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Chapter 20: Problem 52

Calculate the equivalent inductance \(L_{\mathrm{eq}}\) of two ideal inductors, \(L_{1}\) and \(L_{2},\) connected in series in a circuit. Assume that their mutual inductance is negligible. [Hint: Imagine replacing the two inductors with a single equivalent inductor \(L_{\mathrm{cq}} .\) How is the emf in the series equivalent related to the emfs in the two inductors? What about the currents?]

Short Answer

Expert verified
Answer: The equivalent inductance of two ideal inductors connected in series is the sum of their individual inductances, \(L_{\mathrm{eq}} = L_{1} + L_{2}\).

Step by step solution

01

Write down the expression for the emf induced in each inductor and Faraday's law

The emf induced in an inductor is given by \(E = -L\frac{dI}{dt}\), where \(L\) is the inductance and \(\frac{dI}{dt}\) is the rate of change of current with time. According to Faraday's law, the total emf \(E_{\mathrm{tot}}\) is the sum of the emfs induced by the two inductors. So, we have: \(E_{\mathrm{tot}} = E_{1} + E_{2} = -L_{1}\frac{dI}{dt} - L_{2}\frac{dI}{dt}\)
02

Write down the expression for the emf induced in the equivalent inductor

If we replace the two inductors with a single equivalent inductor \(L_{\mathrm{eq}}\), the emf induced in this inductor is given by: \(E_{\mathrm{eq}} = -L_{\mathrm{eq}}\frac{dI}{dt}\)
03

Relate the emf in the equivalent inductor to the emfs in the two inductors

Since the mutual inductance between the two inductors is negligible, the emf induced in each inductor doesn't affect the other, which means that we can safely replace the inductors with an equivalent one, as long as the total emf remains unchanged. So, we have: \(E_{\mathrm{eq}} = E_{\mathrm{tot}}\)
04

Write down the expression for the current through each inductor

Since the inductors are connected in series, the current flowing through them is the same, which means that: \(I_{1} = I_{2} = I\)
05

Find the equivalent inductance

Now, we can use the relationship between emfs to find the equivalent inductance: \(-L_{\mathrm{eq}}\frac{dI}{dt} = -L_{1}\frac{dI}{dt} - L_{2}\frac{dI}{dt}\) Divide both sides by \(- \frac{dI}{dt}\): \(L_{\mathrm{eq}} = L_{1} + L_{2}\) So, the equivalent inductance \(L_{\mathrm{eq}}\) of the two ideal inductors connected in series is the sum of their individual inductances.

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Most popular questions from this chapter

In this problem, you derive the expression for the self inductance of a long solenoid [Eq. \((20-15 a)] .\) The solenoid has \(n\) turns per unit length, length \(\ell,\) and radius \(r\) Assume that the current flowing in the solenoid is \(I\) (a) Write an expression for the magnetic field inside the solenoid in terms of \(n, \ell, r, I,\) and universal constants. (b) Assume that all of the field lines cut through each turn of the solenoid. In other words, assume the field is uniform right out to the ends of the solenoid-a good approximation if the solenoid is tightly wound and sufficiently long. Write an expression for the magnetic flux through one turn. (c) What is the total flux linkage through all turns of the solenoid? (d) Use the definition of self-inductance [Eq. \((20-14)]\) to find the self inductance of the solenoid.
A TV tube requires a 20.0 -kV-amplitude power supply. (a) What is the turns ratio of the transformer that raises the 170 -V-amplitude household voltage to \(20.0 \mathrm{kV} ?\) (b) If the tube draws 82 W of power, find the currents in the primary and secondary windings. Assume an ideal transformer.
In Section \(20.9,\) in order to find the energy stored in an inductor, we assumed that the current was increased from zero at a constant rate. In this problem, you will prove that the energy stored in an inductor is \(U_{\mathrm{L}}=\frac{1}{2} L I^{2}-\) that is, it only depends on the current \(I\) and not on the previous time dependence of the current. (a) If the current in the inductor increases from \(i\) to \(i+\Delta i\) in a very short time \(\Delta t,\) show that the energy added to the inductor is $$\Delta U=L i \Delta i$$ [Hint: Start with \(\Delta U=P \Delta t .]\) (b) Show that, on a graph of \(L i\) versus \(i,\) for any small current interval \(\Delta i,\) the energy added to the inductor can be interpreted as the area under the graph for that interval.(c) Now show that the energy stored in the inductor when a current \(I\) flows is \(U=\frac{1}{2} L I^{2}\)
The time constant \(\tau\) for an \(L R\) circuit must be some combination of $L, R,\( and \)\mathscr{E} .$ (a) Write the units of each of these three quantities in terms of \(\mathrm{V}, \mathrm{A},\) and \(\mathrm{s}\) (b) Show that the only combination that has units of seconds is \(L / R\)
A 100 -turn coil with a radius of \(10.0 \mathrm{cm}\) is mounted so the coil's axis can be oriented in any horizontal direction. Initially the axis is oriented so the magnetic flux from Earth's field is maximized. If the coil's axis is rotated through \(90.0^{\circ}\) in \(0.080 \mathrm{s},\) an emf of $0.687 \mathrm{mV}$ is induced in the coil. (a) What is the magnitude of the horizontal component of Earth's magnetic field at this location? (b) If the coil is moved to another place on Earth and the measurement is repeated, will the result be the same?
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