[SOLVED] What is the response of the given circuit

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iVenky

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There are two capacitors C1 (4 F) and C2 (16 F). They are separated by a switch. The switch closes at t=0.
When t<0 the voltage in C1 is 10V. What will happen when the switch is closed? What will be the voltage across C1 and C2 now (ie t>0) ?

Thanks in advance.
 

iVenky said:
There are two capacitors C1 (4 F) and C2 (16 F). They are separated by a switch. The switch closes at t=0.
When t<0 the voltage in C1 is 10V. What will happen when the switch is closed?
Click to expand...

160 J of energy will be dissipated into heat and/or electromagnetic (radio) energy.
 

Unfortunately, your schematic is different from your previous description.

The original question (shorting a switch beteen two capacitors) involves infinite currents and an insolvable differential equation when assuming ideal components. Doesn't sound like a well considered text book exercise or homework problem.
 

There's an irresistible logic to thinking in terms of 10V on a 4F cap as being a certain number of units of energy.

Then this energy is divided among 20F.

Intuitively it seems as though the answer should be 2V.

My simulator confirms this is the case.
 

If you consider two capacitors C1 and C2 in parallel, separated by a switch, where C1 is charged at Vi volts, and C2 is not charged then:

Charge in C1: q1 = C1*Vi Coulombs
Charge in C1: q2 = 0 Coulombs

When the switch is closed the totale charge (q1+q2) must be conserved because there is no way to make it flows outside the circuit since it is insulated, then (if Vo is the final voltage across the two capacitors):

C1*vi + 0 = C1*Vo + C2*Vo

then:

Vo = C1/(C1+C2)*Vi

If C1=4F, C2=16F and Vi=10V:

Vo = 4/(16+4)*10 = 2V

You can note that the initial energy of the system was:

Ei = 0.5*C1*Vi^2 = 0.5*4*100 = 200 joule

but the final energy is only:

Ef = 0.5*C1*Vo^2 + 0.5*C2*Vo^2 = 0.5*4*4 + 0.5*16*4 = 40 joule

To balance the 200 - 40 = 160 joule, a spike (dirac pulse) will be generated when the switch is closed.
 
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    iVenky

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You can actually do this problem... assume a finite switch resistance and solve the problem. You'll find that the answer does not depend on resistance. You can also do it using charge conservation: Total charge before = total charge after (like I see albbg did).
 
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Hello I understood about the charge conservation. But I couldn't understand that spike. What's that spike? (I know little about this switch thing.)

---------- Post added at 19:07 ---------- Previous post was at 19:04 ----------

How to do this in Pspice. I had attached the image in one of my previous posts and I think that there's something wrong in the schematic. I would be much obliged if you could say where the mistake is?
 

How to do this in Pspice. I had attached the image in one of my previous posts and I think that there's something wrong in the schematic. I would be much obliged if you could say where the mistake is?
Click to expand...
SPICE doesn't know negative times or +0 and -0. It recognizes both switches closed during initial solution, resulting in balanced charges. You should close the switch at a time t1>0, not t=0.
But I couldn't understand that spike. What's that spike?
Click to expand...
To turn the circuit in a real world one with finite currents, the switch must have a series resistance. The PSpice switch component already has, otherwise it would generate a simulation error.
 

FvM said:
SPICE doesn't know negative times or +0 and -0. It recognizes both switches closed during initial solution, resulting in balanced charges. You should close the switch at a time t1>0, not t=0.
Click to expand...

How to change the time? Please help me.

Thanks in advance.
 

FvM said:
Are you joking? How about tclose = 1u?

You need to perform a transient analysis, by the way.
Click to expand...

I know transient analysis. I don't know how to do that in pspice. I will try what you have said.
 

O.K., I have forgotten something. The PSpice switch has a leakage resistance, you can see it in the device parameters. You need a shorting switch across the first capacitor or a similar means, e.g an initial condition statement to keep the capacitor discharged.
 

You'll need to connect the shorting switch in parallel to C1.
 

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