Problem 10
Question
Find the current (in amperes) in each inductive circuit. \(L=30.0 \mathrm{mH}, f=7.00 \mathrm{MHz}, E=75.0 \mathrm{~V}$$L=30.0 \mathrm{mH}, f=7.00 \mathrm{MHz}, E=75.0 \mathrm{~V}\)
Step-by-Step Solution
Verified Answer
The current in the circuit is approximately 56.8 mA.
1Step 1: Determine the Inductive Reactance
First, we need to calculate the inductive reactance \(X_L\) using the formula: \(X_L = 2 \pi f L\). Given:\(L = 30.0 \text{ mH} = 30.0 \times 10^{-3} \text{ H}\) \(f = 7.00 \text{ MHz} = 7.00 \times 10^6 \text{ Hz}\)Now apply the values: \(X_L = 2 \pi \times 7.00 \times 10^6 \times 30.0 \times 10^{-3}\)\(X_L = 2 \pi \times 210.0 \times 10^3\)\(X_L \approx 1319.47 \text{ ohms}\).
2Step 2: Use Ohm's Law for AC Circuits
Next, we need to find the current \(I\) in the circuit using Ohm's Law for AC circuits: \(I = \frac{E}{X_L}\). Given \(E = 75.0 \text{ V}\) and \(X_L \approx 1319.47 \text{ ohms}\), we can calculate:\(I = \frac{75.0}{1319.47}\)\(I \approx 0.0568 \text{ A}\) or \(56.8 \text{ mA}\).
Key Concepts
Ohm's LawInductive CircuitAC Circuit AnalysisElectrical Inductance
Ohm's Law
Ohm's Law is a fundamental principle in electrical engineering that describes the relationship between voltage, current, and resistance. It's widely used in both DC and AC circuits, though its application slightly varies in alternating current scenarios.
For DC circuits, Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points. The simplified form is given by the equation:
For DC circuits, Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points. The simplified form is given by the equation:
- \( V = I \times R \)
- \( I = \frac{E}{X_L} \)
Inductive Circuit
An inductive circuit primarily consists of a coil or inductor. Inductors oppose changes in current flowing through them by generating a magnetic field. This characteristic is crucial in AC circuits, where current continuously changes direction.
Key characteristics of inductive circuits include:
- They store energy in the form of a magnetic field.
- Inductive reactance increases with frequency, meaning the greater the frequency of the voltage applied, the greater the opposition to the flow of current.
- In AC circuits, the current lags behind the voltage because the inductor takes time to build up its magnetic field.
AC Circuit Analysis
AC Circuit Analysis involves examining circuits powered by alternating current to understand their behavior and performance. Unlike direct current (DC), alternating current changes direction, typically in a sinusoidal pattern, which introduces a new set of variables and challenges.
There are several key concepts and tools used to analyze AC circuits:
- Impedance: This is the total opposition a circuit presents to the flow of AC and is a combination of resistance and reactance.
- Phase Angle: In AC circuits, current and voltage are often out of phase, meaning they don't reach their peak values simultaneously. The phase angle quantifies this difference.
- Phasors: These are a graphical representation used to simplify the analysis of sinusoidal signals by converting them into rotating vectors in a complex plane.
Electrical Inductance
Electrical Inductance is the property of a conductor by which a change in current through it induces a voltage across the conductor. This effect arises because of the magnetic field generated around the conductor. Measurement of inductance is in henries (H), and it plays a crucial role in the behavior of AC circuits. A few aspects of inductance to consider are:
- Self-Inductance: This occurs when a changing current within the same circuit induces a voltage in itself.
- Mutual Inductance: This takes place when a changing current in one coil induces a voltage in another nearby coil.
- The inductive reactance \( X_L \) of an inductor is calculated using the formula \( X_L = 2\pi fL \), where \( f \) is the frequency and \( L \) is the inductance.
Other exercises in this chapter
Problem 10
A circuit contains a \(375-\Omega\) resistance, a \(50 \overline{0}-\mu \mathrm{F}\) capacitor, and a \(0.500-\mathrm{H}\) inductance in series with a \(55.0-\m
View solution Problem 10
Find the capacitive reactance of a \(222-\mu \mathrm{F}\) capacitor in a circuit of frequency \(12 \overline{0} \mathrm{~Hz}\).
View solution Problem 10
What power is used by a heater with resistance \(19.5 \Omega\) and that draws a current of \(5.55 \mathrm{~A}\) ?
View solution Problem 11
Find the power factor of a generating station whose actual power is \(55,800 \mathrm{~kW}\) and whose apparent power is \(63,400 \mathrm{kVA}\).
View solution