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RLC Circuit - RF Cafe Forums

The original RF Cafe Forums were shut down in late 2012 due to maintenance issues. Original posts:

Amateur Radio | Antennas | Circuits & Components | Systems | Test & Measurement


adamzaman
Post subject: RLC circuit
Unread postPosted: Wed Mar 16, 2005 8:30 am

Ok, I am having a problem with a RLC circuit and my teacher was unable to offer a satisfactory explanation. Hope that you'll help.
Consider a DC battery, a switch, a resistor, a capacitor and an inductor in series. Before t=0 the switch is open and assume that the circuit has achieved steady state. This means that the capacitor voltage is 0 and so is the inductor current. At t=0 the switch is closed. It follows that at t=0+ (i.e. just after t=0) the capacitor voltage will be 0 and so will the inductor current. It also follows that the voltage across the inductor at t=0+ will be V, i.e. the voltage of the source (the capacitor is acting as short-circuit and the inductor as open circuit). Therefore the rate of change of current in the circuit at t=0+ will be some positive value (from v=L di/dt). But the rate at which the capacitor voltage is increasing is ZERO at t=0+ (from i=C dv/dt; i=0). How could the rate of increase of current in the inductor be non-zero, but the rate of increase of voltage in the capacitor be zero? If one is increasing, shouldn't the other as well?
Please explain in physical terms and not mathematical. I'll be highly grateful.


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Guest
Post subject:
Unread postPosted: Wed Mar 16, 2005 5:28 pm

"Therefore the rate of change of current in the circuit at t=0+ will be some positive value (from v=L di/dt). But the rate at which the capacitor voltage is increasing is ZERO at t=0+ (from i=C dv/dt; i=0)."

How do you get this statement? There is current charging the capacitor the moment the switch is flipped. It is limited by R but it is not 0.







Posted  11/12/2012

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