Analyzing DC Steady State in an RLC Circuit with a Closed and Opened Switch

In summary, the conversation discusses finding the values of vC(t) and iL(t) immediately after a switch was opened in a circuit. The solution involves finding the current in the circuit and using it to calculate the voltage across the capacitor. The confusion arises when considering the effect of a 2A current source on the current in the inductor, and it is suggested to use Thevenin equivalents and consider the circuit via superposition to resolve the issue.
  • #1
galaxy_twirl
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1

Homework Statement



In the Figure Q1 (c), the switch S was closed for very long time before it is opened at time t = 0. Find the value of vC(t) and iL(t) at t = 0+ c L i.e. immediately after the switch was opened.

2ewzh1i.jpg

Homework Equations



V = IR

The Attempt at a Solution



From DC steady state, we know that C acts as an open circuit while L acts as a short wire, hence, we will have:

Current in circuit = 20 / (10+10) = 1A
vC(t) is thus 20 - (10*1) = 10V.

However, for iL(t), I am a bit confused, because since this is a short wire, will the 2A current affect the current in this wire? This is because my answer is 1A, however, when I simulate this circuit in LTSpice, it gave me an answer of 3A, which is a bit puzzling. The direction of current would be different, and thus, even if the 2A current source affects the current in the inductor (which is a short), it should be 1-2 = -1A right?

Thanks for any assistance rendered! :D
 
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  • #2
Try replacing the 2A source and 50Omh resistor with their Thevinin equivalent - see if that clears it up for you.
 
  • #3
Simon Bridge said:
Try replacing the 2A source and 50Omh resistor with their Thevinin equivalent - see if that clears it up for you.

otngo7.jpg


I think I know why now. I must add 2A to the 1A already present from the 20V source. Thanks! So that means, unless the right of the inductor are purely resistors, I cannot just ignore the right branches even though the inductor is shorting, right?
 
  • #4
In general you can't ignore any sources that might contribute to the voltage or current you're interested in. When in doubt, try looking at the circuit via superposition (consider the contributions of each source individually).
 
  • #5
gneill said:
In general you can't ignore any sources that might contribute to the voltage or current you're interested in. When in doubt, try looking at the circuit via superposition (consider the contributions of each source individually).

I see. I will keep that in mind. Thank you! :D
 

1. What is a DC steady state RLC circuit?

A DC steady state RLC circuit is a type of electrical circuit that contains a resistor (R), inductor (L), and capacitor (C) connected in series or parallel. It is powered by a direct current (DC) source and reaches a steady state where the voltage and current remain constant over time.

2. How does a DC steady state RLC circuit differ from an AC circuit?

Unlike an AC circuit, which constantly alternates between positive and negative voltage, a DC steady state RLC circuit has a constant voltage source. This means that the voltage and current in the circuit will not change over time, resulting in a steady state.

3. What are the applications of DC steady state RLC circuits?

DC steady state RLC circuits have various applications, including in power supplies, filters, and oscillators. They are also commonly used in electronic devices such as radios, televisions, and computers.

4. How is the behavior of a DC steady state RLC circuit affected by changes in the values of R, L, and C?

Changing the values of R, L, and C in a DC steady state RLC circuit can affect its behavior in different ways. Increasing the resistance will result in a decrease in current, while increasing the inductance or capacitance will increase the time it takes for the circuit to reach a steady state.

5. What is the formula for calculating the impedance of a DC steady state RLC circuit?

The impedance of a DC steady state RLC circuit can be calculated using the formula Z = sqrt(R^2 + (Xl - Xc)^2), where R is the resistance, Xl is the inductive reactance, and Xc is the capacitive reactance. This formula takes into account the effects of both the inductor and capacitor in the circuit.

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