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CHAPTER 7: MAGNETICALLY COUPLED CIRCUIT 1 SUB - TOPICS SELF AND MUTUAL INDUCTANCE. COUPLING COEFFICIENT (K) DOT DETERMINATION 2 OBJECTIVES To understand the basic concept of self inductance and mutual inductance. To understand the concept of coupling coefficient and dot determination in circuit analysis. 3 SELF AND MUTUAL INDUCTANCE When two loops with or without contacts between them affect each other through the magnetic field generated by one of them, it called magnetically coupled. Example: transformer An electrical device designed on the basis of the concept of magnetic coupling. Used magnetically coupled coils to transfer energy from one circuit to another. 4 a) Self Inductance It called self inductance because it relates the voltage induced in a coil by a time varying current in the same coil. Consider a single inductor with N number of turns when current, i flows through the coil, a magnetic flux, Φ is produces around it. + i(t) Φ V _ Fig. 1 5 According to Faraday’s Law, the voltage, v induced in the coil is proportional to N number of turns and rate of change of the magnetic flux, Φ; d vN .......(1) dt But a change in the flux Φ is caused by a change in current, i. Hence; d d di .......(2) dt di dt 6 Thus, (2) into (1) yields; d di vN .......(3) di dt or di v L .......(4) dt From equation (3) and (4) the self inductance L is define as; d H........(5) LN di The unit is in Henrys (H) 7 b) Mutual Inductance When two inductors or coils are in close proximity to each other, magnetic flux caused by current in one coil links with the other coil, therefore producing the induced voltage. Mutual inductance is the ability of one inductor to induce a voltage across a neighboring inductor. 8 Consider the following two cases: Case 1: two coil with self – inductance L1 and L2 which are in close proximity which each other (Fig. 2). Coil 1 has N1 turns, while coil 2 has N2 turns. + i1(t) L1 Φ12 V1 _ L2 + V2 Φ11 _ N1 turns N2 turns Fig. 2 9 Magnetic flux Φ1 from coil 1 has two components; * Φ11 links only coil 1. * Φ12 links both coils. Hence; Φ1 = Φ11 + Φ12 ……. (6) Thus; Voltage induces in coil 1 d11 di1 di1 v1 N1 L1 .......(7) di1 dt dt 10 Voltage induces in coil 2 d12 di1 di1 v2 N 2 M 21 .......(8) di1 dt dt Subscript 21 in M21 means the mutual inductance on coil 2 due to coil 1 11 Case 2: Same circuit but let current i2 flow in coil 2. + Φ21 L1 L2 + V1 _ N1 turns V2 Φ22 i2(t) _ N2 turns Fig. 3 The magnetic flux Φ2 from coil 2 has two components: * Φ22 links only coil 2. * Φ21 links both coils. Hence; Φ2 = Φ21 + Φ22 ……. (9) 12 Thus; Voltage induced in coil 2 d22 di2 di2 v2 N 2 L2 .......(10) di2 dt dt Voltage induced in coil 1 d21 di2 di2 v1 N1 M 12 .......(11) di2 dt dt Subscript 12 in M12 means the Mutual Inductance on coil 1 due to coil 2 13 Since the two circuits and two current are the same: M 21 M12 M Mutual inductance M is measured in Henrys (H) 14 COUPLING COEFFICIENT (k) It is measure of the magnetic coupling between two coils. Range of k : 0 ≤ k ≤ 1 • k = 0 means the two coils are NOT COUPLED. • k = 1 means the two coils are PERFECTLY COUPLED. • k < 0.5 means the two coils are LOOSELY COUPLED. • k > 0.5 means the two coils are TIGHTLY COUPLED. 15 k depends on the closeness of two coils, their core, their orientation and their winding. The coefficient of coupling, k is given by; k M L1 L2 or M k L1L2 16 DOT DETERMINATION Required to determine polarity of “mutual” induced voltage. A dot is placed in the circuit at one end of each of the two magnetically coupled coils to indicate the direction of the magnetic flux if current enters that dotted terminal of the coil. 17 Φ12 Φ21 Φ11 Coil 1 Φ22 Coil 2 18 Dot convention is stated as follows: if a current ENTERS the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is