Question

Consider the \(R L\) and \(R C\) networks shown, with the associated equations for the current \(i(t)\). Assume that the network element values are \(R=1 \mathrm{k} \Omega, L=1 \mathrm{H}, C=1 \mu \mathrm{F}\) and that \(e(t)\), given in volts, is $$ e(t)=\left\\{\begin{array}{lr} 0, & 0 \leq t \leq 0.5, \\ 1, & 0.5

Short answer

Question: Determine the current i(t) in both the RL and RC circuits when the voltage source e(t) is given as a piecewise function: \(e(t) = \left\\{\begin{array}{lr} 0, & 0 \leq t \leq 0.5 V\\\ 1, &0.5 < t < 1 V\end{array} \right.\) with \(R = 1 k\Omega\), \(L = 1 H\), and \(C = 1 \mu F\). Answer: The resulting current i(t) for both circuits is given as: For RL circuit: $$ i(t) = \left\\{\begin{array}{lr} 0, & 0 \leq t \leq 0.5, \\\ \frac{(1-e^{-1000(t-0.5)})}{1000} + e^{-1000 t}, & 0.5<t<1, \end{array} \right. $$ For RC circuit: $$ i(t) = \left\\{\begin{array}{lr} 0, & 0 \leq t \leq 0.5, \\\ \frac{1}{1000} - C\cdot e^{-t/RC}, & 0.5<t<1, \end{array} \right. $$

Step-by-step solution

Analyze the RL network

An RL network is a series R-L circuit, with the resistance R and the inductance L connected in series. The voltage across both elements is equal to e(t).

Write the equation for RL network

Using Kirchhoff's voltage law, we can write the following equation: \(e(t) = R\cdot i(t) + L \frac{di(t)}{dt}\) Given \(R = 1 k\Omega\), \(L = 1 H\), and \(e(t)\) as a piecewise function.

Solve the differential equation for RL network

With the given values of e(t), we can solve the differential equation: Case 1: \(0 \leq t \leq 0.5\) \(0 = 1000 i(t) + \frac{di(t)}{dt}\) Solve this linear differential equation for i(t) to get: \(i(t) = A e^{-1000 t}\) Case 2: \(0.5 < t < 1\) \(1 = 1000 i(t) + \frac{di(t)}{dt}\) Solve this linear differential equation for i(t) to get: \(i(t) = \frac{(1-e^{-1000(t-0.5)})}{1000} + B e^{-1000 t}\) We have two unknown constants A and B, which can be found by considering the initial conditions and continuity. At \(t=0\), \(i(0)=0\), so \(A = 0\). At \(t=0.5\), \(i_1(0.5) = i_2(0.5)\), where \(i_1(t)\) is the solution for case 1 and \(i_2(t)\) is the solution for case 2. So the complete solution for the RL network is: $$ i(t) = \left\\{\begin{array}{lr} 0, & 0 \leq t \leq 0.5, \\\ \frac{(1-e^{-1000(t-0.5)})}{1000} + e^{-1000 t}, & 0.5<t<1, \end{array} \right. $$ Now let's deal with the RC network:

Analyze the RC network

An RC network is a parallel R-C circuit, with the resistor R and the capacitor C connected in parallel. The voltage across both elements is equal to e(t).

Write the equation for RC network

Using Kirchhoff's current law, we can write the following equation: \(i(t) = C \frac{de(t)}{dt} + \frac{e(t)}{R}\) Given \(R = 1 k\Omega\), \(C = 1\mu F\) and \(e(t)\) as a piecewise function.

Solve the differential equation for RC network

With the given values of e(t), we can solve the differential equation: Case 1: \(0 \leq t \leq 0.5\) \(0 = 0 + \frac{0}{1000}\) \(i(t) = 0\) Case 2: \(0.5 < t < 1\) \(0 = \frac{de(t)}{dt} + \frac{1}{1000}\) \(i(t) = \frac{1}{1000} - C\cdot e^{-t/RC}\) We have an unknown constant C which can be found by considering the continuity at \(t=0.5\). At \(t=0.5\), \(i_1(0.5) = i_2(0.5)\), where \(i_1(t)\) is the solution for case 1 and \(i_2(t)\) is the solution for case 2. So the complete solution for the RC network is: $$ i(t) = \left\\{\begin{array}{lr} 0, & 0 \leq t \leq 0.5, \\\ \frac{1}{1000} - C\cdot e^{-t/RC}, & 0.5<t<1, \end{array} \right. $$

Key concepts

RL Network Analysis

RL network analysis involves studying the behavior of circuits containing resistors (R) and inductors (L) when they are subjected to a voltage source. The fundamental principle behind analyzing these networks is to apply Kirchhoff's voltage law, which leads to the formation of differential equations that describe the current in the circuit. The key is to solve these equations within the specified time domains, often requiring initial conditions to determine any constants that arise during integration. In practice, the response of an RL circuit to a change in voltage over time provides insights into important characteristics such as transient and steady state behavior.

In the given exercise, a piecewise function for the voltage source was provided, which led to two different cases in the analysis. This is common when the input takes different forms at different intervals of time, like a switching action in the circuit. Understanding how to solve the differential equations for each interval and ensuring the continuity of the solution across the intervals is essential for accurate prediction of the circuit's behavior.

RC Network Analysis

Just like RL networks, RC network analysis deals with circuits containing resistors (R) and capacitors (C) and focuses on determining the current flow or voltage across elements in response to an applied voltage. In the case of RC circuits, Kirchhoff's current law is the tool of choice, resulting in a different form of the differential equation than the one for RL circuits. For RC analysis, understanding the charging and discharging processes of the capacitor is critical, as these define the time-varying behavior of the circuit.

Once again, the piecewise nature of the input voltage necessitates a segmented approach, with the circuit's response in each time segment relying on initial conditions and the nature of the capacitor's exponential charge and discharge curves. The exercise highlights the necessity of continuity at transition points to achieve an accurate piecewise solution that complies with the physical behavior of capacitors in real circuits.

Kirchhoff's Laws

Kirchhoff's laws are the cornerstone of circuit analysis, consisting of the voltage law (KVL) and the current law (KCL). KVL states that the sum of electrical potential differences around any closed network is zero. It's akin to saying what goes up must come down; if a voltage is increased across a component, it must drop somewhere else in the loop. On the other hand, KCL asserts that the total current entering a junction must equal the total current leaving, reflecting the principle of conservation of electric charge.

When analyzing circuits, these laws enable us to form equations that govern the behavior of currents and voltages across various elements. This foundational knowledge is applied in both RL and RC network analysis. As seen in the textbook solutions, Kirchhoff's laws are employed implicitly when setting up the differential equations that must be solved to find the currents in the respective networks.

Piecewise Functions

A piecewise function is a mathematical expression defined by multiple sub-functions, where each sub-function applies to a certain interval in the function's domain. In the context of electrical circuits, piecewise functions are used to model inputs or driving functions that change behavior over time, such as a voltage that switches between different values.

In the exercise provided, the voltage source is described by a piecewise function that alters its value at a specific point in time. This illustrates the real-world scenario where a voltage may not be constant and requires segment-wise analysis using differential equations. Understanding how to work with piecewise functions allows students to analyze circuits that experience changes in input signals over time. Properly applying continuity conditions at the transition points between segments is also significant to ensure that the overall solution models the circuit correctly, as it ensures physical feasibility and adherence to the actual workings of electronic components.