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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.
Top 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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