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(a) What is the rms current in an \(R C\) circuit if \(R=5.70 \mathrm{k} \Omega\), \(C=1.80 \mu \mathrm{F},\) and the rms applied voltage is \(120 \mathrm{~V}\) at \(60.0 \mathrm{~Hz} ?\) (b) What is the phase angle between voltage and current? (c) What is the power dissipated by the circuit? (d) What are the voltmeter readings across \(R\) and \(C\) ?

Short Answer

Expert verified
RMS current: 0.0203 A; Phase angle: 14.5°; Power: 2.34 W; Voltages: 115.7 V (R), 29.9 V (C).

Step by step solution

01

Calculate the capacitive reactance (Xc)

The capacitive reactance is given by the formula \(X_c = \frac{1}{2\pi f C}\). First, convert the capacitance to farads: \(C = 1.80\, \mu F = 1.80 \times 10^{-6}\, F\). Then use the frequency \(f = 60.0\, Hz\).\[X_c = \frac{1}{2 \pi \times 60.0 \times 1.80 \times 10^{-6}} \approx 1474.7\, \Omega\]
02

Determine the total impedance (Z) of the circuit

The total impedance in an RC circuit is calculated using the formula \(Z = \sqrt{R^2 + X_c^2}\). Let \(R = 5.70 \times 10^{3}\, \Omega\).\[Z = \sqrt{(5.70 \times 10^{3})^2 + (1474.7)^2} \approx 5912.4\, \Omega\]
03

Calculate the rms current (Irms) in the circuit

Use Ohm's Law to find the rms current: \(I_{rms} = \frac{V_{rms}}{Z}\), where \(V_{rms} = 120\, V\).\[I_{rms} = \frac{120}{5912.4} \approx 0.0203\, A\]
04

Find the phase angle (φ) between voltage and current

The phase angle can be calculated from \( \tan(\phi) = \frac{X_c}{R} \).\[\phi = \arctan \left( \frac{1474.7}{5.70 \times 10^{3}} \right) \approx 14.5^\circ\]
05

Calculate the power dissipated by the circuit

The power dissipated in the circuit is given by \( P = I_{rms}^2 \cdot R \).\[P = (0.0203)^2 \times 5700 \approx 2.34 \text{ W}\]
06

Determine the voltmeter readings across R and C

The voltage across the resistor is \(V_R = I_{rms} \times R\) and across the capacitor \(V_C = I_{rms} \times X_c\).For \(R\):\[V_R = 0.0203 \times 5700 \approx 115.7 \text{ V}\]For \(C\):\[V_C = 0.0203 \times 1474.7 \approx 29.9 \text{ V}\]

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Key Concepts

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

Capacitive Reactance
Capacitive reactance is a critical concept in AC (alternating current) circuits, especially those involving capacitors, like in an RC circuit. It represents the opposition a capacitor provides to the change in voltage across it. The capacitive reactance is inversely proportional to the frequency of the applied AC signal and the capacitance of the capacitor itself.
Using the formula \(X_c = \frac{1}{2\pi f C}\), where:
  • \(X_c\) is the capacitive reactance
  • \(f\) is the frequency of the AC source
  • \(C\) is the capacitance
It's important to note that capacitive reactance decreases with increasing frequency or capacitance. Thus, a higher frequency or larger capacitance will result in less opposition to the current, further influencing the overall behavior of the circuit.
This concept is pivotal when analyzing how circuits respond to different frequencies, especially in filtering applications or designing circuits that require precise timing characteristics.
Impedance in RC Circuit
Impedance is a measure of how much a circuit resists the flow of alternating current, combining both resistance and reactance components. In an RC circuit, which includes a resistor (R) and a capacitor (C), the total impedance can be calculated using the formula:\[Z = \sqrt{R^2 + X_c^2}\]where \( Z \) is the impedance, \( R \) is the resistance, and \( X_c \) is the capacitive reactance. This formula shows that impedance is a combination of the resistance squared and the capacitive reactance squared, both under a square root.
The significance of impedance lies in its role in determining the RMS (root mean square) current in the circuit. Using Ohm's Law for AC circuits, the RMS current \( I_{rms} \) is the applied RMS voltage \( V_{rms} \) divided by the impedance \( Z \). This relationship helps us understand how much current will flow through the circuit for a given voltage.
In practical applications, knowing the impedance is crucial for ensuring that circuits perform efficiently. It affects how much power is delivered to the circuit and how different components interact with one another, making it a fundamental concept for both circuit design and analysis.
Phase Angle in AC Circuit
The phase angle in an AC circuit describes the difference in phase between the voltage across the circuit and the current through it. For an RC circuit, the phase angle \( \phi \) can be determined using the tangent function:\[ \tan(\phi) = \frac{X_c}{R} \]which gives \( \phi = \arctan\left(\frac{X_c}{R}\right) \).
Phase angle is essential because it provides insight into power factor, which is a measure of how effectively the circuit converts electrical power into useful work. A phase angle of 0° means that the voltage and current are in phase, and power conversion is at its most efficient.
In an RC circuit, the phase angle isn't zero. It indicates a lag of current behind voltage due to the capacitive nature of the circuit. This is vital for applications requiring precise control of the timing between voltage and current, such as in signal processing or in circuits involving complex load interactions.
Understanding phase angle allows engineers to design circuits that achieve desired performance characteristics and operate efficiently within their given electrical systems.

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Most popular questions from this chapter

(II) In some experiments, short distances are measured by using capacitance. Consider forming an \(L C\) circuit using a parallel-plate capacitor with plate area \(A ,\) and a known inductance \(L . ( a )\) If charge is found to oscillate in this circuit at frequency \(f = \omega / 2 \pi\) when the capacitor plates are separated by distance \(x ,\) show that \(x = 4 \pi ^ { 2 } A \epsilon _ { 0 } f ^ { 2 } L\) . (b) When the plate separation is changed by \(\Delta x ,\) the circuit's oscillation frequency will change by \(\Delta f .\) Show that \(\Delta x / x \approx 2 ( \Delta f / f ) . ( c )\) If \(f\) is on the order of 1\(\mathrm { MHz }\) and can be measured to a precision of \(\Delta f = 1 \mathrm { Hz } ,\) with what percent accuracy can \(x\) be determined? Assume fringing effects at the capacitor's edges can be neglected.

(I) The variable capacitor in the tuner of an AM radio has a capacitance of \(1350 \mathrm{pF}\) when the radio is tuned to a station at \(550 \mathrm{kHz}\). ( \(a\) ) What must be the capacitance for a station at \(1600 \mathrm{kHz} ?\) (b) What is the inductance (assumed constant)? Ignore resistance.

(II) A 425 -pF capacitor is charged to 135\(\mathrm { V }\) and then quickly connected to a 175 -mH inductor. Determine \(( a )\) the frequency of oscillation, \(( b )\) the peak value of the current, and \(( c )\) the maximum energy stored in the magnetic field of the inductor.

What is the inductance of a coil if the coil produces an emf of \(2.50 \mathrm{~V}\) when the current in it changes from \(-28.0 \mathrm{~mA}\) to \(+25.0 \mathrm{~mA}\) in \(12.0 \mathrm{~ms} ?\)

(II) \(( a )\) What is the reactance of a \(0.086 - \mu F\) capacitor connected to a \(22 - \mathrm { kV } ( \mathrm { rms } ) , 660 - \mathrm { Hz }\) line? \(( b )\) Determine the frequency and the peak value of the current.

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