Solving Transient Analysis Homework Problem

[ranju]

Homework Statement

In the given circuit (voltage source is V) , we have to find initial and final values of current and also di(0+)/dt , di₂)0+)/dt , d²i1(0+)/dt² .
Now , by intial and values of currents , does they mean i(0-) and i(0+) or the discharging or charging current??
nd how to form the equations to solve it out??

Homework Equations

The Attempt at a Solution

beforing closing the switch , the current across the inductor , say i(0-) = (R1+R2)i2 = i(0+)
when we close the switch , capacitor will be shorted , and inductor will be open-circuited , so the KVL equation in the first loop will be V-i1R=0 by differentiating we get di1(0+)/dt = 0 similarly , d²i1(0+)/dt² =0
how we will solve for di₂(0+)/dt and d²i₂(0+)/dt² then..??

 

[NascentOxygen]

You don't differentiate your expression for V(0⁺) to get dV/dt (0⁺). There is no time information in the expression. You have to arrive at the derivatives by a different technique.

 

[ranju]

Which technique..?? Please elaborate...

 

[NascentOxygen]

Which technique..?? Please elaborate...

How has this been done in worked examples in your class notes?

 

[ranju]

If you are asking about class..please don't ask because there's nothing productive being taught..I am studying this topic from book..& in the book they are just using KVL.. & then differentiating..and using the corresponding conditions for inductor and capacitor.. but that is for simple circuits.. but I am not getting how to solve such circuits which I'hv attached..!

 

[NascentOxygen]

You can look at it like this...

Immediately after the switch is closed, with as yet no current through the inductor and its branch, all current from the source is taking the path through the capacitor and R1. With this current charging the capacitor, you can determine what that capacitor's dV/dt would be with such a level of current.

when we close the switch , capacitor will be shorted , and inductor will be open-circuited , so the KVL

Bear in mind that in another circuit, for example, if there was initially current through the inductor before the switch is closed, then that same level of inductor current will continue at t=0+ so that needs to fit in with your concept of the inductor and it being "open circuited".

 

[ranju]

Actually I worked on this problem and got the correct values... and worked on some more problems..and in of the problem..I just wanted to make it confirm whether my answer is correct or not..Can you please help out in getting me confirmed about it..so that I can move further..
In the circuit (voltage source V0 and current i is flowing across capcitor) ) , we have to find i(0+) , di(0+)/dt , d²i(0+)/dt².
I got i(0+) =V0/2 , di(0+)/dt = 2V₀ & d²i(0+)/dt² = -3V₀/2

 

[NascentOxygen]

Explain your thinking that led you to decide I(0+) will be Vo/2 .

Connection to the source doesn't occur until t=0. Is that what we are told?

 

[NascentOxygen]

Actually I worked on this problem and got the correct values ...

What were your answers for the first circuit?

 

[ranju]

for the first circuit I got di1 (0+)/dt = V/L-V/R₁²C and di2(0+)/dt=V/L..
And in the second circuit...actually I did a mistake the current i(0+) is coming V0/L & I used the conditions at t0+ , inductor open & capacitor short-circuit,, so the current in 1st loop will be zero and in 2nd loop is also zero..
then I wrote the KVL equations in the 2 loops & since these equ. hold in general so they'll hold for t=0+ also.. so at t=0+ I put the values and found it out..
Should I write the equ. also??

 

[NascentOxygen]

for the first circuit I got di1 (0+)/dt = V/L-V/R₁²C and di2(0+)/dt=V/L.✔ [/size][/color]

You need a pair of these ( )[/size][/color] around a denominator in that expression.

And in the second circuit...actually I did a mistake the current i(0+) is coming V0/L & I used the conditions at t0+ , inductor open & capacitor short-circuit,, so the current in 1st loop will be zero and in 2nd loop is also zero..

That's di/dt(0+) = V0/L

 

[ranju]

You need a pair of these ( ) around a denominator in that expression.That's di/dt(0+) = V0/L

I am really sorry , I am doing blunders in typing , Yes it is di(0+)/dt = V0/L

 

[ranju]

so Is that correct..??

 

[NascentOxygen]

In the circuit (voltage source V0 and current i is flowing across capcitor) ) , we have to find i(0+) , di(0+)/dt , d²i(0+)/dt².
I got i(0+) =V0/2 , di(0+)/dt = 2V₀ & d²i(0+)/dt² = -3V₀/2

Even if there was no inductor, the capacitor current at t=0+ could not be as high as V0/2

 

[ranju]

I guess You did'nt see the earlier posts.. I said that I did mistake in finding the currents in the 2 loops.. I am getting i(0+) = 0 & di(0+)/dt = V0/L & not V0/2...

 

[NascentOxygen]

I guess You did'nt see the earlier posts.. I said that I did mistake in finding the currents in the 2 loops.. I am getting i(0+) = 0 & di(0+)/dt = V0/L & not V0/2...

V0/L is di/dt for inductor current. This then halves as it divides equally between the two subsequent branches. I'd like to see how you are working these out.

You are not attempting the second derivatives, d²i / dt² ?

 

[ranju]

previously I'hv done many mistakes in posting my solutions.. and evverything got messed up ..Now I explain the whole thing...
first of all at t=0+ , the inductor is open ckt & capacitor is shorted..so the current in 1st loop will be zero and in 2nd loop is also zero..
then I wrote the KVL equations in the 2 loops & since these equ. hold in general so they'll hold for t=0+ also.. so at t=0+
For 1st loop KVL equ. : V0-2i₁R+iR -Ldi₁/dt = 0 (I'hv assumed current in 1st loop as i1 )
for t=0+ , V0-2i₁(0+)R +i(0+)R=Ldi₁(0+)/dt ______(1)
di₁(0+)/dt = V0 (R=1ohm & L=1henry )
Again for 2nd loop KVL equ. : -2Rdi/dt -i/C +Rdi₁/dt=0
for t=0+ , -i(0+)/C+di₁(0+)/dt =2di(0+)/dt _____(2)
di(0+)/dt = V0/2
for d²i(0+)/dt² , I differentiated equ.2 & 1 once again...
got d²i₁(0+)/dt² = -3V0/2 & d²i(0+) /dt² = -V0
Now You check what is wrong..??

 

[NascentOxygen]

got d2i1(0+)/dt2 = -3V0/2 & d2i(0+) /dt2 = -V0

Those are the values I calculate, too. http://imageshack.com/a/img29/6853/xn4n.gif

 

[ranju]

so now can we conclude that it is right..?

 

[NascentOxygen]

so now can we conclude that it is right..?

Within a 98% confidence interval.