INDUCTANCE Self Inductance

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Millicent Grant
5 years ago
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1 NDUTANE 3. Self nductance onsider the circuit shown in the Figure. When the switch is closed the current, and so the magnetic field, through the circuit increases from zero to a specific value. The increasing magnetic flux induces an S R emf. By enz's law, this induced emf opposes the change in flux. The effect of this induced emf is to retard the change of the original current, that is, retard its increasing. The same phenomena occurred when the switch is opened where the current in this case decreases from a specific value to zero. The emf induced due to the decreasing of the magnetic flux now ε tends to oppose the decreasing of the original current. This phenomena is called the self induction since the changing flux through the circuit arises from the circuit itself. The emf induced due to this phenomena is called the self-induced emf. f the emf induced in a circuit is due to the changing of the magnetic flux set up by another circuit we have the mutual induction phenomena. To obtain a quantitative description of the self induction, we know from Faraday's law that the induced emf is proportional to the time rate of the magnetic flux, i.e., dφ ε N m 3. But Φ m B and B ε d 3. The proportionality constant is called the self-inductance, or simply the inductance of the coil. The S unit of inductance is Henry (H), which, from Equation 3., is equivalent to V.s H A

2 Now comparing Eqs. 3. & 3. NΦm 3.3 As it is clear from Eq. 3.3 depends on the geometric features of the coil. t should be noted that all elements in a circuit have some inductance but it is too small to be significant except that of a coil. A coil that has significant inductance is called inductor, and is represented in the circuits by the symbol Example 3. Find the inductance of an ideal solenoid of N turns and length l. N Solution Knowing that, inside the solenoid B is uniform and given by B µ o n µ o l Φ BAcos 0 o m µ Now using Eq. 3.3 N N µ µ o A l N l o N l A A 3. R ircuits To study explicitly the effect of self induction in a circuit we refer to the circuit shown. Suppose that the switch is thrown tom point at t 0. Applying Kirchhoff's loop rule to the circuit at time t we get ε S R d ε R t is not difficult to verify that the solution of the differential equation given in Equation 3.4 is t e τ max 3.5 with the maximum current is

3 ε max 3.6 R and the time constant of the R circuit is τ 3.7 R From Equation 3.5 we conclude that at t 0, 0, while max as t. This means that: the inductor acts as an open circuit at t 0 and acts as an ordinary wire after a long time. f the battery is suddenly removed, by throwing the switch to point in the circuit and applying Kirchhoff's rule again we get d R + 0 t τ max e 3.8 The relations of Equations 3.5 and 3.8 are plotted in as a function of time. As it is clear from the graph (a), the current takes some time to reach its maximum value. The graph of Figure (b) tells that the current takes some time to reach it zero value. n another word, the inductor has the effect to hinder the current from reaching its final value for some time. m m t t (a) (b) (a) The current versus time in an R circuit when connected to a battery. (b) The current versus time in an R circuit when the battery is disconnected.

4 Example 3.3 onsider the circuit shown, find a) the time constant of the circuit, b) the current in the circuit at t. 0 ms, and c) compare the P.D across the resistor with that across the inductor. S 6 Ω V 30 mh Solution a) the Equation The time constant is given by τ R 6.0 b) The current is ms 5 max t e τ e 0.66A 6 c) The P.D. across the resistor is given by V t R R e τ o R While the P.D. across the inductor is given by V d o e τ t τ o Re t τ VR+ V or ε V

5 Example (problem 3). The switch in the figure is open for t <0 and then closed at time t 0. Find the currents if the circuit at t0 and a long time after closing the switch Ω 8 Ω 0 V S 4 Ω Ω H Solution At t0, the inductor treated as an open circuit A After a long time the inductor treated as a wire R eq Ω 0 eq. 5 A A and 0. 5 A 3

6 3.3 Energy in Magnetic Field Multiplying Eq. 3.4 by ε R d 0 The st term represents the power of the battery, while the nd term represents the power delivered to the resistor the 3 rd term represents the power delivered to the inductor, i.e., du d P du 0 d U m Oscillations in an ircuit onsider the circuit shown with the capacitor is charged with Q max. After closing S the charge will flow through the inductor. At some time let the charge in the capacitor to be q and the current in the inductor to be. The total energy in the circuit at this time is S U total U + U q + Deriving the above Eq. with respect to time But du total + dq d du total 0 q dq d + 0 Knowing that dq d d q and

7 q d q + 0 d q + q 0 q Q cos t max ( ω + γ ) To find the constant γ we know that q Q at t 0 γ 0 max ( ωt ) q Q cos 3.9 max With ω 3.0 Now With dq Qmaxω sin ωt max sin ωt 3. ω 3. max Q max Example 3.8 onsider the ircuit show. First S is open and S is closed such that the capacitor is charged. Now if S is opened to remove the battery and then S is closed to connect the capacitor with the inductor. a) Find ω of the circuit. b) Find Q max and max. c) Find (t) and Q(t). Solution a) The frequency is given by V 9 pf S ω 3 (.8 0 )( 9 0 ) b) The maximum charge on the capacitor is the initial charge before opening S, i.e., 6 Hz.8 mh S

8 Now Q max ( 9 0 ) ε 6 max ω Q 0 max c) q( t) Q cos( t) 0 max ω.08 0 cos( t) 6 q( t) sin( t ) max ω 0 sin( t) A

Inductance, RL Circuits, LC Circuits, RLC Circuits

Inductance, RL Circuits, LC Circuits, RLC Circuits Inductance, R Circuits, C Circuits, RC Circuits Inductance What happens when we close the switch? The current flows What does the current look like as a function of time? Does it look like this? I t Inductance