Simple AC Bridge Circuit Formula and Theory of Operation

An ac bridge circuit is one of the simple application of RLC ac circuits along with phase shifter.

This bridge circuit is used in measuring the inductance L of an inductor or the capacitance C of a capacitor.

Make sure to read what is ac circuit first.

Make sure to read:

1. What is phasor
2. Impedance and admittance
3. Kirchhoff’s laws for ac circuit
4. Power calculation in ac circuit
5. Three phase circuit

And its applications:

1. Phase shifter circuit and formula
2. AC op-amp
3. Capacitance multiplier circuit
4. Wien bridge oscillator

AC Bridge Circuit

It is similar in form to the Wheatstone bridge for measuring an unknown resistance. To measure L and C, however, an ac source is required as well as an ac meter instead of the galvanometer.

Consider the general ac bridge circuit displayed in Figure.(1).

Figure 1. A general ac bridge

The bridge is balanced when no current flows through the meter. This means that V₁ = V₂. Applying the voltage division principle,

(1)

Hence,

(2)

or

(3)

This is the balanced equation for the ac bridge.

Specific ac bridges for measuring L and C are drawn in Figure.(2), where L_x and C_x are the unknown inductance and capacitance to be measured while L_s and C_s are a standard inductance and capacitance (the values of which are known to great precision).

In each case, two resistors, R₁ and R₂, are varied until the ac meter reads zero. Then the bridge is balanced. From Equation.(3), we get

(4)

and

(5)

Notice that the balancing of the ac bridges in Figure.(2) does not depend on the frequency f of the ac source since f does not appear in the relationships in Equations.(4) and (5).

Figure 2. Spesific ac bridges: (a) for measuring L, (b) for measuring C

 

Read also : laplace transform circuit element

AC Bridge Circuit Example

1. The ac bridge circuit of Figure.(1) balances when Z₁ is a 1 kΩ resistor, Z₂ is a 4.2 kΩ resistor, Z₃ is a parallel combination of a 1.5 MΩ resistor and a 12 pF capacitor, and f = 2kHz. Find: (a) the series components that makeup Z_x, and (b) the parallel components that makeup Z_x.

Solution :

From Equation.(3),

(1.1)

where Z_x = R_x + jX_x,

(1.2)

and

or

(1.3)

(a) Assuming that Z_x is made up of series components, we substitute Equations.(1.2) and (1.3) in (1.1) and get

(1.4)

Equating the real and imaginary parts yields R_x = 5.993 MΩ and a capacitive reactance

or

(b) Z_x remains the same as in Equation.(1.4) but R_x and X_x are in parallel. Assuming an RC parallel combination,

By equating the real and imaginary parts, we get