Find heat produced on closing the switch S

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SUMMARY

The discussion centers on calculating the heat produced when closing switch S in a circuit involving capacitors and resistors. The initial charge on the 4 microfarad capacitor is determined to be 80 microcoulombs. After closing the switch, the work done by the battery is calculated as -35.56 microjoules, while the energy absorbed by the capacitors is -355.56 microjoules. The resulting heat produced is 320 microjoules. However, it is concluded that after the switch is closed, no current flows, leading to a final heat dissipation of 0 microjoules.

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Sanchayan Dutta
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Homework Statement


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Homework Equations

HEAT PRODUCED=WORK DONE BY BATTERIES - ENERGY ABSORBED BY CAPACITORS

The Attempt at a Solution


INITIALLY BEFORE SWITCH CLOSED :Initially a current flows only in the left most loop involving 4 microfarad capacitor.The charged on the capacitor during steady state is (20*4)=80 microcoulumbs.

FINALLY AFTER SWITCH IS CLOSED:Now a decaying current current flows in all the three loops -left,right, and bottom.However we are concerned only with steady state.So suppose the potential at the topmost junction (between the two capacitors and 2 ohm resistor) is x volts.Assuming that the node on the left (connected to 20 V cell,resistance R and resistance 4 ohm) is 0 volts.So we can write the nodal equation at x. (x-20)4+(x-0)5=0.Hence,x=80/9 volts.

Now work done by cell= (Charge Flown )* Cell Voltage = Final Charge On Left Plate Of Capacitor 4 microfarads - Initial Potential On Left Plate Of 4 microfarad= 4(20-80/9) - 4*20=-35.56 microjoule(negative).

Energy Absorbed By Capacitor = 1/2(4)(20-80/9)^2 + 1/2(5)(80/9)^2 - 1/2(4)(20)^2= -355.56 microjoule

HEAT PRODUCED=WORK DONE BY BATTERIES - ENERGY ABSORBED BY CAPACITORS=-35.56 + 355.56=320 microjoules

But this does not match with any of the options.Please help.
 
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What is the energy stored in the circuit before the switch is closed? What will be the energy in the circuit (in steady state) after the switch is closed?
 
Please read my solution which I posted.
 
Sanchayan Dutta said:
Initially a current flows only in the left most loop involving 4 microfarad capacitor.The charged on the capacitor during steady state is (20*4)=80 microcoulumbs.
Right.
Sanchayan Dutta said:
FINALLY AFTER SWITCH IS CLOSED:Now a decaying current current flows in all the three loops
Are you sure? Think in terms of the steady state.
 
Is the final charge on 4 microfarad capacitor same as initial because no current flows finally in steady state?Does the other capacitor remain uncharged?
 
Sanchayan Dutta said:
Is the final charge on 4 microfarad capacitor same as initial because no current flows finally in steady state?Does the other capacitor remain uncharged?
No current flows even in the transient state. You can verify this by applying KVL when the switch is closed.
 
I agree that current will flow as transient current initially but finally the capacitors will behave as open circuits and no current will flow in any of the loops.There will be p.d. only across the left capacitor.
 
Sanchayan Dutta said:
I agree that current will flow as transient current initially
After the switch is closed, there's no transient. And the transient before closing the switch should be ignored since the switch is open for a sufficiently long time.
 
Ok I get that.So should the heat dissipated be 0 after switch is closed?
 
  • #10
Sanchayan Dutta said:
Ok I get that.So should the heat dissipated be 0 after switch is closed?
Yes. The 4uF capacitor retains all the charge and no current flows through the circuit after the switch is closed.
 
  • #11
Alright.Thank You.You are from india?Which university? =) Thanks a lot for your help!
 
  • #12
Sanchayan Dutta said:
Thank You
You're welcome!:smile:
Sanchayan Dutta said:
Which university?
Pune..
 
  • #13
Nice to meet you.Bye :-)
 

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