POSITIVE at the dotted terminal of the second coil. Conversely, Dot convention may also be stated as follow: if a current LEAVES the dotted terminal of one coil, the reference polarity of the mutual voltage in the second coil is NEGATIVE at the dotted terminal of the second coil. 19 The following dot rule may be used: i. when the assumed currents both entered or both leaves a pair of couple coils by the dotted terminals, the signs on the L – terms. ii. if one current enters by a dotted terminals while the other leaves by a dotted terminal, the sign on the M – terms will be opposite to the signs on the L – terms. 20 Once the polarity of the mutual voltage is already known, the circuit can be analyzed using mesh method. Application of the dot convention Example 1 M i1(t) + V1 _ + L1 L2 V2 (t) = M di1/dt _ The sign of the mutual voltage v2 is determined by the reference polarity for v2 and the direction of i1. Since i1 enters the dotted terminal of coil 1 and v2 is positive at the dotted terminal of coil 2, the mutual voltage is M di1/dt 21 Example 2 M i1(t) + V1 _ + L1 L2 V2 (t) = -M di1/dt _ Current i1 enters the dotted terminal of coil 1 and v2 is negative at the dotted terminal of coil 2. the mutual voltage is –M di1/dt 22 Same reasoning applies to the coil in example 3 and example 4. Example 3 M i2(t) + + V1= -M di2/dt L1 L2 _ V2 (t) _ Example 4 M i2(t) + V1= M di2/dt _ + L1 L2 V2 (t) _ 23 Dot convention for coils in series M i i L1 (+) Series – aiding connection L2 L L1 L2 2M M Series – opposing connection i i L1 (-) L2 L L1 L2 2M 24 Below are examples of the sets of equations derived from basic configurations involving mutual inductance M Circuit 1 R1 ja Vs + I1 jb R2 I2 R3 Solution: KVL I1 : ( R1 R2 ja) I1 MI 2 Vs .......(1) KVL I 2 : R2 I1 ( R2 R3 jb) I 2 MI1 0.......(2) 25 Circuit 2 ja R1 R2 M Vs + I1 jb I2 -jc Solution: KVL I1 : ( R1 ja jb) I1 jbI 2 M ( I1 I 2 ) MI1 Vs .......(1) KVL I 2 : jbI1 ( R2 jb jc) I 2 MI1 0.......(2) 26 Circuit 3 R1 jb M Vs + I1 R2 ja I2 Solution: KVL I1 : ( R1 ja) I1 jaI 2 MI 2 Vs .......(1) KVL I 2 : jaI1 ( R2 ja jb) I 2 MI 2 M ( I 2 I1 ) 0.......(2) 27 Circuit 4 -jc R1 Vs + I1 ja I2 R2 jb M Solution: KVL I1 : ( R1 R2 jc) I1 R2 I 2 Vs .......(1) KVL I 2 : R2 I1 ( R2 ja jb) I 2 2MI 2 0.......(2) 28 Circuit 5 M3 R1 ja jb jc Vs + M2 M1 -jd I1 R2 I2 Solution: KVL I1 : ( R1 R2 ja jc) I1 ( R2 jc) I 2 M 3 I 2 M 1 ( I1 I 2 ) M 1 I1 M 2 I 2 Vs .......(1) KVL I 2 : ( R2 jc) I1 ( R2 jc jb jd ) I 2 M 1 I1 M 2 I 2 M 3 I1 M 2 ( I 2 I1 ) 0.......(2) 29 Example 1 Calculate the mesh currents in the circuit shown below -j3Ω 4Ω j8Ω j2Ω 100V + I1 5Ω j6Ω I2 30 Solution KVL I1 : (4 j 3) I1 j 6 I 2 j 2 I 2 100 (4 j 3) I1 j8I 2 100.......(1) KVL I 2 : j 6 I1 (5 j14) I 2 j 2 I 2 j 2( I 2 I1 ) 0 j8I1 (5 j18) I 2 0.......(2) In matrix form; 4 j3 j8 I1 100 j8 5 j18 I 0 2 31 The determinants are: 4 j3 j8 j8 5 j18 100 j8 0 5 j18 1 2 4 j 3 100 j8 0 30 j87 500 j1800 j800 Hence: 1 I1 20.33.5 A 2 I2 8.719 A 32 Example 2 Determine the voltage Vo in the circuit shown below. 5Ω j3Ω + 10V + j2Ω I1 j6Ω Vo -j4Ω I2 _ 33 Solution KVL I1 : (5 j9) I1 j 6 I 2 j 2( I1 I 2 ) j 2 I1 10 (5 j5) I1 j 4 I 2 10.......(1) KVL I 2 : j 6 I1 j 2 I 2 j 2 I1 0 j 4 I1 j 2 I 2 0.......(2) In matrix form; 5 j5 j 4 I1 10 j4 j 2 I 2 0 34 Answer: I1 1.47 j 0.88 I 2 2.94 j1.76 V0 j 6( I1 I 2 ) j 2 I1 or Vo j 6( I 2 I1 ) j 2 I1 or Vo j 4 I 2 hence, Vo 7.04 j11.76 35 Example 3 Calculate the phasor currents I1 and I2 in the circuit below. j3Ω -j4Ω 120 V + I1 j5Ω j6Ω I2 12Ω 36 Solution For coil 1, KVL gives -12 + (-j4+j5)I1 – j3I2 = 0 Or 1 jI1 – j3I2 = 12 For coil 2, KVL gives -j3I1 + (12 + j6)I2 = 0 Or I1 = (12 + j6)I2 = (2 – j4)I2 j3 2 37 Substituting 2 into 1 : (j2 + 4 – j3)I2 = (4 – j)I2 = 12 Or 12 I2 2.91 14.04 A 4- j From eqn. 2 and 3 3 , I1 (2 - j4)I 2 (4.472 - 63.43 ) (2.9114.04 ) 13.01 - 49.39 A 38