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Chapter 3: Problem 2

An RC circuit with a $$1-\Omega \text { resistor and a } 0.000001-\mathrm{F}$$ capacitor is driven by a voltage $$E(t)=\sin 100 t V$$ If the initial capacitor voltage is zero, determine the sub-sequent resistor and capacitor voltages and the current.

Short Answer

Expert verified
The sub-sequent resistor voltage is \(VR(t) = 0.1 \cos(100t) V\), the capacitor voltage is \(Vc(t) = \sin(100t) - 0.1 \cdot \cos(100t) V\), and the current is \( I(t) = 0.1 \cos(100t) A\).

Step by step solution

01

Write out the given details and the formula we will be using

We know that the resistor is given as \(1-\Omega\), the capacitor is given as \(0.000001-F\), the initial voltage across the capacitor \(Vc(0) = 0V\) (as it is initially uncharged), and the voltage is given as \(E(t) = \sin(100t) V\).\n\nLet's also recall the formula used to calculate the voltage across the capacitor in an RC circuit, for any time t: \(Vc(t) = E(t) - I(t)R\). We use this equation because according to Kirchhoff's voltage law, the sum of the electrical potential differences (voltages) around any closed loop or mesh in a network is always equal to zero. This formula is derived from that law.\n\nWe also know that the current I(t) in a capacitor is given by \(I(t) = C dv/dt\), where \(dv/dt\) is the derivative of the voltage with respect to time.
02

Calculate the derivative of the voltage with respect to time

First, we need to find the derivative of the given voltage function, \(E(t) = \sin(100t)\). So, \(E'(t) = dE(t)/dt = 100 \cos(100t)\) V/s.
03

Determine the current I(t) using the Capacitor's current-voltage relationship

Next, we can insert the derivative of the voltage in the equation for the current in the capacitor we mentioned above: \(I(t) = C \cdot E'(t) = 0.000001 \cdot 100 \cos(100t) = 0.1 \cos(100t) A\).
04

Determine the voltage across the resistor

The voltage across the resistor \(VR(t)\) can be determined from Ohm's law, which states that \(I = V/R\). So, rearranging the formula gives \(V = I \cdot R\). Substituting \(I(t)\) and R into the equation gives: \(VR(t) = I(t) \cdot R = 0.1 \cos(100t) \cdot 1 = 0.1 \cos(100t) V\).
05

Determine voltage across the capacitor

Finally, we use the Kirchhoff's voltage law equation we mentioned earlier to calculate the voltage over the capacitor. Rearranging the formula gives: \(Vc(t) = E(t) - VR(t) = \sin(100t) - 0.1 \cdot \cos(100t) V\).

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Key Concepts

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

RC Circuits
RC circuits, also known as resistive-capacitor circuits, are basic electrical circuits consisting of a resistor (R) and a capacitor (C) connected in series or parallel to a power source. These circuits are important in electronics because they can filter signals, manage the charging and discharging of capacitors, and shape the timing of circuit responses.

In an RC circuit, the resistor controls the flow of current, while the capacitor stores and releases electrical energy. Upon connecting to a power source, the circuit will initially allow the capacitor to charge through the resistor. Over time, the capacitor builds up voltage, which approaches the voltage of the power source, and the current decreases exponentially.

The equations governing RC circuits highlight the time-dependent behavior of voltages and currents. The time constant of the circuit, denoted by \(\tau = R \cdot C\), determines how quickly the circuit responds. It expresses the time it takes for the capacitor to charge up to about 63% of its maximum voltage. A small time constant means the capacitor charges quickly, while a larger time constant indicates a slower charge.
Capacitor Voltage
The voltage across a capacitor in an RC circuit is crucial for understanding how energy storage in the circuit works over time. When a capacitor is connected in an RC circuit, it initially opposes changes in voltage by charging up or discharging through the resistor.

For a given voltage source \(E(t)\), the voltage across the capacitor, denoted as \(Vc(t)\), can be determined using the equation \(Vc(t) = E(t) - I(t)R\), where \(I(t)\) is the current through the circuit. The capacitor voltage builds over time but never exceeds the source voltage. In situations where the source voltage is time-dependent, as in the given problem, \(Vc(t)\) will also change over time according to the governing equations.

