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Alternating-Current Bridges

Alternating-current bridges16 are used for measuring the effective resistance and inductance of inductors, the effective resistance and reactance of capacitors, and the input impedance of lines and networks. The condition of bridge balance usually is determined with a telephone receiver if the test frequency is audible. Above the audible-frequency range the null detector often is an amplifier and a rectifier. At higher frequencies, a radio-receiving set makes an excellent detector.17, 18

In the alternating-current bridge of Fig. 19 R1 and R2 are resistors with negligible reactance. Impedance Zs may be a standard inductor or capacitor, and Zx is the unknown inductor or capacitor to be measured.

Figure 19. An alternating-current bridge.

At balance, negligible current flows through the receiver, minimum tone is heard in the receiver, and the points across which the receiver is connected are at the same potential. Thus I1R1 = I2R2 in both magnitude and phase, and I1Zs = I2Zx in both magnitude and phase. Then,

When this equation is generalized,

Because resistances and reactances produce different effects, in balancing an alternating-current bridge, first one quantity, and then the other, is varied, until minimum tone of the test frequency is heard. The in-phase and out-of-phase terms of equation 39a must be separated as follows: RxR1 + jXxR1 = R2Rs + jXsR2, RxR1 = R2Rs, and XxR1 = XsR2. Solving for the unknown terms gives

where Rx is the effective resistance and Xx is the reactance of the unknown. When a standard inductor is used to measure an unknown inductor, equation 40 becomes

When a standard capacitor is used to measure an unknown capacitor, equation 40 becomes

in which the ratio R1/R2 is opposite from equation 41. In the preceding equations the units are ohms, henrys, farads, and cycles per second.

Bridge with Standard Inductor. The unknown inductor or capacitor of Fig. 20 is measured in terms of the standard inductor Ls and standard resistor Rs. The effective resistance of the standard inductor necessitates a correction. If Fig. 20 (a) is used, then Rs of equation 40 equals Rs of Fig. 20(a) plus the effective resistance of the standard inductor. If Fig. 20(b) is used, then the effective resistance of the unknown capacitor is Rx of equation 40 minus the effective resistance of the standard inductor.

Often, in bridges, such as Figs. 20(a) and (b), R1 and R2 each equal some value, such as 1000 ohms, and are fixed. Also, a resistor, equal to the effective resistance of the standard inductor, is connected in the arm opposite the standard variable inductor. The bridge of Fig. 20(a) then becomes direct reading; at balance the setting of Rs gives the effective resistance of the unknown inductor, and the setting of Ls gives the inductance of the unknown.

Figure 20. An alternating-current bridge using standard inductor Ls.

Figure 21. An alternating-current bridge using standard capacitor Cs.

The bridge of Fig. 20(6) is adjusted until the inductive reactance 2πfLs equals the capacitive reactance l/(2πfC). For this condition,

As will be noted, the frequency must be known.

Bridge with Standard Capacitor. For many purposes the standard capacitor of Fig. 21 has negligible effective resistance and no resistance corrections are necessary. For the circuit of Fig. 21(a), when resonance is obtained the value of the unknown inductor is

If the circuit of Fig. 21 (b) is used, then the capacitance of the unknown is given by equation 42.

Figure 22. Connections for determining mutual inductance.

Bridge Measurements of Mutual Inductance. The circuits of Figs. 20(a) and 21(a) are used for measuring mutual inductance. Suppose that the two coils of Fig. 22 are connected aiding so that the magnetic effects add; then, because of the mutual inductance between the coils, the back voltage between terminals 1 and 4 may be considered as composed of four components: (1) the back voltage caused by the self-inductance of coil A, (2) the voltage induced in coil A by the current in coil B and the mutual inductance between B and A, (3) the back voltage caused by the self-inductance of coil B, and (4) the voltage induced in coil B by the current in coil A and the mutual inductance between A and B. This total back voltage appears as an inductive effect at the terminals, and hence the equivalent self-inductance as measured by a bridge is

But if the coils are connected opposing so that the magnetic effects subtract, then the induced voltages caused by the mutual inductances subtract from those caused by the self-inductances, and

Subtracting equation 45a from equation 45 gives

The mutual inductance between two coils is, accordingly, one-fourth the difference in the inductance measured with the two coils aiding and the two coils opposing.

Figure 23. Circuit for measuring the incremental inductance Lx and the effective resistance Rx of a coil with a core of iron or other ferromagnetic material.

Bridge Measurements of Incremental Inductance. The magnitudes of the incremental inductance and effective resistance of an inductor with a ferromagnetic core will vary with the magnitude of both the direct and the alternating currents through the coil (see Self-Inductance). The bridge of Fig. 23 provides both direct current and alternating current. The direct current from the battery is regulated by resistance r and measured with the milliammeter. It is difficult to measure directly the magnitude of the alternating-current component, although it can be done. A high-impedance vacuum-tube voltmeter connected as indicated can be used to maintain constant the alternating voltage drop across the inductor. This is accomplished by varying the voltage divider R. Both the direct current and the alternating voltage must be kept constant as the bridge is balanced.

If the inductor has many turns (a filter choke for instance), then a low frequency of about 100 cycles must be used to avoid the effects of the distributed capacitance. For this reason, null detector D is often a tunable vacuum-tube amplifier and detector, or some similar device. It should be noted that resistor S must pass the direct-current component. At balance,16

where all values are in henrys, farads, and ohms, and ω equals 2π times the frequency.

Last Update: 2011-05-30