Problem 49 - Advanced Engineering Mathematics

Question

A series circuit contains an inductance of \(L=1 \mathrm{~h}\), a capacitance of \(C=10^{-4} f\), and an electromotive force of \(E(t)=\) \(100 \sin 50 t \mathrm{~V}\). Initially the charge \(q\) and current \(i\) are zero. (a) Find the equation for the charge at time \(t\). (b) Find the equation for the current at time \(t\). (c) Find the times for which the charge on the capacitor is zero.

Step-by-Step Solution

Answer

(a) The charge equation: \( q(t) = -\frac{1}{150} \sin(100t) + \frac{1}{75} \sin(50t) \). (b) The current equation: \( i(t) = -\frac{2}{3} \cos(100t) + \frac{2}{3} \cos(50t) \). (c) Charge is zero when \( \sin(50t) = \frac{1}{2} \sin(100t) \).

Step 1: Setting up the Differential Equation

The circuit is governed by the second-order differential equation derived from Kirchhoff's voltage law: \[ \frac{d^2q}{dt^2} + \frac{1}{C}q = E(t) \]Given, \( L=1 \), \( C=10^{-4} \), and \( E(t) = 100 \sin 50t \). Substitute these values:\[ \frac{d^2q}{dt^2} + 10000q = 100 \sin 50t \]

Step 2: Solving the Homogeneous Equation

First, solve the homogeneous equation:\[ \frac{d^2q}{dt^2} + 10000q = 0 \]The characteristic equation is:\[ r^2 + 10000 = 0 \]Solving for \( r \): \[ r = \pm 100i \]Thus, the homogeneous solution is:\[ q_h(t) = A \cos(100t) + B \sin(100t) \]

Step 3: Solving the Particular Solution

Assume a particular solution of the form:\[ q_p(t) = C \sin(50t) \]Substitute \( q_p(t) \) and \( q_p'(t) \) into the non-homogeneous equation:\[ (C(-50^2) + 10000C) \sin(50t) = 100 \sin(50t) \]Simplify:\[ 7500C \sin(50t) = 100 \sin(50t) \]Thus, \( C = \frac{1}{75} \) and\[ q_p(t) = \frac{1}{75} \sin(50t) \]

Step 4: Finding the General Solution for Charge

Combine the homogeneous and particular solutions:\[ q(t) = q_h(t) + q_p(t) = A \cos(100t) + B \sin(100t) + \frac{1}{75} \sin(50t) \]

Step 5: Applying Initial Conditions for Charge

The initial conditions are \( q(0) = 0 \) and \( i(0) = 0 \). i.e., \( \frac{dq}{dt}(0) = 0 \). Substitute into the general solution:1. \( q(0) = A + 0 + 0 = 0 \Rightarrow A = 0 \)2. \( \frac{dq}{dt} = -100A\sin(100t) + 100B\cos(100t) + \frac{50}{75}\cos(50t) \)Using \( \frac{dq}{dt}(0) = 0 \):\[ 100B + \frac{2}{3} = 0 \Rightarrow B = -\frac{1}{150} \]The charge equation:\[ q(t) = -\frac{1}{150} \sin(100t) + \frac{1}{75} \sin(50t) \]

Step 6: Deriving the Current Equation

The current is the derivative of the charge:\[ i(t) = \frac{dq}{dt} = -\frac{100}{150}\cos(100t) + \frac{50}{75}\cos(50t) \]Simplify:\[ i(t) = -\frac{2}{3}\cos(100t) + \frac{2}{3}\cos(50t) \]

Step 7: Finding the Times When Charge Is Zero

Set the expression for charge to zero:\[ -\frac{1}{150} \sin(100t) + \frac{1}{75} \sin(50t) = 0 \]Simplify:\[ \sin(50t) = \frac{1}{2} \sin(100t) \]Use trigonometric identities to solve for \( t \). This occurs at regular intervals determined by the frequency of the combined sinusoidal components.

Key Concepts

Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives. In the context of RLC circuit analysis, it is used to model the behavior of electric circuits with resistors, inductors, and capacitors. This is crucial for understanding how the voltage, current, and charge change over time in the circuit.

For the given RLC circuit, Kirchhoff's Voltage Law helps us establish a differential equation: \[ \frac{d^2q}{dt^2} + \frac{1}{C}q = E(t) \]where \(q(t)\) is the charge on the capacitor at time \(t\) and \(E(t)\) represents the electromotive force applied. By substituting the known values of inductance \(L=1\), capacitance \(C=10^{-4}\), and electromotive force \(E(t) = 100 \sin 50t\), our equation becomes:\[ \frac{d^2q}{dt^2} + 10000q = 100 \sin 50t \]

These equations often require solving both homogeneous and particular solutions to find how the system behaves overall. Combining these solutions allows us to determine the complete equation describing the system's behavior over time.

Kirchhoff's Voltage Law

Kirchhoff's Voltage Law (KVL) states that the total voltage around a closed loop in a circuit is zero. This principle is vital in circuit analysis because it allows us to write equations that describe the voltages in a circuit.

Applying KVL to our RLC circuit means summing the voltages across the inductor, the capacitor, and any external emf sources. For our particular RLC circuit problem, we have:- Voltage across the inductor: \(L \frac{di}{dt}\)- Voltage across the capacitor: \(\frac{1}{C}q\)- External emf: \(E(t)\)

When applying KVL, the sum is set to the external emf:\[ L\frac{di}{dt} + \frac{1}{C}q = E(t) \]

This results in the differential equation that allows us to analyze the behavior of the circuit as a function of time.

Characteristic Equation

When solving a differential equation for an RLC circuit, the characteristic equation is central to finding the solution. For a homogeneous differential equation, it determines the natural response of the circuit, ignoring any external forces.

Our circuit's homogeneous equation was:\[ \frac{d^2q}{dt^2} + 10000q = 0 \]

To solve it, we use the characteristic equation:\[ r^2 + 10000 = 0 \]

Solving this gives us complex roots:\[ r = \pm 100i \]

These roots suggest oscillatory solutions, which makes sense for circuits involving capacitors and inductors. The general solution to this homogeneous equation is a combination of sinusoidal functions:\[ q_h(t) = A \cos(100t) + B \sin(100t) \]

Here, \(A\) and \(B\) are constants determined by the initial conditions of the circuit.

Initial Conditions

Initial conditions are used to determine the specific solution to a differential equation that fits a real-world scenario. For an RLC circuit, they are usually the initial charge \(q(0)\) and initial current \(i(0)\).

In our problem, the initial conditions are:

• Initial charge: \(q(0) = 0\)
• Initial current: \(i(0) = 0\)

These conditions are substituted into the general solution to find the specific constants in the equations, ensuring the solution accurately reflects how the circuit operates from the start.

For instance, when we substitute these into our solution formulas, we find that:- \( A = 0 \)- \( B = -\frac{1}{150} \)

These constants allow us to develop specific equations for charge and current over time, such as:\[ q(t) = -\frac{1}{150} \sin(100t) + \frac{1}{75} \sin(50t) \]
These particular solutions give us insight into the dynamic behavior of the circuit starting from those specific initial moments.