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Ignoring any mutual inductance, what is the equivalent inductance of two inductors connected \((a)\) in series, \((b)\) in parallel?

Short Answer

Expert verified
Series: Add inductances. Parallel: Use \(L_{eq, parallel} = \frac{L_1 \cdot L_2}{L_1 + L_2}\).

Step by step solution

01

Understanding Inductors in Series

When inductors are connected in series, their total (equivalent) inductance is simply the sum of their individual inductances. This is analogous to resistors in series. Thus, if two inductors with inductances \(L_1\) and \(L_2\) are connected in series, the equivalent inductance \(L_{eq, series}\) is given by: \[ L_{eq, series} = L_1 + L_2 \]
02

Understanding Inductors in Parallel

For inductors in parallel, the calculation is similar to that of resistors in parallel. The reciprocal of the equivalent inductance is the sum of the reciprocals of the individual inductances. If two inductors with inductances \(L_1\) and \(L_2\) are connected in parallel, the equivalent inductance \(L_{eq, parallel}\) is given by: \[ \frac{1}{L_{eq, parallel}} = \frac{1}{L_1} + \frac{1}{L_2} \]To find \(L_{eq, parallel}\), you can solve the above equation: \[ L_{eq, parallel} = \frac{L_1 \cdot L_2}{L_1 + L_2} \]

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Key Concepts

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

Inductors in Series
When dealing with inductors in series, think of them just like resistors or any other components that are connected end-to-end along a single path for current to flow. This is a straightforward concept because the total or equivalent inductance, noted as \( L_{eq, series} \), simply becomes the sum of all individual inductances in that series.
For instance, if you have two inductors, with inductances \( L_1 \) and \( L_2 \), and they are connected in series, the formula to determine the equivalent inductance is fairly simple: \\[ L_{eq, series} = L_1 + L_2 \]
This principle applies regardless of how many inductors you have in series. Just like adding more units in a line, the total inductance grows longer, allowing for greater energy storage in an electrical circuit. This can be useful when you want to increase the time it takes for a circuit to respond to changes in voltage.
Inductors in Parallel
In the context of parallel inductors, the behavior changes as compared to when they are in series. Imagine stacking two books side by side rather than end-to-end. For inductors in parallel, the overall inductance, thought of as \( L_{eq, parallel} \), decreases. This is similar to how resistors behave in parallel circuits. The concept might seem a bit tricky at first because we're dealing with the inverses of the inductances.
The equation that helps us find the equivalent inductance in a parallel configuration is:\[ \frac{1}{L_{eq, parallel}} = \frac{1}{L_1} + \frac{1}{L_2} \]
To deduce \( L_{eq, parallel} \), you would eventually find it as:\[ L_{eq, parallel} = \frac{L_1 \, \cdot \, L_2}{L_1 + L_2} \]
When you connect inductors in parallel, you effectively create more pathways for the magnetic fields they generate. The total magnetic field lines are then shared among several pathways, causing the equivalent inductance to decrease, which means the circuit can respond more quickly to changes.
Electrical Circuits
Understanding the role of inductors within electrical circuits helps grasp why series and parallel configurations matter. In simple terms, inductors are components that store energy in the form of magnetic fields, and their behavior is dependent on how they're connected within the circuit.
  • Series Configuration: In this setup, the inductors' magnetic fields accumulate, leading to increased energy storage capacity, useful for applications needing slow response to changes.
  • Parallel Configuration: Conversely, this setup divides the current among multiple pathways, thereby reducing effective inductance and allowing for faster circuit responses.
This fundamental knowledge assists in designing circuits to achieve desired electrical behaviors. Whether to store energy efficiently or to react swiftly to changes, choosing between series or parallel configurations directly impacts your electric circuit’s performance.

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Most popular questions from this chapter

(II) A 2.5 -k\Omega resistor in series with a \(420 - \mathrm { mH }\) inductor is driven by an ac power supply. At what frequency is the impedance double that of the impedance at 60\(\mathrm { Hz }\) ?

(I) Calculate the reactance of, and rms current in, a 36.0 -mH radio coil connected to a \(250 - \mathrm { V } ( \mathrm { rms } ) 33.3 - \mathrm { kHz }\) ac line. Ignore resistance.

At \(t=0,\) the current through a \(60.0-\mathrm{mH}\) inductor is \(50.0 \mathrm{~mA}\) and is increasing at the rate of \(78.0 \mathrm{~mA} / \mathrm{s}\). What is the initial energy stored in the inductor, and how long does it take for the energy to increase by a factor of 5.0 from the initial value?

Show that the fraction of electromagnetic energy lost (to thermal energy) per cycle in a lightly damped \(\left(R^{2} \ll 4 L / C\right) L R C\) circuit is approximately $$\frac{\Delta U}{U}=\frac{2 \pi R}{L \omega}=\frac{2 \pi}{Q}$$ The quantity \(Q\) can be defined as \(Q=L \omega / R,\) and is called the \(Q\) -value, or quality factor, of the circuit and is a measure of the damping present. A high \(Q\) -value means smaller damping and less energy input required to maintain oscillations.

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