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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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