Question

What is the steady-state voltage across the 1mic.F capacitor ?(see attachment)

Answer 1

what are the answer choices?
Seems like this is a second year electrical eng problem just before "delta - y" transforms or just after.

Answer 2

http://en.wikipedia.org/wiki/Y-%CE%94_transform

Answer 3

Steady state huh.... Then there will be no voltage drop across the inductor (assuming no resistance of coil wire), the capacitor will be fully charged assuming no leakage and this is DC. My guess is 10 volts with the polarity as shown.

Answer 4

@radar: Steady state needn't necessarily mean constant or zero voltage across components. It just means the behavior of the circuit, a very long time after the input to it has undergone a change. Many components - especially groups of capacitors and inductors go into sustained oscillations (assuming no energy loss) in response to transient inputs. That is, they continue to oscillate indefinitely and these oscillations are their steady state response.

Answer 5

Those were the assumptions I made that the inductive opposition from a changing current no longer applied, the capacitor had charged to it final value etc. Do you still see some dynamic action still occurring?

Answer 6

Or are you assuming the 1 micro farad - 1 H inductor is in some kind of oscillator arrangement?

Answer 7

Remember how some steady state analysis using Laplace domains give sinusoids as solution? Sinusoids can be steady state response. However, there must be no loss of energy in the system. That can be ensured only with analysis.

Answer 8

Looking at the circuit strictly from a "practical" point of view, considering the circuit is powered from a steady DC power of 30 volts, I doubt very seriously that this circuit continues to "ring" at 160 kHz especially with the resistance shown, there is no obvious no feedback, no negative resistance devices, in fact no active devices at all. If you think detailed analysis is required, please assist alma1969 in that effort. That is above my "pay grade" I am only basing my solution on 38 1/2 years working in the electronics field.

Answer 9

Agreed! But, theoretical questions are hardly practical. I was merely implying that a Laplace analysis need to be done before commenting about steady state. I tried that, but found myself too tired to solve the simplified s-domain equations.

Answer 10

On a side note, I was thinking about a strange condition where the resistances in a circuit may not face any voltage (ie, voltage on either side of resistance balance in steady state). I have never seen such a circuit. But that can't rule out that condition - especially with a circuit that resembles a bridge. If that happens, the circuit will have no energy loss.

Answer 11

it does appear to be a bridge circuit, but it does not appear to be a "balanced" bridge.
@alma, sorry I can't do any more on it.

Answer 12

I redraw circuit and it is a possible solution,see attachment.
Assuming no resistance from inductors.Power is from 30 V battery.The only opposition to current flow in the circuit are the resistors and capacitive reactance.R1 and R2 acting as voltage dividers and as voltage supply to the loops.