Take a two-terminal network of \(R^{\prime}\) s, \(L^{\prime} s\) and \(C^{\prime}\) of your own choosing with a voltage \(V\) between the terminals and a curreat \(I\) flowing into one and out of the other. Find the differential equation relating \(V\) and \(I_{2}\) and show that if \(V\) is given by \(V_{0}\) sin wt, the equation takes the form of \((10.1)\). For example, you could take \(L\) in series with \(R_{1}\) both in series with a parallel combination of \(C\) and \(R_{2}\).

Short Answer

Expert verified
Question: Write the differential equation relating the voltage V across the terminals and the current I2 through the two-terminal network consisting of resistances, inductances, and capacitances, and show that if the voltage V is given by \(V_0\sin(\omega t)\), the equation takes the form of equation (10.1). Answer: The differential equation relating the voltage V and the current I2 is given by: \(L\frac{d^2V}{dt^2} + (R_1 + R_2)\frac{dV}{dt} + \frac{1}{C}V = (R_1 + R_2)I_2\). When the voltage V is given by \(V_0\sin(\omega t)\), the equation takes the form of equation (10.1), which is: \(L\omega^2 - (R_1 + R_2)\omega\frac{\cos(\omega t)}{\sin(\omega t)} - \frac{1}{C} = -(R_1 + R_2)Y\).

Step by step solution

01

Draw the circuit

Draw a circuit diagram with an inductor L in series with a resistor R1, both in series with a parallel combination of a capacitor C and a resistor R2.
02

Write the voltage equations

For the inductor and capacitor, we should write the voltage in terms of their currents: Inductor: \(V_L = L\frac{dI(t)}{dt}\) Capacitor: \(V_C = \frac{1}{C}\int I(t) dt\)
03

Calculate currents in the elements

Let's denote the current through the inductor and resistor R1 as I1, and the current through the capacitor and resistor R2 as I2. Then we can write the following equations for the currents: \(I(t) = I_1(t) + I_2(t)\) \(I_1(t) = \frac{V}{R_1 + L\frac{d}{dt}}\) \(I_2(t) = C\frac{dV}{dt} + \frac{V}{R_2}\)
04

Write the differential equation

Substitute the expressions for I1 and I2 from step 3 into the equation for current I: \(I(t) = \frac{V}{R_1 + L\frac{d}{dt}} + C\frac{dV}{dt} + \frac{V}{R_2}\) Combine the terms and rearrange the equation to form a differential equation relating V and I2: \(L\frac{d^2V}{dt^2} + (R_1 + R_2)\frac{dV}{dt} + \frac{1}{C}V = (R_1 + R_2)I_2\)
05

Show that the equation takes the form of equation (10.1)

Let's consider the case when the voltage V is given by \(V_0\sin(\omega t)\). Then: \(V = V_0\sin(\omega t)\) Find the first and second derivatives of V with respect to time: \(\frac{dV}{dt} = \omega V_0\cos(\omega t)\) \(\frac{d^2V}{dt^2} = -\omega^2 V_0\sin(\omega t)\) Substitute the expressions for V, dV/dt, and d^2V/dt^2 into the differential equation derived in step 4: \(-L\omega^2 V_0\sin(\omega t) + (R_1 + R_2)\omega V_0\cos(\omega t) + \frac{1}{C}V_0\sin(\omega t) = (R_1 + R_2)I_2\) Divide by \(V_0\sin(\omega t)\): \(-L\omega^2 + (R_1 + R_2)\omega\frac{\cos(\omega t)}{\sin(\omega t)} + \frac{1}{C} = (R_1 + R_2)\frac{I_2}{V_0\sin(\omega t)}\) With rearrangement, if we denote \(\frac{I_2}{V_0\sin(\omega t)}\) as Y and multiply both sides by -1, we will have an equation with the form of equation (10.1), which is: \(L\omega^2 - (R_1 + R_2)\omega\frac{\cos(\omega t)}{\sin(\omega t)} - \frac{1}{C} = -(R_1 + R_2)Y\) We have shown that the equation takes the form of equation (10.1) when the voltage V across the terminals is given by \(V_0\sin(\omega t)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

RLC Circuit Analysis
Understanding the dynamics of an RLC circuit, which includes resistors (R), inductors (L), and capacitors (C) is fundamental to circuit analysis. The behavior of currents and voltages in this type of circuit can be easily understood by applying Kirchhoff's laws, which can help derive the circuit's differential equations. For example, to analyze currents within an RLC circuit consisting of specific arrangements of resistors, inductors, and capacitors, one can follow certain steps to determine the relationship between voltage and current.

First, a circuit diagram is necessary to visualize the arrangement and to identify the relationships between different components. Next, voltage equations for inductors and capacitors are written by considering their reactive properties; inductors' voltage is the product of inductance (L) and the rate of change of current, while for capacitors, voltage is the integral of current over time scaled by the capacitance (C). Using these principles and Ohm's law, the total current and the individual currents through the different branches of the circuit can be calculated. Through algebraic manipulation and substitution, a differential equation relating voltage and a specific current in the circuit is derived, capturing the essence of RLC circuit analysis.
Sinusoidal Steady-State
When an RLC circuit is subjected to a sinusoidal voltage input, such as one of the form \(V_0\sin(\omega t)\), where \(V_0\) is the amplitude and \(\omega\) is the angular frequency, it eventually reaches a sinusoidal steady-state. This means that after transient behaviors have decayed, the circuit's response will also be sinusoidal and will oscillate at the same frequency as the input. The goal of sinusoidal steady-state analysis is to determine the amplitude and phase relationships between the sinusoidal inputs and the resulting sinusoidal outputs.

In this state, the relationship between voltage and current for the inductor and capacitor is the key to solving circuit problems. The inductor's voltage leads its current by 90 degrees, and for the capacitor, the current leads the voltage. By deriving the first and second derivatives of a sinusoidal function, the differential equation of the circuit can be converted into a form that represents the behavior of the circuit in this steady state. This is crucial because it simplifies understanding the circuit's performance under the conditions commonly found in AC power systems, signal processing, and other electronics applications.
Circuit Differential Equations
Circuit differential equations are at the heart of understanding how electrical circuits respond over time. They are derived from the physical laws governing the components in the circuit — Ohm's law for resistors, Faraday's law for inductors, and the definition of capacitance for capacitors. These equations relate the circuit's current, voltages, and their derivatives and can be thought of as quantitatively capturing the 'rules' that voltages and currents follow in an electrical circuit.

In our exercise example, after laying out the necessary equations for each component and adding them together, we get a differential equation relating the voltage across the network to current \(I_2\). When dealing with sinusoidal inputs, calculus is used to find derivatives that, when substituted back into the equation, provide insights into how the system behaves specifically under this condition. The manipulation of these equations will result in a format such as equation (10.1), which represents a standard form to express the circuit's response in a more generalizable way. This form is very useful for engineers and physicists, as it can be more readily analyzed to predict the circuit's behavior or to design circuits with desired characteristics.

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Most popular questions from this chapter

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