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Chapter 10: Problem 44

An \(8 \mu \mathrm{F}\) capacitor is connected across \(220 \mathrm{~V}, 50 \mathrm{~Hz}\) line. What is the peak value of charge through the capacitor? a. \(2.5 \times 10^{-3} \mathrm{C}\) b. \(2.5 \times 10^{-4} \mathrm{C}\) c \(5 \times 10^{-5} \mathrm{C}\) d. \(7.5 \times 10^{-2} \mathrm{C}\)

Short Answer

Expert verified
The peak charge is approximately \(2.5 \times 10^{-3} \mathrm{C}\), matching option (a).

Step by step solution

01

Identify the Given Values

We are given the capacitance of the capacitor as \( C = 8\, \mu F = 8 \times 10^{-6} \ F\) and the voltage across the capacitor as \( V = 220 \ V \). We need to find the peak charge on the capacitor.
02

Calculate the Peak Voltage

The peak voltage \( V_0 \) can be calculated from the RMS voltage using the relationship \( V_0 = \sqrt{2} \cdot V_{RMS}\). Plug in the given value to get \( V_0 = \sqrt{2} \cdot 220 \approx 311 \ V\).
03

Use the Capacitance Formula

The peak charge \( Q_0 \) on a capacitor is given by the formula \( Q_0 = C \cdot V_0 \). Substitute \( C = 8 \times 10^{-6} \ F \) and \( V_0 = 311 \ V \) into the formula to calculate \( Q_0 \).
04

Perform the Calculation

Substitute the values into the formula: \[ Q_0 = 8 \times 10^{-6} \cdot 311 \]This gives us \[ Q_0 = 2.488 \times 10^{-3} \ C \].
05

Choose the Closest Option

The calculated peak charge \( Q_0 \approx 2.5 \times 10^{-3} \ C \), which matches option (a).

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Key Concepts

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

RMS Voltage
RMS voltage stands for "Root Mean Square" voltage. It is an important concept in AC (Alternating Current) circuits, particularly because it provides a way to express AC voltages that can be compared to DC (Direct Current) voltages in terms of power delivery.
In AC circuits, the voltage is constantly changing its value and direction, making it more complex to analyze compared to constant DC voltage. The RMS voltage is defined as the equivalent DC voltage that would deliver the same power to a load as the AC voltage does over a cycle.
  • Mathematically, RMS voltage ( V_{RMS} ) is derived from the peak voltage ( V_0 ) using the formula: V_{RMS} = rac{V_0}{ ext{鈭�2}} .

  • This means that the RMS voltage is approximately 0.707 times the peak voltage.

  • In practice, most electrical appliances and devices are rated in terms of their RMS voltage.
Understanding how RMS voltage works allows you to handle AC circuit problems more effectively, as calculations often start from knowing either the RMS voltage or deriving the peak voltage from it.
Peak Voltage
The peak voltage is the maximum instantaneous value of a waveform. In an AC circuit, this is the highest value the voltage reaches during a cycle.
Finding the peak voltage is especially useful for calculating other parameters, such as the peak charge in capacitors, or for performing safety assessments since circuits can be at risk if components can only handle voltages below the peak value.
  • In a sinusoidal AC circuit, the peak voltage is related to the RMS voltage by the formula: V_0 = ext{鈭�2} imes V_{RMS} .

  • This math expression shows why peak voltage is always larger than RMS voltage, specifically by a factor of 鈭�2, or about 1.414.

  • In the exercise, the RMS voltage given was 220 V, so using the formula, the peak voltage was calculated to be approximately 311 V.
Understanding the peak voltage is crucial for designing and working with circuits as it ensures that all elements within the system are rated to withstand the maximum voltage they might encounter.
Charge on Capacitor
The charge on a capacitor is an essential part of how capacitors store energy and manage power in circuits. When a capacitor is connected to an AC supply, the charge on the capacitor continuously changes as the voltage alternates.
To find the peak charge on a capacitor, you use its capacitance and the peak voltage across it:
  • The capacitance ( C ) describes the capacitor's ability to store charge for a given voltage. It's measured in farads ( F ).

  • The peak charge ( Q_0 ) on a capacitor is calculated using the formula Q_0 = C imes V_0 , where V_0 is the peak voltage.

  • In this exercise, with a capacitance of 8 imes 10^{-6} F and peak voltage of 311 V, the peak charge was found to be approximately 2.5 imes 10^{-3} C.
Grasping the concept of charge on a capacitor can help you understand how they work within AC circuits. They regularly charge and discharge, storing energy while leveling electrical signals.

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Most popular questions from this chapter

A dc ammeter and a hot wire ammeter are connected to a circuit in series. When a direct current is passed through circuit, the dc ammeter shows \(6 \mathrm{~A}\). When ac current flows through circuit, the ac ammeter shows \(8 \mathrm{~A}\). What will be reading of each ammeter, if dc and ac currents flow simultaneously through the circuit? a. \(d c=6 A, a c=10 A\) b. \(d c=3 A, a c=5 A\) c. \(d c=5 A, a c=8 A\) d. \(d c=2 A, a c=3 A\)

When an AC source of e.m.f. \(e=E_{0} \sin (100 t)\) is connected across a circuit, the phase diffcrence between the c.m.f. \(e\) and the current \(i\) in the circuit is observed to be \(\pi / 4\) as shown in Fig. 10.54. If the circuit consists possibly only of \(R-C\) or \(R-C\) or \(L-C\) series, find the relationship between the two elements. a. \(R=1 \mathrm{k} \Omega, C=10 \mu \mathrm{F}\) b \(R=1 \mathrm{k} \Omega, C=1 \mu \mathrm{F}\) c \(R=1 \mathrm{k} \Omega, L=10 \mathrm{H}\) d \(R=1 k \Omega, L=1 \mathrm{H}\)

Determine the rms value of a semi-circular current wave which has a maximum value of \(a\). a. \((1 / \sqrt{2}) a\) b \(\sqrt{(3 / 2)} a\) c. \(\sqrt{(2 / 3)} a\) d \((1 / \sqrt{3}) a\)

An inductive coil and resistance of \(100 \Omega\). When an ac signal of frequency \(1000 \mathrm{~Hz}\) is freed to the coil, the applied voltage leads the current by \(45^{\circ}\). What is the inductance of the coil? a. \(2 \mathrm{mH}\) b \(3.3 \mathrm{mH}\) c. \(16 \mathrm{mH}\) d. \(\sqrt{5} \mathrm{mH}\)

In a black box of unknown clements \((L\), or \(R\) or any other combination), an ac voltage \(E=E_{0} \sin (\omega i+\phi)\) is applied and current in the circuit was found to be \(l=i_{0} \sin \langle\omega t+\phi\) \(+\pi / 4)\). Then the unknown clements in the box naly be a. only capacitor b. indutor and resister both c. cither capacitor, resistor and inductor or only capacitor and rcsistor d only resistor
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