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The maximum value obtained by connecting 4 identical inductances in series is \(16 \mathrm{H}\). What is the value of each inductor? What is the minimum value achieved when they are connected in parallel?

Short Answer

Expert verified
Each inductor has inductance 4 H, and the minimum inductance in parallel is 1 H.

Step by step solution

01

Understanding Series Connection

When inductors are connected in series, their inductances add up. If you have 4 inductors each with inductance \(L\), then the total inductance \(L_{total}\) in series is given by \(L_{total} = 4L\). We are told this total is \(16\) H.
02

Finding Inductance of One Inductor

From the equation \(4L = 16\), we solve for \(L\) by dividing both sides by \(4\). This gives \(L = \frac{16}{4} = 4\) H. Therefore, each inductor has an inductance of \(4\) H.
03

Understanding Parallel Connection

When inductors are connected in parallel, the reciprocal of the total inductance is the sum of reciprocals of the individual inductances. If each inductor has inductance \(L\), then for 4 inductors connected in parallel, the total inductance \(L_{parallel}\) is found using \(\frac{1}{L_{parallel}} = \frac{1}{L} + \frac{1}{L} + \frac{1}{L} + \frac{1}{L} = \frac{4}{L}\).
04

Finding the Total Inductance in Parallel

Using the expression from parallel connection, \(\frac{1}{L_{parallel}} = \frac{4}{4}\) because each inductor has an inductance of \(4\) H. Simplifying gives \(\frac{1}{L_{parallel}} = 1\), hence \(L_{parallel} = 1\) H.

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Key Concepts

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

Series Connection of Inductors
When you connect inductors in series, it's similar to adding up lengths of a rope. The total inductance is simply the sum of the individual inductances of each inductor. This happens because, in a series connection, the same amount of magnetic flux links through all inductors. Here's how to make a quick calculation for series inductors:
  • If you have multiple inductors and you know the inductance of each one, simply add them together. For example, if you have four inductors, each with an inductance of 4 H, your series inductance is calculated as 4H + 4H + 4H + 4H = 16H.
  • This series connection of inductors is analogous to connecting resistors in series, where you simply add their resistances as well.
Understanding this concept is also crucial in circuit design, helping ensure the correct amount of inductance in your electrical system.
Parallel Connection of Inductors
Connecting inductors in parallel is a bit more complex than connecting them in series. When inductors are in parallel, their total inductance is calculated via the reciprocals of their individual inductances. This is different from series connections because you subtract the inductive opposition instead of adding it. Here's a simple way to understand and calculate parallel inductance:
  • For multiple inductors in parallel, the formula to calculate the total inductance is the reciprocal of the sum of the reciprocals of each individual inductance. So, that looks like: \( \frac{1}{L_{total}} = \frac{1}{L_1} + \frac{1}{L_2} + \dots \)
  • For four identical inductors each with 4H connected in parallel, this becomes \( \frac{1}{L_{total}} = \frac{4}{4} \), which simplifies to \( L_{parallel} = 1H \).
This means that even though you have several inductors, the total inductance is surprisingly less than any of the individual inductors’ inductance on their own. As strange as it seems, this technique is used to decrease the total inductance in a circuit.
Electrical Engineering Problem Solving
Problem solving in electrical engineering often requires a systematic approach, and the exercise of calculating inductance is no exception.
To start, recognize whether you're dealing with a series or parallel configuration, as this significantly influences the calculation.
  • With series connections, remember to simply add up the individual values to get the total.
  • For parallel connections, don't forget to flip the equation around and deal with it in terms of reciprocals.
This exercise demonstrates the importance of understanding what configuration you’re working with. With this understanding, breaking down problems into simpler tasks can make complex calculations much more manageable.
Step-by-step techniques also help in reducing the chance of errors, as each step builds logically on the last. Practicing these techniques consistently is key to mastering electrical engineering problem solving, as it brings clarity and understanding to complex problems.

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Most popular questions from this chapter

A \(10000-\mu \mathrm{F}\) capacitor is initially in a discharged condition. How long will it take to charge it to a value of \(10 \mathrm{~V}\) if a steady current of \(100 \mu \mathrm{A}\) is used?

The voltage across a \(10-\mu \mathrm{H}\) inductance is given by \(v_{L}(t)=5 \sin \left(10^{6} t\right) \mathrm{V}\). The argument of the sine function is in radians. The initial current is \(i_{L}(0)=-0.5 \mathrm{~A}\). Find expressions for the current, power, and stored energy for \(t>0\). Sketch the waveforms to scale for time ranging from zero to \(2 \pi \mu \mathrm{s}\).

One type of microphone is formed from a parallel-plate capacitor arranged so the acoustic pressure of the sound wave affects the distance between the plates. Suppose we have such a microphone in which the plates have an area of \(10 \mathrm{~cm}^{2}\), the dielectric is air and the distance between the plates is a function of time given by $$ d(t)=100+0.3 \cos (1000 t) \mu \mathrm{m} $$ A constant voltage of \(200 \mathrm{~V}\) is applied to the plates. Determine the current through the capacitance as a function of time by using the approximation \(1 /(1+x) \cong 1-x\) for \(x<<1\). (The argument of the sinusoid is in radians.)

A \(20-\mu \mathrm{F}\) capacitor has a voltage given by $$ v(t)=3 \cos \left(10^{5} t\right)+2 \sin \left(10^{5} t\right) $$ Assume that the arguments of the sine and cosine functions are in radians. Find the power at \(t=0\) and state whether the power flow is into or out of the capacitor. Repeat for \(t_{2}=(\pi / 2) \times 10^{-5} \mathrm{~s}\).

We need to combine (in series or in parallel) an unknown inductance \(L\) with a second inductance of \(4 \mathrm{H}\) to attain an equivalent inductance of \(3 \mathrm{H}\). Should \(L\) be placed in series or in parallel with the original inductance? What value is required for \(L\) ?

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