 AC Bridges # AC Bridges

In this chapter, let us discuss about the AC bridges, which can be used to measure inductance. AC bridges operate with only AC voltage signal. The circuit diagram of AC bridge is shown in below figure. As shown in above figure, AC bridge mainly consists of four arms, which are connected in rhombus or square shape. All these arms consist of some impedance.

The detector and AC voltage source are also required in order to find the value of unknown impedance. Hence, one of these two are placed in one diagonal of AC bridge and the other one is placed in other diagonal of AC bridge. The balancing condition of Wheatstone’s bridge as −

R4=R2R3R1R4=R2R3R1

We will get the balancing condition of AC bridge, just by replacing R with Z in above equation.

Z4=Z2Z3Z1Z4=Z2Z3Z1

Z1Z4=Z2Z3⇒Z1Z4=Z2Z3

Here, Z1Z1 and Z2Z2 are fixed impedances. Whereas, Z3Z3 is a standard variable impedance and Z4Z4 is an unknown impedance.

Note − We can choose any two of those four impedances as fixed impedances, one impedance as standard variable impedance & the other impedance as an unknown impedance based on the application.

Following are the two AC bridges, which can be used to measure inductance.

• Maxwell’s Bridge
• Hay’s Bridge

Now, let us discuss about these two AC bridges one by one.

## Maxwell’s Bridge

Maxwell’s bridge is an AC bridge having four arms, which are connected in the form of a rhombus or square shape. Two arms of this bridge consist of a single resistor, one arm consists of a series combination of resistor and inductor & the other arm consists of a parallel combination of resistor and capacitor.

An AC detector and AC voltage source are used to find the value of unknown impedance. Hence, one of these two are placed in one diagonal of Maxwell’s bridge and the other one is placed in other diagonal of Maxwell’s bridge.

Maxwell’s bridge is used to measure the value of medium inductance. The circuit diagram of Maxwell’s bridge is shown in the below figure. In above circuit, the arms AB, BC, CD and DA together form a rhombus or square shape. The arms AB and CD consist of resistors, R2R2 and R3R3 respectively. The arm, BC consists of a series combination of resistor, R4R4 and inductor, L4L4. The arm, DA consists of a parallel combination of resistor, R1R1 and capacitor, C1C1.

Let, Z1,Z2,Z3Z1,Z2,Z3 and Z4Z4 are the impedances of arms DA, AB, CD and BC respectively. The values of these impedances will be

Z1=R1(1jωC1)R1+1jωC1Z1=R1(1jωC1)R1+1jωC1
Z1=R11+jωR1C1⇒Z1=R11+jωR1C1

Z2=R2Z2=R2

Z3=R3Z3=R3

Z4=R4+jωL4Z4=R4+jωL4

Substitute these impedance values in the following balancing condition of AC bridge.

Z4=Z2Z3Z1Z4=Z2Z3Z1
R4+jωL4=R2R3(R11+jωR1C1)R4+jωL4=R2R3(R11+jωR1C1)

R4+jωL4=R2R3(1+jωR1C1)R1⇒R4+jωL4=R2R3(1+jωR1C1)R1

R4+jωL4=R2R3R1+jωR1C1R2R3R1⇒R4+jωL4=R2R3R1+jωR1C1R2R3R1

R4+jωL4=R2R3R1+jωC1R2R3⇒R4+jωL4=R2R3R1+jωC1R2R3

By comparing the respective real and imaginary terms of above equation, we will get

R4=R2R3R1R4=R2R3R1Equation 1

L4=C1R2R3L4=C1R2R3Equation 2

By substituting the values of resistors R1R1R2R2 and R3R3 in Equation 1, we will get the value of resistor, R4R4. Similarly, by substituting the value of capacitor, C1C1 and the values of resistors, R2R2 and R3R3 in Equation 2, we will get the value of inductor, L4L4.

The advantage of Maxwell’s bridge is that both the values of resistor, R4R4 and an inductor, L4L4 are independent of the value of frequency.

## Hay’s Bridge

Hay’s bridge is a modified version of Maxwell’s bridge, which we get by modifying the arm, which consists of a parallel combination of resistor and capacitor into the arm, which consists of a series combination of resistor and capacitor in Maxwell’s bridge.

Hay’s bridge is used to measure the value of high inductance. The circuit diagram of Hay’s bridge is shown in the below figure. In above circuit, the arms AB, BC, CD and DA together form a rhombus or square shape. The arms, AB and CD consist of resistors, R2R2 and R3R3 respectively. The arm, BC consists of a series combination of resistor, R4R4 and inductor, L4L4. The arm, DA consists of a series combination of resistor, R1R1 and capacitor, C1C1.

Let, Z1,Z2,Z3Z1,Z2,Z3 and Z4Z4 are the impedances of arms DA, AB, CD and BC respectively. The values of these impedances will be

Z1=R1+1jωC1Z1=R1+1jωC1

Z1=1+jωR1C1jωC1⇒Z1=1+jωR1C1jωC1

Z2=R2Z2=R2

Z3=R3Z3=R3

Z4=R4+jωL4Z4=R4+jωL4

Substitute these impedance values in the following balancing condition of AC bridge.

Z4=Z2Z3Z1Z4=Z2Z3Z1

R4+jωL4=R2R3(1+jωR1C1jωC1)R4+jωL4=R2R3(1+jωR1C1jωC1)

R4+jωL4=R2R3jωC1(1+jωR1C1)R4+jωL4=R2R3jωC1(1+jωR1C1)

Multiply the numerator and denominator of right hand side term of above equation with 1jωR1C11−jωR1C1.

R4+jωL4=R2R3jωC1(1+jωR1C1)×(1jωR1C1)(1jωR1C1)⇒R4+jωL4=R2R3jωC1(1+jωR1C1)×(1−jωR1C1)(1−jωR1C1)

R4+jωL4=ω2C12R1R2R3+jωR2R3C1(1+ω2R12C12)⇒R4+jωL4=ω2C12R1R2R3+jωR2R3C1(1+ω2R12C12)

By comparing the respective real and imaginary terms of above equation, we will get

R4=ω2C12R1R2R3(1+ω2R12C12)R4=ω2C12R1R2R3(1+ω2R12C12)Equation 3

L4=R2R3C1(1+ω2R12C12)L4=R2R3C1(1+ω2R12C12)Equation 4

By substituting the values of R1,R2,R3,C1R1,R2,R3,C1 and ωω in Equation 3 and Equation 4, we will get the values of resistor, R4R4 and inductor, L4L4.  