01 Self-Inductance in AC Circuit

01 Self-Inductance in AC Circuit#

Aim#

To show how alternating current depends on the value of self-inductance.

Subjects#

  • 5J10 (Self Inductance) 5L20 (LCR Circuits – AC)

Diagram#

../../../../../_images/figure_0213.png

Fig. 522 .#

Equipment#

  • Lamp, \(220\mathrm{~V}/200\mathrm{~W}\).

  • Coil, \(\mathrm{n}=500 ; \mathrm{R}=2.5 \Omega\).

  • U-core with bar.

  • 2 Demonstration meters.

  • Safety connection box .

  • Measuring junction box (See Figure 526).

  • Net-adapter for mobile telephone (or other appliance).

Safety#

  • It’s a circuit connected to mains voltage (\(220\mathrm{~V}/50\mathrm{~Hz}\)). That’s why we use a safety connection box. This box shows a green light when the mains is disconnected and a red light when the mains is connected. Self-inductance in AC-circuit

Presentation#

The circuit is build as shown in Figure 523 and in Diagram. First we show the circuit setup to the students and then connect the two Voltmeters.

../../../../../_images/figure_1168.png

Fig. 523 .#

  1. Connecting the \(220 \mathrm{~V}\) to the circuit makes the lamp glows strongly (see Figure 524A ). The Voltmeter connected to the lamp reads almost \(220 \mathrm{~V}\) : All voltage appears across the lamp; just a very little voltage is read across the coil.

    Conclusion is that only a very small emf of self-inductance is generated in the coil.

  2. The bar is partly shifted on to the U-core. As soon as the bar touches the second leg of the \(U\)-core the lamp dims (see figure \(2 B\) ). the Voltmeter across the lamp shows a lower voltage now and at the same time we observe an increase in voltage across the coil.

    Conclusion is that there is now a higher emf of self-inductance that opposes the \(220 \mathrm{~V}\).

../../../../../_images/figure_293.png

Fig. 524 .#

  1. When the bar is shifted completely on to the U-core, the lamp only glows very faintly. The voltage read across it is very low. The voltage across the coil is almost \(220 \mathrm{~V}\) now!

    Conclusion is that the emf of self-inductance generated in the coil is almost \(220 \mathrm{~V}\) now.

    Shifting the bar back and forth across the U-core makes the lamp dim less or more.

  2. Finally we disconnect the lamp. Now only the self-inductance is connected to the \(220 \mathrm{~V}\) (see Figure 525).

../../../../../_images/figure_324.png

Fig. 525 .#

Now the effect of self-inductance is most clear: the voltmeter reads \(220 \mathrm{~V}\) across the coil, and only a small current is flowing (we measure \(0.4 \mathrm{~A}\) ). When there would be no self-inductance, the current would be \(220 \mathrm{~V} / 2.5 \Omega=88 \mathrm{~A}\) !

Conclusion is that the emf of self-inductance really opposes the applied voltage. 5. The same demonstration is performed with a commercial net-adapter (used as charger for a mobile telephone; see Figure 526). Here also only the primary coil of the adapter is connected to the mains. We read a current of only \(0.3 \mathrm{~mA}\)!

../../../../../_images/figure_48.png

Fig. 526 .#

Explanation#

The emf induced in a coil is, from Faraday’s law: \(E=-N \frac{d \Phi_{B}}{d t}=-L \frac{d I}{d t} ; L\) being the coefficient of self-inductance. For a solenoid with a core \(\left(\mu_{r}\right)\) this is:

\(L=\frac{\mu_{r} \mu_{0} N^{2} A}{l} L=\frac{\mu_{r} \mu_{0} N^{2} A}{l}\). This shows that the higher \(L\), the higher the emf of self-inductance. Shifting the bar across the core changes \(L\), and so the induced emf.

Remarks#

  • The core on the bar makes a lot of noise. This is a \(100 \mathrm{~Hz}\) mains hum due to the mains frequency \((50 \mathrm{~Hz})\).

  • The effect of self-inductance can also be translated into impedance of the circuit. In our demonstration 4. the circuit shows an impedance of \(220 \mathrm{~V} / 0.4 \mathrm{~A}=550 \Omega\) instead of the \(2.5 \Omega\) of the copper coil.

  • In figure B we read \(V_{\text {coil }}=130 \mathrm{~V}\) and \(V_{\text {lamp }}=110 \mathrm{~V}\). Students easily read this as a total of \(240 \mathrm{~V}\), so higher than the applied \(220 \mathrm{~V}\). Phase-shift between these two voltages is responsible for that. The situation must be something like Figure 527 below shows.

../../../../../_images/figure_54.png

Fig. 527 .#

Sources#

  • Giancoli, D.G., Physics for scientists and engineers with modern physics, pag. 758-759 and 773-774.

  • Wolfson, R., Essential University Physics, pag. 474-477 and 491-492. 110V

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