Find v(t) across the resistor in first order circuit

Question

I found this problem and it confused me a bit and I was hoping someone can clear me on this.

Basically as I said we have to find the voltage across the resistor when the switch turns.

This is a picture of the exercise (sorry for using paint)

The thing that confuses me is that the voltage source is initially separated from the rest of the circuit and I know that we have a dependent voltage source but since that depends on the ix current and we dont have a "full" capacitor that can serve as a voltage source can we assume that the voltage across the capacitor and the resistance in series with the capacitor initially is 0?

So all in all what I'm asking is :what is the voltage v(t) across the resistor R1 in series with the capacitor for t<0

Answer 1

Let us initially address your inquiry: The initial currents and voltages across all components are equal to zero for \$t<0\$, as \$2\text{I}_x\$ is contingent upon \$\text{I}_x\$, and it becomes evident that \$\text{I}_x=0\$ for \$t<0\$.

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Well, we are trying to analyze the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \begin{alignat*}{1} \text{I}_\text{i}&=\text{I}_1+\text{I}_2\\ \\ \text{I}_3&=\text{I}_0+\text{I}_2\\ \\ \text{I}_1&=\text{I}_\text{i}+\text{I}_4\\ \\ \text{I}_0&=\text{I}_3+\text{I}_4 \end{alignat*} \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \begin{alignat*}{1} \text{I}_1&=\frac{\displaystyle\text{V}_\text{i}-0}{\displaystyle\text{R}_1}\\ \\ \text{I}_2&=\frac{\displaystyle\text{V}_\text{i}-\text{n}\cdot\text{I}_2}{\displaystyle\text{R}_2}\\ \\ \text{I}_3&=\frac{\displaystyle\text{n}\cdot\text{I}_2-\text{V}}{\displaystyle\text{R}_3}\\ \\ \text{I}_3&=\frac{\displaystyle\text{V}-0}{\displaystyle\text{R}_4} \end{alignat*} \end{cases}\tag3 $$

Substitute \$(2)\$ into \$(1)\$, in order to get:

$$ \begin{cases} \begin{alignat*}{1} \text{I}_\text{i}&=\frac{\displaystyle\text{V}_\text{i}-0}{\displaystyle\text{R}_1}+\frac{\displaystyle\text{V}_\text{i}-\text{n}\cdot\text{I}_2}{\displaystyle\text{R}_2}\\ \\ \frac{\displaystyle\text{n}\cdot\frac{\displaystyle\text{V}_\text{i}-\text{n}\cdot\text{I}_2}{\displaystyle\text{R}_2}-\text{V}}{\displaystyle\text{R}_3}&=\text{I}_0+\frac{\displaystyle\text{V}_\text{i}-\text{n}\cdot\text{I}_2}{\displaystyle\text{R}_2}\\ \\ \frac{\displaystyle\text{V}_\text{i}-0}{\displaystyle\text{R}_1}&=\text{I}_\text{i}+\text{I}_4\\ \\ \text{I}_0&=\frac{\displaystyle\text{n}\cdot\frac{\displaystyle\text{V}_\text{i}-\text{n}\cdot\text{I}_2}{\displaystyle\text{R}_2}-\text{V}}{\displaystyle\text{R}_3}+\text{I}_4\\ \\ \frac{\displaystyle\text{V}-0}{\displaystyle\text{R}_4}&=\text{I}_0+\frac{\displaystyle\text{V}_\text{i}-\text{n}\cdot\text{I}_2}{\displaystyle\text{R}_2}\\ \\ \text{I}_0&=\frac{\displaystyle\text{V}-0}{\displaystyle\text{R}_4}+\text{I}_4 \end{alignat*} \end{cases}\tag3 $$

Now, let's solve for \$\text{V}\$:

$$\text{V}=\frac{\displaystyle\text{V}_\text{i}\text{n}\text{R}_4}{\displaystyle\left(\text{n}+\text{R}_2\right)\left(\text{R}_3+\text{R}_4\right)}\tag4$$

