500}{s}$$

where \$\small I_1(s)\$ is the source current.

Solving simutaneously: \$\small I(s)=\large\frac{1.727}{s}+\frac{7.72}{s+861.8}-\frac{7.72}{s+638.2}\$

Giving the time function: \$\small I(t)=1.727+7.72e^{-861.8t}-7.72e^{-638.2t}\$.

By inspection of the diagram, \$\small I(t) =1.727 A\$ is the correct current for \$\small t=0 \$ and \$\small t=\infty\$.

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  • \$\begingroup\$ Have you applied KVL after the convertion? \$\endgroup\$ – Federico Zucchi Aug 28 '18 at 0:18
  • \$\begingroup\$ I've written three expressions for the voltage at the top node, then solved the three simultaneous equations. Note: The inductor current must be 1.727A at t=0 and at t=infinity. Are you writing the initial condition as 0.017/s? It should be just 0.017. \$\endgroup\$ – Chu Aug 28 '18 at 8:11
  • \$\begingroup\$ Thank you very much! I have a last question. I imposed the initial condition with the voltage generator for the inductor but I can't figure out how to impose the condition for t=infinity. I mean how can the Laplace method takes in consideration the final condition, when I solve the same circuit with the differential equations I have to use the condition at t=infinity \$\endgroup\$ – Federico Zucchi Aug 28 '18 at 8:18
  • \$\begingroup\$ There is no need to impose a condition at infinity, I only used this to verify the equations - see my updated answer. \$\endgroup\$ – Chu Aug 28 '18 at 8:56
  • \$\begingroup\$ If the answer is useful, it's usual to upvote. \$\endgroup\$ – Chu Aug 28 '18 at 11:30

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