Lecture 7: 1st Order RLC Circuits

1st Order RLC Circuits

Objectives

• Understand transient responses in first-order circuits.
• Analyze 1st order RLC circuits.

First Order Circuits

• A first-order circuit contains only one energy storage element (capacitor or inductor) and a resistor.
• Two types:
  • RC circuit
  • RL circuit

Source-Free Circuits

• A source-free circuit has all independent sources disconnected after a switch action.
• Voltages and currents exhibit transient response due to initial conditions (initial capacitor voltages or inductor currents).

Example 1.1: RC Circuit

• Given: $i(10^{-2}) = 3.68$ mA
• a) Find $V(0)$.
• For t > 0, $i<em>c = -i</em>R$.
• $\text{From } C \frac{dv}{dt} = -\frac{V}{R}$, we get $\frac{1\mu FdV}{dt} = - \frac{V}{10k}$.
• $\frac{dv}{V} = -100 dt$
• Integrate both sides: $\int \frac{dv}{V} = \int -100 dt$
• $\ln V = -100t + k$
• Take exponential: $e^{\ln V} = e^{-100t + k}$
• $V = e^{-100t} e^k$
• $V(t) = V_0 e^{-100t}$
• Thus, $V(0) = V_0$.
• b) Find $V(t)$.
• $V(10^{-2}) = i(10^{-2}) \times R = 3.68m \times 10k = 36.8 V$
• $V_0 e^{-100(10^{-2})} = 36.8$
• $V_0 = 100 V$
• $V(t) = 100e^{-100t}$
• c) Sketch $V(t)$ for t > 0.
• $V(0) = 100$
• $V(\infty) = 0$

RC Circuit Response

• From Example 1.1:
• $V(t) = V_0 e^{-\frac{t}{RC}}$
• $V(t) = V_0 e^{-\frac{t}{\tau}}$
• where $\tau = RC$ is the time constant.
• $V_0 = V(0)$
• RC time constant ($\tau$): Time to charge the capacitor from zero to approximately 63.2% of the applied DC voltage or discharge to 36.8% of its initial voltage.
• $63.2\% = 1 - e^{-1}$
• $36.8\% = e^{-1}$

Exercise 1.1

• Determine the time constant.
• $R_{eq} =$
• $C_{eq} =$
• $\tau = R<em>{eq} \cdot C</em>{eq} =$

Exercise 1.2

• Select C so that the time constant is 14 s.

Exercise 1.3

• Given $v(0) = 9 V$, find $v(t)$.

Exercise 1.4

• Given $v(0) = 9 V$, find $v(t)$.
• $v(t) = 9e^{-\frac{t}{60 \times 10^{-6}}} V$

Exercise 1.5

• Given $V(0) = 5 V$, find current $i(t)$ and the maximum power dissipated by the resistor.
• $V(t) = V_0 e^{-\frac{t}{RC}}$
• $i(t) =$
• The power dissipated by the resistor is $P(t) =$
• The maximum power dissipated is

Example 2.1: RL Circuits

• Given $i(0) = 3 A$,
• a) Find $i(t)$.
• Apply KVL: $L \frac{di}{dt} + iR = 0$
• Try natural solution: $i(t) = Ke^{st}$, where K and s are constants.
• $L \frac{dKe^{st}}{dt} + Ke^{st} R = 0$
• $sLKe^{st} + Ke^{st} R = 0$
• $(sL + R)Ke^{st} = 0$
• So, $sL + R = 0$ or $Ke^{st} = 0$
• But $i = Ke^{st} \neq 0$ when $t \geq 0$, thus,
• Solve for s: $s = -\frac{R}{L}$
• Solve for K: $i(0) = Ke^{s \times 0} = 3 \implies K = 3$
• Thus, $i(t) = 3e^{st}$
• $i(t) = 3e^{-2t} A$

RL Circuit Natural Response

• $i(t) = i_0 e^{-\frac{t}{\tau}}$
• where $\tau = \frac{L}{R}$
• and $i_0 = i(0)$

Exercise 2.1

• Given $i<em>1 = 2A$ at $t = 0$, find $i</em>1(t)$ for t > 0 and find $i_1$ at $t = 200 \mu s$.

Exercise 2.2

• Given $i_R(0) = 6 A$, find $i$ at $t = 1 ns$.

Exercise 2.3

• Find the response for t > 0.
• $i(0) = 6 mA$