# RL circuit unit impulse response

Consider this circuit.

We apply a unit impulse $$\\delta(t)\$$ that is $$\v_{IN}(t) = 1 \delta(t)\$$ volt-seconds. That is, $$\v_{IN}\$$ is a unit impulse at time t = 0. Since this circuit is driven by an impulse and there is no other source of energy, the capacitor voltage and inductor current at $$\t\$$ = 0 will be zero.

The problem asks value of $$\v_{R_1}(t)\$$ at $$\t=0^+\$$ and $$\t=1ms\$$ also value of $$\v_C(t)\$$ at $$\t=0^+\$$ and $$\t=1ms\$$

Since a huge current is applied should we assume inductor behaves an open circuit for a short period of time.

If there was only $$\R_2C\$$ part of the circuit $$\v_C(0^+)\$$ would be $$\\frac{V_{IN}}{\tau}\$$ right. And $$\v_C(t)\$$ would be $$\\frac{V_{IN}}{R_2C}e^{-t/\tau}u(t)\$$

How should I calculate the value of components asked in the problem.

Also should these kind of problems solved using heaviside operator.

• Your third to the last sentence ends in an expression that fails dimensional analysis.
– jonk
May 23, 2019 at 1:39

It's not complicated. You have two unknown voltage nodes:

simulate this circuit – Schematic created using CircuitLab

Note that I've made the two legs independent of each other, because they are. $$\V_\text{IN}\cdot u_t\$$ is just a fixed value, $$\V_\text{IN}\$$, multiplied by $$\u_t\$$ where \\begin{align*}u_t&\left\{\begin{array}{l} t \lt0& u_t=0\\t\ge0&u_t=1\end{array}\right.\end{align*}\.

So it should be pretty clear from the above that $$\V_\text{X}\$$ is independent of $$\V_\text{Y}\$$. So your questions can be treated, separately.

The initial conditions at $$\t=0^-\$$ will be that that current in $$\L_1\$$ is zero and that the voltage across $$\C_1\$$ will also be zero.

The nodal equations at $$\t=0^+\$$ are:

\begin{align*} \frac{V_\text{X}}{R_1}+\frac{1}{L_1}\int V_\text{X}\:\text{d}t &= \frac{V_\text{IN}}{R_1}&\therefore\quad \frac{\text{d}V_\text{X}}{\text{d}t}+\frac{R_1}{L_1}V_\text{X}&=0\:\text{A}\\\\ \frac{V_\text{Y}}{R_2}+C_1\frac{\text{d}V_\text{Y}}{\text{d}t}&=\frac{V_\text{IN}}{R_2}&\therefore\quad \frac{\text{d}V_\text{Y}}{\text{d}t}+\frac{1}{C_1\,R_2}V_\text{Y}&=\frac{V_\text{IN}}{C_1\,R_2} \end{align*}

The solutions, using the initial conditions of $$\V_\text{X}=V_\text{IN}\$$ at $$\t=0^+\$$ and $$\V_\text{Y}=0\:\text{V}\$$ at $$\t=0^+\$$, are:

\begin{align*} V_\text{X} &= V_\text{IN}\cdot e^{\frac{-t}{\left[\frac{L_1}{R_1}\right]}}\\\\ V_\text{Y} &= V_\text{IN}\cdot \left(1-e^{\frac{-t}{R_2\,C_1}}\right) \end{align*}

These are solved using the integrating factor for a 1st order ordinary differential equation.

To compute the values at $$\t=1\:\text{ms}\$$ you will need to have specific values for $$\L_1\$$, $$\R_1\$$, $$\R_2\$$, and $$\C_1\$$. You didn't provide those.

You may have questions about how to proceed with such general approaches. If so, feel free to ask.