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2021-10-21 17:42:05

Introduction to Inductive Reactance

**Catalog**

Ⅰ Introduction |

Ⅱ Detailed description |

Ⅲ Calculation formula |

Ⅳ Role of inductive reactance in the circuit |

When a current flows through the coil, an induced electromagnetic field will be formed in the coil, and the induced electromagnetic field will generate an induced current in the coil to resist the current passing through the coil. Therefore, we call the interaction between this current and the coil the **electrical inductance**, which is the inductance in the circuit.

Alternating current can also pass through the coil, but the inductance of the coil has an obstructive effect on the alternating current. This obstruction is called **inductive reactance**. The more difficult it is for alternating current to pass through the coil, the greater the inductance, the greater the obstructive effect of the inductance. The higher the frequency of the alternating current, the more difficult it is to pass through the coil, and the obstructive effect of inductance is also greater. Experiments have proved that inductive reactance is proportional to inductance and also proportional to frequency. If the inductive reactance is represented by XL, the inductance is represented by L, and the frequency is represented by f, then the calculation formula is:

*XL= 2πfL=ωL*

The unit of inductive reactance is the ohm. Knowing the frequency f (Hz) of the alternating current and the inductance L (H) of the coil, the inductive reactance can be calculated using the above formula. The unit of inductance is "Henry (H)". We can use the special properties of current and coil to make inductance devices of different sizes and values to form a circuit system network with different functions.

XL is the inductive reactance, the unit is ohms, ω is the angular velocity of the alternator, the unit is radians/second, f is the frequency, the unit is Hertz, and L is the coil inductance, the unit is Henry.

① When the alternating current passes through the circuit of the inductance coil, the self-induced electromotive force is generated in the circuit, which hinders the change of current and forms inductive reactance. The greater the self-inductance coefficient, the greater the self-inductance electromotive force, and the greater the inductive reactance. If the frequency of the alternating current is large, the rate of change of the current is also large, and the self-induced electromotive force must also be large, so the inductive reactance also increases with the frequency of the alternating current. The inductance in the alternating current is proportional to the frequency of the alternating current and the self-inductance of the inductance coil. In practical applications, the inductance plays the role of "blocking and passing through", so the characteristics of inductance are often used in AC circuits to bypass low-frequency and direct current and prevent high-frequency alternating current.

*The curve of inductive reactance and capacitive reactance*

② In a purely inductive circuit, the relationship between the alternating voltage (u) at both ends of the inductor and the self-induced electromotive force (εL) is u=-εL and εL =-Ldi/dt, so u=Ldi/dt. The sinusoidal alternating current changes periodically, and the self-induced electromotive force in the coil is constantly changing. When the current of sinusoidal alternating current is zero, the current rate of change is the largest, so the voltage is the largest. When the current is at the maximum, the current rate of change is the smallest, so the voltage is zero. It can be concluded that the voltage phase across the inductor leads the current phase by π/2.

In a purely inductive circuit, the frequency of current and voltage are the same. The impedance of an inductive component is the inductive reactance (XL=ωL=2πfL), which is proportional to both ω and L. When ω=0, XL =0, so the inductor plays the role of "passing DC and blocking AC" or "passing low frequency and blocking high frequency".

③ In a purely inductive circuit, the inductive reactance does not consume electric energy, because during any 1/4 cycle when the current increases from zero to the maximum value, the current in the circuit will generate a magnetic field near the coil. And the electric energy will be converted into magnetic field energy storage stored in a magnetic field. But in the next 1/4 cycle, the current changes from large to small, the magnetic field gradually weakens, and the stored magnetic field energy is converted into electrical energy and returned to the power supply, so the inductive reactance does not consume electrical energy (the resistance heating is ignored ).

Winding small voltage transformer, the calculation formula of inductive reactance is derived as follows:

*2πfL=R primary load (1)*

The primary load of R includes the impedance and inductive reactance of the primary coil of the transformer. Because I only need to wind about 10 turns, the impedance can be regarded as approximately 0; so the primary load of R is mainly caused by inductive reactance. Know the magnitude of R primary load and f (the frequency is known to be 500KHz), then:

*L = R primary load/(2πf) (2)*

So how to get the value of R primary load? This value is derived from the quiescent current and the primary voltage:

*R primary load = V primary / I static (3)*

The primary voltage is known, and the empirical value of the quiescent current (the current in the primary coil when the secondary is open) is:

*I static=5%*I primary full load (4)*

*I primary full load * V primary = I secondary full load * V secondary (5)*

Because the primary and secondary voltage ratios are known quantities, the full load value of the I primary can be known as long as the full load value of the I secondary is known. The voltage ratio between the primary and secondary voltages of the transformer I want to do is 1:1.2, and the I secondary full load is 200 mA. Then I primary full load = 240 mA, put this value into equation (4), you can find that I static is about 10 mA. V primary is a known quantity, where the primary voltage of my transformer is V primary = 5V. Substituting V primary = 5V and I static = 10 mA into equation (3), we get R primary load = 500 ohms. Substituting R primary load=500 ohms into equation (2), you can find:

**L=500/(2πf)=500/(2π*500000)=159 (microhenry)**

Inductance: "pass DC, block AC; pass low frequency, block high frequency"

From the causes of inductive reactance, we know: the inductance coil has no hindrance to the DC current, that is, it "passes DC and blocks AC".

From the expression of inductive reactance XL= 2πfL, it is known that an inductance coil with a large self-inductance coefficient will have obvious inductive reactance to an alternating current with a small frequency, let alone a high-frequency alternating current. We call this kind of inductance coil a low-frequency choke. As long as the AC passes through the low-frequency choke, there will be a greater inductance, and there is no hindrance to the DC. That is, the low-frequency choke "passes DC and blocks AC".

Inductance coils with a small self-inductance coefficient have very small inductance to alternating currents with small frequencies. Only when high-frequency alternating currents pass through can they have obvious inductive reactance. We call this type of coil a high-frequency choke. The high-frequency choke "passes low frequency and blocks high frequency".