DC circuit analysis: conceptual mistake?

In summary, the conversation discusses a circuit with three components in parallel - a current source Ia, a resistor R, and a capacitor C. The correct equation for this circuit is given by option (a), while option (b) is incorrect. The validity of option (c) depends on whether Ia remains constant or not, and since this information is not given, it cannot be concluded as correct. The conversation also touches upon the application of KCL and the concept of steady state in analyzing the circuit.
  • #1
Granger
168
7

Homework Statement


Homework Equations

[/B]
I have a circuit that consists on 3 components in parallel: a current source Ia, a resistor R and a capacitor C. Ia is directed upwards, while the currents through the capacitor and the resistor are both directed downwards.
Select the correct option in the following three:
(a) The equation $$\frac{i_a(t)}{C}=\frac{v_2(t)}{RC} + \frac{dv_2(t)}{dt}$$ is valid to this circuit.
(b) The equation $$\frac{Ri_a(t)}{C}= \frac{dv_2(t)}{dt}$$ is valid to this circuit
(c) If the circuit is working for a long time, after dome time we will have $$i_a(t)=i_1(t)$$ and $$i_2(t)=0$$ whatever is the value of $$i_a(t)$$.

The Attempt at a Solution


[/B]
So I had no problem concluding that (a) is correct and (b) is not correct.

First, I applied KCL to the upper node

$$i_a=i1+i2$$

We have that:
$$i1=\frac{u1}{R}$$
$$i2=C\frac{du2}{dt}$$

Substituting:

$$i_a=\frac{u1}{R}+C\frac{du2}{dt}$$

Because the components are in parallel: u1=u2. Dividing everything by C we get to the expected expression.

My doubt in the c statement. I don't understand why is it wrong. I thought well if the circuit is working for a long time then we can admit that it starts working in steady state (or is this an incorrect statement only valid if $$i_a(t)$$ remains constant?). Because of this, the voltage across the capacitor will remain constant and because of the expression $$i2=C\frac{du2}{dt}$$ we will have i2=0 (the derivative of a constant is zero).
Because of that we can exchange the capacitor to an open circuit. Therefore all the current Ia(t) will go through the resistor. Or does this depend on Ia(t)?
I'm probably making a conceptual mistake somewhere in here.

Thanks!
 
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  • #2
Granger said:
or is this an incorrect statement only valid if
ia(t)ia(t)​
i_a(t) remains constant?
Yes. Since they didn't mention whether it's a constant current source or not in the question, you can't go for option c. It would be correct only if Ia is constant.
 

FAQ: DC circuit analysis: conceptual mistake?

1. What is a DC circuit analysis?

A DC circuit analysis is a method used to analyze and understand the behavior of direct current (DC) circuits. It involves using mathematical equations and principles to determine the voltage, current, and power in a circuit.

2. What is a conceptual mistake in DC circuit analysis?

A conceptual mistake in DC circuit analysis refers to an error in understanding or applying the fundamental concepts and principles of circuit analysis. This can lead to incorrect calculations and interpretations of circuit behavior.

3. How can a conceptual mistake affect the accuracy of DC circuit analysis?

A conceptual mistake can significantly impact the accuracy of DC circuit analysis as it can result in incorrect calculations and predictions of circuit behavior. This can lead to faulty designs and potential safety hazards in real-world applications.

4. What are some common examples of conceptual mistakes in DC circuit analysis?

Some common examples of conceptual mistakes in DC circuit analysis include incorrect application of Kirchhoff's laws, misinterpretation of voltage and current relationships, and neglecting to account for the effects of resistance and capacitance in a circuit.

5. How can I avoid making conceptual mistakes in DC circuit analysis?

To avoid making conceptual mistakes in DC circuit analysis, it is essential to have a strong understanding of the fundamental principles and equations used in circuit analysis. Practicing problem-solving and double-checking calculations can also help ensure accuracy in analysis.

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