RLC circuit and rate of increase of current

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In an RLC circuit with a DC battery, when the switch is closed at t=0, both the capacitor voltage and inductor current are initially zero. At t=0+, the inductor experiences a voltage equal to the source voltage, leading to a positive rate of change of current. However, the misconception arises in assuming the capacitor's voltage increase is also zero at this moment; the current through the capacitor can change instantaneously, allowing its voltage to rise. Therefore, while the inductor current increases, the capacitor voltage also begins to increase, contradicting the initial assumption. Understanding this distinction clarifies the behavior of current and voltage in the circuit at the moment the switch is closed.
hell-hawk
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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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I will try to explain as best that I can


hell-hawk said:
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)..

Up till here, you are correct!

hell-hawk said:
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.

There is a slight misconception here. You are saying that the current i at time t=0 is zero, using the current continuity of the inductor i(0+)=0, i.e. physically, the current cannot have a discontinuity jump at time t=0. But there is no current continuity at the capacitor, i.e. at time t=0, i(0-)=0 AND
i(0+)=some value. So the rate of increase of capacitor voltage is NOT zero.
Hence the capacitor voltage is increasing at a positive rate together with the inductor current at time t=0+

I may be wrong, because it has been one year ago since I touch on this concept. If I am wrong, can someone please advice. Thanks.
 
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