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An ac generator has a frequency of \(2.2 \mathrm{kHz}\) and a voltage of \(240 \mathrm{V} .\) An inductance \(L_{1}=6.0 \mathrm{mH}\) is connected across its terminals. Then a second inductance \(L_{2}=9.0 \mathrm{mH}\) is connected in parallel with \(L_{1} .\) Find the current that the generator delivers to \(L_{1}\) and to the parallel combination.

Short Answer

Expert verified
The current to L1 is approximately 2.89 A, and to the parallel combination is 4.81 A.

Step by step solution

01

Calculate the Inductive Reactance of L1

The inductive reactance, denoted as \(X_L\), for an inductor is calculated using the formula \(X_L = 2 \pi f L\). For \(L_1 = 6.0\mathrm{\,mH}\), we have \(f = 2.2 \mathrm{\,kHz}\):\[X_{L1} = 2 \pi \times 2200 \times 0.006 = 83.04 \mathrm{\,ohms}\]
02

Calculate the Current through L1

Ohm's law for AC circuits involving inductance can be written as \(I = \frac{V}{X_L}\). Using the voltage \(V = 240\mathrm{\,V}\) and \(X_{L1} = 83.04\mathrm{\,ohms}\):\[I_{L1} = \frac{240}{83.04} \approx 2.89 \mathrm{\,A}\]
03

Calculate the Inductive Reactance of L2

Again, use the formula \(X_L = 2 \pi f L\) for \(L_2 = 9.0\mathrm{\,mH}\):\[X_{L2} = 2 \pi \times 2200 \times 0.009 = 124.56 \mathrm{\,ohms}\]
04

Calculate Combined Inductive Reactance

For inductors in parallel, the total inductive reactance \(X_{T}\) is given by \(\frac{1}{X_{T}} = \frac{1}{X_{L1}} + \frac{1}{X_{L2}}\):\[\frac{1}{X_{T}} = \frac{1}{83.04} + \frac{1}{124.56} \X_{T} \approx 49.86 \mathrm{\,ohms}\]
05

Calculate Total Current through Parallel Combination

Using Ohm's law with the combined inductive reactance and the voltage:\[I_{T} = \frac{240}{49.86} \approx 4.81 \mathrm{\,A}\]

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Key Concepts

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

Inductive Reactance
Inductive reactance is a concept that describes how an inductor resists the flow of alternating current (AC) due to its inductance. It is denoted by the symbol \(X_L\) and is directly related to the frequency of the AC signal and the inductance of the coil.
The formula used to calculate inductive reactance is \(X_L = 2 \pi f L\), where \(f\) is the frequency in hertz (cycles per second) and \(L\) is the inductance in henrys. This means that higher frequencies or greater inductance will result in higher inductive reactance.
  • For example, when calculating for an inductor \(L_1\) with an inductance of 6.0 mH at a frequency of 2.2 kHz, the inductive reactance is calculated as \(X_{L1} = 2 \pi \times 2200 \times 0.006 = 83.04 \text{ ohms}\).
  • Similarly, for another inductor \(L_2\) with 9.0 mH, the reactance would be \(X_{L2} = 2 \pi \times 2200 \times 0.009 = 124.56 \text{ ohms}\).
Understanding inductive reactance is crucial in designing and analyzing circuits with coils and inductors, as it impacts how much current can flow based on the resistance offered by the inductor.
Ohm's Law in AC Circuits
Ohm's Law in AC circuits extends the basic concept of Ohm's Law, which relates voltage, current, and resistance in DC circuits. In AC circuits, instead of simple resistance, we deal with impedance which includes both resistive and reactive components like inductive reactance.
The AC version of Ohm’s Law is expressed as \(I = \frac{V}{Z}\), where \(I\) is the current, \(V\) is the voltage, and \(Z\) is the impedance. For purely inductive circuits, the impedance is essentially the inductive reactance \(X_L\).
Using this concept:
  • For the inductor \(L_1\) with an inductive reactance \(X_{L1} = 83.04 \, \text{ohms}\), and a voltage \(V = 240 \, \text{V}\), the current is calculated as \(I_{L1} = \frac{240}{83.04} \approx 2.89 \, \text{A}\).
  • Similarly, for the total impedance of the parallel inductors \(X_T = 49.86 \, \text{ohms}\), the current becomes \(I_T = \frac{240}{49.86} \approx 4.81 \, \text{A}\).
This law helps in understanding how much current will flow in circuits with various forms of resistance, providing valuable insight into circuit behavior and design.
Parallel Inductors
In AC circuits, inductors can be connected in parallel to influence the total inductance and reactive behavior of the circuit. Parallel inductors are akin to resistors in parallel, but instead, they deal with inductive reactance.
For two inductors connected in parallel, the total inductive reactance \(X_T\) is found by the reciprocal formula:
\[\frac{1}{X_T} = \frac{1}{X_{L1}} + \frac{1}{X_{L2}}\]
  • For instance, consider \(L_1\) with \(X_{L1}\) of 83.04 ohms and \(L_2\) with \(X_{L2}\) of 124.56 ohms, we calculate \(X_T\) as:
    \[\frac{1}{X_T} = \frac{1}{83.04} + \frac{1}{124.56}\]
  • The result \(X_T \approx 49.86 \text{ ohms}\) shows the effective reduction in reactance when inductors are placed in parallel.
Understanding how parallel inductors work is essential, as it allows for precise control of circuit characteristics, enabling engineers to tailor the behavior of their designs by adjusting the total reactance seen by the source.

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Most popular questions from this chapter

When only a resistor is connected across the terminals of an ac generator (112 V) that has a fixed frequency, there is a current of \(0.500 \mathrm{A}\) in the resistor. When only an inductor is connected across the terminals of this same generator, there is a current of \(0.400 \mathrm{A}\) in the inductor. When both the resistor and the inductor are connected in series between the terminals of this generator, what are (a) the impedance of the series combination and (b) the phase angle between the current and the voltage of the generator?

An \(8.2-\mathrm{mH}\) inductor is connected to an ac generator \((10.0 \mathrm{V} \mathrm{rms},\) \(620 \mathrm{Hz}\) ). Determine the peak value of the current supplied by the generator.

A series \(\mathrm{RCL}\) circuit contains a \(47.0-\Omega\) resistor, a \(2.00-\mu \mathrm{F}\) capacitor, and a \(4.00-\mathrm{mH}\) inductor. When the frequency is \(2550 \mathrm{Hz},\) what is the power factor of the circuit?

A capacitor is connected to an ac generator that has a frequency of \(3.4 \mathrm{kHz}\) and produces a voltage of \(2.0 \mathrm{V}\). The current in the capacitor is \(35 \mathrm{mA}\). When the same capacitor is connected to a second ac generator that has a frequency of \(5.0 \mathrm{kHz},\) the current in the capacitor is \(85 \mathrm{mA}\). What voltage does the second generator produce?

In one measurement of the body's bioelectric impedance, values of \(Z=4.50 \times 10^{2} \Omega\) and \(\phi=-9.80^{\circ}\) are obtained for the total impedance and the phase angle, respectively. These values assume that the body's resistance \(R\) is in series with its capacitance \(C\) and that there is no inductance \(L .\) Determine the body's resistance and capacitive reactance.

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