Calculating the voltage across the capacitor involves evaluating the integrative behavior of the circuit. Initially uncharged capacitors start from a voltage of zero, and, depending on the circuit configuration and source voltage, they charge at a rate defined by the circuit's time constant \(\tau\). Understanding capacitor voltage behavior is crucial for applications like signal processing and timing circuits.
Kirchhoff's Voltage Law
Kirchhoff's Voltage Law (KVL) is a fundamental concept in circuit theory that allows for the analysis of multi-loop circuits. This law states that the sum of the electrical potential differences (voltages) around any closed loop or mesh in a network equals zero. It essentially means that energy is conserved in electrical circuits.

In analyzing the RC circuit from the problem, KVL helps us derive the relationship between different voltages in the circuit, such as the source voltage \(E(t)\), the voltage across the resistor \(VR(t)\), and the voltage across the capacitor \(Vc(t)\). By applying KVL, one can write the equation \(Vc(t) = E(t) - VR(t)\), ensuring that the sum of the voltages across the loop returns to zero around a complete path in the circuit.

KVL is essential for verifying circuit behavior and calculating unknown quantities within the circuit. By using it in combination with other circuit laws, engineers and physicists can solve complex circuit problems and ensure that devices function correctly and safely.
Ohm's Law
Ohm's Law is a fundamental principle used to relate voltage, current, and resistance in electrical circuits. It states that the current \(I\) through a conductor between two points is directly proportional to the voltage \(V\) across the two points and inversely proportional to the resistance \(R\), with the equation \(I = V/R\).

In the context of our RC circuit problem, Ohm's Law is applied to determine the voltage across the resistor excreted by the flowing current. Since the circuit uses a 1-ohm resistor, we use Ohm鈥檚 Law to calculate \(VR(t) = I(t) \cdot R\), where \(I(t)\) is the current flowing through the resistor.

This relationship is essential for calculating the distribution of voltage within the circuit and understanding how resistors limit current flow. Ohm's Law makes analyzing electrical networks easier by simplifying how we compute parts of the circuit, especially in linear circuits like RC circuits, where the resistance is constant.

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Most popular questions from this chapter

Escape Velocity. According to Newton's law of gravitation, the attractive force between two objects varies inversely as the square of the distances between them. That is, $$F_{g}=G M_{1} M_{2} / r^{2}$$, where $$M_{1}$$ and $$M_{2}$$ are the masses of the objects, $$r$$ is the distance between them (center to center). $$F_{g}$$ is the attractive force, and $$G$$ is the constant of proportionality. Consider a projectile of constant mass $$m$$ being fired vertically from Earth (see Figure 3.12). Let $$t$$ represent time and $$v$$ the velocity of the projectile. (a) Show that the motion of the projectile, under Earth's gravitational force. is governed by the equation $$\frac{d v}{d t}=-\frac{g R^{2}}{r^{2}}$$ where $$r$$ is the distance between the projectile and the center of Earth, $$R$$ is the radius of Earth, $$M$$ is the mass of Earth, and $$g=G M / R^{2}$$. (b) Use the fact that $$d r / d t=v$$ to obtain $$v \frac{d v}{d r}=-\frac{g R^{2}}{r^{2}}$$ (c) If the projectile leaves Earth's surface with velocity $$v_{0}$$, show that $$v^{2}=\frac{2 g R^{2}}{r}+v_{0}^{2}-2 g R$$ (d) Use the result of part (c) to show that the velocity of the projectile remains positive if and only if $$v_{0}^{2}-2 g R>0$$. The velocity $$v_{e}=\sqrt{2 g R}$$ is called the escape velocity of Earth. (e) If $$g=9.81 \mathrm{m} / \mathrm{sec}^{2}$$ and $$R=6370 \mathrm{km}$$ for Earth, what is Earth's escape velocity? (f) If the acceleration due to gravity for the moon is $$g_{m}=g / 6$$ and the radius of the moon is $$R_{m}= 1738 km$$, what is the escape velocity of the moon?

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