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If there is no mutual inductance among two or more parallel-connected inductors, their values add up like the values of resistors in parallel. Suppose you have inductances L1, L2, L3, . . . , Ln all connected in parallel. Then you can find the reciprocal of the total inductance, 1/L, using the following formula: 1/L = 1/L1 + 1/L2 + 1/L3 + . . . + 1/Ln The total inductance, L, is found by taking the reciprocal of the number you get for 1/L. Again, as with inductances in series, it s important to remember that all the units have to agree during the calculation process. Once you have completed the calculation, you can convert the result to any inductance unit.
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Problem 10-3 Suppose there are three inductors, each with a value of 40 H, connected in parallel with no mutual inductance, as shown in Fig. 10-4. What is the net inductance of the combination Let s call the inductances L1 = 40 H, L2 = 40 H, and L3 = 40 H. Use the preceding formula to obtain 1/L = 1/40 + 1/40 + 1/40 = 3/40 = 0.075. Then L = 1/0.075 = 13.333 H. This should be rounded off to 13 H, because the original inductances are specified to only two significant digits. Problem 10-4 Imagine four inductors in parallel, with no mutual inductance and values of L1 = 75.0 mH, L2 = 40.0 mH, L3 = 333 H, and L4 = 7.00 H. What is the net inductance of this combination
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10-4 Inductances in parallel.
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You can use henrys, millihenrys, or microhenrys as the standard units in this problem. Suppose you decide to use henrys. Then L1 = 0.0750 H, L 2 = 0.0400 H, L 3 = 0.000333 H, and L4 = 7.00 H. Use the preceding formula to obtain 1/L = 13.33 + 25.0 + 3003 + 0.143 = 3041.473. The reciprocal of this is the inductance L = 0.00032879 H = 328.79 H. This should be rounded off to 329 H. This is only a little less than the value of the 333 H inductor alone. If there are several inductors in parallel, and one of them has a value that is much smaller than the values of all the others, then the total inductance is a little smaller than the value of the smallest inductor.
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In real-world circuits, there is almost always some mutual inductance between or among solenoidal coils. The magnetic fields extend significantly outside such coils, and mutual effects are difficult to avoid or eliminate. The same is true between and among lengths of wire, especially at high ac frequencies. Sometimes, mutual inductance has no detrimental effect, but in some situations it is not wanted. Mutual inductance can be minimized by using shielded wires and toroidal inductors. The most common shielded wire is coaxial cable. Toroidal inductors are discussed later in this chapter.
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Coefficient of Coupling The coefficient of coupling, symbolized k, is an expression of the extent to which two inductors interact. It is specified as a number ranging from 0 (no interaction) to 1 (the maximum possible interaction). Two coils separated by a sheet of solid iron, or by a great distance, have a coefficient of coupling of zero (k = 0); two coils wound on the same form, one right over the other, have the maximum possible coefficient of coupling (k = 1). Sometimes, the coefficient of coupling is multiplied by 100 and expressed as a percentage from 0 to 100 percent. Mutual Inductance The mutual inductance between two inductors is symbolized M, and is expressed in the same units as inductance: henrys, millihenrys, microhenrys, or nanohenrys. The value of M is a function of the values of the inductors, and also of the coefficient of coupling. In the case of two inductors having values of L1 and L2 (both expressed in the same size units), and with a coefficient of coupling equal to k, the mutual inductance M is found by multiplying the inductance values, taking the square root of the result, and then multiplying by k. Mathematically:
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M = k(L1L 2 )1/2 where the 1 2 power represents the square root. The value of M thus obtained will be in the same size unit as the values of the inductance you input to the equation.
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