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An ac generator with emf \(\mathcal{G}={8}_{\mathrm{m}} \sin \omega_{2} t\), where \({B}_{m}=18.0 \mathrm{~V}\) and \(\omega_{d}=377 \mathrm{rad} / \mathrm{s}\), is connected to a \(4.15 \mu \mathrm{F}\) capacitor. (a) What is the maximum value of the current? (b) When the current is a maximum, what is the emf of the generator? (c) When the emf of the generator is \(-12.5 \mathrm{~V}\) and increasing in magnitude, what is the current?

Short Answer

Expert verified
(a) Maximum current is approximately 0.121 A. (b) Emf at maximum current is 0 V. (c) Current is approximately 0.084 A (increasing).

Step by step solution

01

Find the Capacitive Reactance

The capacitive reactance, \(X_c\), is calculated by the formula \(X_c = \frac{1}{\omega_d C}\), where \(\omega_d = 377\, \text{rad/s}\) and \(C = 4.15\, \mu\text{F} = 4.15 \times 10^{-6}\, \text{F}\).\(X_c = \frac{1}{377 \times 4.15 \times 10^{-6}}\). Compute this to find \(X_c\).
02

Calculate Maximum Current

The maximum current, \(I_{max}\), is given by \(I_{max} = \frac{B_m}{X_c}\), where \(B_m = 18.0\, \text{V}\). Use the \(X_c\) found in Step 1 to calculate \(I_{max}\).
03

Determine EMF at Maximum Current

When the current is at its maximum, the voltage across the capacitor, \(V_C = B_m\). Given that \(\mathcal{G} = B_m\), use the known parameters to conclude that the generator's emf \(\mathcal{G}\) is zero at maximum current.
04

Calculate Current at Given EMF

Using the formula for current \(i(t) = \frac{\mathcal{G}}{X_c}\), substitute \(\mathcal{G} = |-12.5|\, \text{V}\) (as it increases in magnitude) and \(X_c\) calculated in Step 1 to find the current. Remember the sign of \(i(t)\) depends on the direction of current.

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Key Concepts

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

Capacitive Reactance
When dealing with AC circuits, capacitive reactance is a crucial concept that describes how a capacitor impedes the flow of alternating current. It is akin to resistance in a DC circuit, but depends on the frequency of the AC signal. The formula to calculate capacitive reactance, denoted by \(X_c\), is given by:\[X_c = \frac{1}{\omega C}\]where \(\omega\) is the angular frequency in radians per second, and \(C\) is the capacitance in farads. This tells us that the capacitive reactance decreases as the frequency or the capacitance increases. This inverse relationship implies that higher frequencies or larger capacitors will result in lower reactance, allowing more AC signal to pass through. In the exercise, finding the capacitive reactance is the first step to understand the behavior of the system and to calculate further parameters like current.
Maximum Current Calculation
The maximum current in an AC circuit with a capacitive element can be determined once the capacitive reactance is known. Using the formula:\[I_{max} = \frac{B_m}{X_c}\]where \(B_m\) is the peak voltage provided by the EMF, and \(X_c\) is the capacitive reactance. This formula is derived from Ohm’s Law, adapted for AC circuits. Here's how it works:
  • First, calculate \(X_c\) using the capacitive reactance formula.
  • Substitute the values for \(B_m\) and \(X_c\) into the equation for \(I_{max}\).
By using the calculated \(X_c\), the peak current can be determined, which indicates the maximum flow of charge through the capacitor. This step is crucial in analyzing circuit performance under maximum load.
Electromotive Force (EMF)
Electromotive force, or EMF, refers to the energy source in a circuit that provides the voltage. However, unlike direct current circuits, the EMF in AC circuits varies over time, usually in a sinusoidal waveform. The sinusoidal EMF is expressed as:\[\mathcal{G} = B_m \sin(\omega t)\]Here, \(B_m\) is the maximum voltage of the generator, \(\omega\) is the angular frequency, and \(t\) is the time. This equation captures how EMF changes continuously and forms the core driving force of AC circuits. In relation to the exercise, understanding when the current reaches its maximum is interlinked with the generator's EMF values, as the capacitor's voltage directly varies with the EMF. Understanding these time-dependent changes helps in determining circuit behavior at different times.
Alternating Current (AC)
Alternating current (AC) is a type of electric current where the flow of electric charge periodically reverses direction. In contrast, direct current (DC) flows only in one direction. AC is typically expressed as a sine wave, symbolizing the cyclical nature of its flow:\[i(t) = I_m \sin(\omega t + \phi)\]where \(I_m\) is the maximum current amplitude, \(\omega\) is the angular frequency, \(t\) is the time, and \(\phi\) is the phase angle. This versatility is particularly useful for power distribution because it can be easily transformed between different voltages. Additionally, it enables efficient long-distance power transmission.
In the context of the exercise, AC governs how the reactance and current interact, since both the EMF and capacitive effects depend on the oscillations of AC. Understanding AC's cyclic pattern is essential when calculating phase relationships in complex circuits.

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Most popular questions from this chapter

An oscillating \(L C\) circuit consisting of a \(1.0 \mathrm{nF}\) capacitor and a \(9.0 \mathrm{mH}\) coil has a maximum voltage of \(3.0 \mathrm{~V}\). What are (a) the maximum charge on the capacitor, (b) the maximum current through the circuit, and (c) the maximum energy stored in the magnetic field of the coil?

In an oscillating series \(R L C\) circuit, show that \(\Delta U / U\), the fraction of the energy lost per cycle of oscillation, is given to a close approximation by \(2 \pi R / \omega L\). The quantity \(\omega L / R\) is often called the \(Q\) of the circuit (for quality). A high- \(Q\) circuit has low resistance and a low fractional energy loss \((=2 \pi / Q)\) per cycle.

In an oscillating \(L C\) circuit, \(L=25.0 \mathrm{mH}\) and \(C=2.89 \mu \mathrm{F}\). At time \(t=0\) the current is \(9.20 \mathrm{~mA}\), the charge on the capacitor is \(3.80 \mu \mathrm{C}\), and the capacitor is charging. What are (a) the total energy in the circuit, (b) the maximum charge on the capacitor, and (c) the maximum current? (d) If the charge on the capacitor is given by \(q=Q \cos (\omega t+\phi)\), what is the phase angle \(\phi\) ? (e) Suppose the data are the same, except that the capacitor is discharging at \(t=0\). What then is \(\phi\) ?

An inductor is connected across a capacitor whose capacitance can be varied by turning a knob. We wish to make the frequency of oscillation of this \(L C\) circuit vary linearly with the angle of rotation of the knob, going from \(2 \times 10^{5}\) to \(4 \times 10^{5} \mathrm{~Hz}\) as the \(\mathrm{knob}\) turns through \(180^{\circ}\). If \(L=2.0 \mathrm{mH}\), plot the required capacitance \(C\) as a function of the angle of rotation of the knob.

(a) At what frequency would a \(12 \mathrm{mH}\) inductor and a \(10 \mu \mathrm{F}\) capacitor have the same reactance? (b) What would the reactance be? (c) Show that this frequency would be the natural frequency of an oscillating circuit with the same \(L\) and \(C\).

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