Yesterday I tried to solve a question in a second order circuit , but I couldn't ! because I have problem understanding the concept of 0+ , 0- ! So let me try it in a first order circuit , it supposed to be much easier :
Is there any thing wrong !?
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2 Answers
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Understand the behavior of Capacitor. see this
- In your circuit, Initially Switch was closed for the long time, nothing but capacitor will act as open circuit. Now reduce the circuit using series and parallel combination concept. You can easily find the Voltage across the circuit. This will give you Vc(0-).
- Since Capacitor will not allow sudden changes in voltage. So Vc(0−)=Vc(0+).
Not sure about remaining questions but Generally you have to find the Time constant(RC) to find the voltage across the capacitor and substitute in this formula,
\$V(t)=V(1-e^{(-t/RC)})\$
Change the values of 't' according to your requirement.
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\$\begingroup\$ Thanks a lot .. I really appreciate your help ❤️ But i have a question about the formula you gave me !! In my book .. We use another formula witch is V(t)=V.e^(-t/RC) .. Is there any different or both are the same ? \$\endgroup\$– HashimCommented Aug 20, 2016 at 12:00
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\$\begingroup\$ @Hashim Welcome to EE.SE. The formula which you are talking about is for the Resistor, and which I gave is for Capacitor, see here. \$\endgroup\$– user105009Commented Aug 20, 2016 at 13:57
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\$\begingroup\$ @Hashim No problem. And I'm glad I was helpful. If you have satisfied with my answer you can click at Tick mark at my answer :) \$\endgroup\$– user105009Commented Aug 20, 2016 at 17:41
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\$\begingroup\$ @Hashim Have you solved your problem? Did you get the correct answer? \$\endgroup\$– user105009Commented Aug 20, 2016 at 17:45
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\$\begingroup\$ yeah i did .. Thank you ❤️ but what is (Tick mark ) 👀 \$\endgroup\$– HashimCommented Aug 20, 2016 at 18:02
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The usual use is that \$ t=0^- \$ represents the instant before the event. In this case the switch is closed and has been for some considerable time while \$t=0^+ \$ is the instant after the event i.e the switch has just opened.
if we consider \$ t=0^-\$ then we can ignore the capacitor as it is charged to some voltage and therefore has zero current flowing through it.
We therefore have 2A flowing into 10R is parallel with 15R which is 6R so we have 12V across the current source. Now considering \$R_1\$ and \$R_2\$ as a potential divider this gives us 8V across \$R_1 \$ and hence \$C_1\$.
Now you can't change the voltage across a capacitor in zero time so
\$V_c = 8\text{V}\$ when \$t=0^-\$ and \$t=0^+\$
Once the switch is open the capacitor starts to discharge via \$R_1\$ giving
\$V_c = 8 \cdot e^{-\frac{t}{C_1 \cdot R_1}}\text{V}\$ for \$ t \ge 0.\$