Now, applying this to your circuit we need to use (from now on I use the lower case letters for the function in the 'complex' s-domain where I used Laplace transform):

• $$\text{R}_3=\frac{\displaystyle1}{\displaystyle\text{sC}}\tag5$$
• The input voltage is a stable DC voltage equal to \$\hat{\text{u}}\$, so: $$\text{v}_\text{i}\left(\text{s}\right)=\frac{\displaystyle\hat{\text{u}}}{\displaystyle\text{s}}\tag6$$

So, we get:

$$\text{v}\left(\text{s}\right)=\frac{\displaystyle\frac{\displaystyle\hat{\text{u}}}{\displaystyle\text{s}}\cdot\text{n}\text{R}_4}{\displaystyle\left(\text{n}+\text{R}_2\right)\left(\frac{\displaystyle1}{\displaystyle\text{sC}}+\text{R}_4\right)}\tag7$$

Taking the inverse Laplace transform, leads to:

$$\text{V}\left(t\right)=\frac{\displaystyle\text{n}\hat{\text{u}}}{\displaystyle\text{n}+\text{R}_2}\cdot\exp\left(-\frac{\displaystyle t}{\displaystyle\text{CR}_4}\right)\tag8$$

Answer 2

Here is my proposed solution

Lets pose that R1 = R2 = R3. Pose that R3 current is not part of i (as per your circuit). For sake of simplification we can then take R3 out of the circuit [fig.C] because iR3 is determined by iR3= V1/R3, and this will not change since V1 is theoretically zero ohm impedance. Pose that, with the switch OPEN, all voltages and currents are zero, except for V1. Also, we know that C1 act as a short circuit when v/t is infinite, which is the case when the switch is momentarily closed.

When the switch CLOSE the voltage on R3 is V, but we don’t care since iR3 is of no interest for our analysis. The voltage between R2\C1\R1 = V (assuming that V2 is a current source with infinite impedance). Since C1 is a short, then i = V/(R1+R2) [fig.D]. Now that i is calculated, we know that 2i = 2 x i. If C1 is short we have to split the current 2i between R1 and R2. Since i2 is double of i then V(R1) will very quickly attain a value equal to V1 since R1=R2 and i2 (which is double the value of i) will induce into R1 the same voltage originally equal to V1.

For sake of explanation let’s utilize some values to better understand what’s happening. Say that V1 = 10 volts and R1=R2=5 ohms. 10V/(5Ω+5Ω)=1A. 2i=2A. 2A on 5Ω = 10Volts. If we now have 10V on R1 and 10V as V1 then R2 will see a voltage of zero volt, hence no current, i now equal zero. But by that time the capacitor will have attain 10 volts and i will now remain at zero since the capacitor is now charged to 10V. No more current will flow in R2 and R1. Since i is now zero then 2i is also zero. As for R3, its current will be 10V/5Ω = 2A.

In essence, that circuit is an RC circuit which the time constant equal to (R1+R2)xC1 except that the value of the time constant must be divided by some factor since the total current imposed on R1C1 is 3 time more than if there was no V2 source. So, instead of considering the C1(R1+R2) time constant one should consider that time constant equal to 0.33C1*(3R1+R2) because the V2 source modify the R1C1 impedance by multiplying it by a factor of 3 (which is determined by 1i+2i=3i.

Since the original question is: What is the voltage on R1 when t=0 (right at the exact moment when the switch is closed)? The answer is V(R1) = half of V1. Because the instantaneous voltage seen by the circuit is V. Since R1=R2 and C1 is a short circuit, then we will see on R1 a value of 0.5V1 because R1\R2 is a simple voltage divider. The mind trick here is when you start to consider the source V2. Here, V2 will only lower the time constant by a factor of 3, but since we are not interested about what’s happening after the switch is closed but rather immediately when the switch is closed, we do not have to consider V2. So, we are left with the simplified circuit such as in fig.D, only at t(0).