## Heaviside-Campbell Equal Ratio Bridge

It is a mutual induction bridge and is used for measuring self-inductance over a wide range in terms of mutual inductometer readings. The connections for Heaviside’s bridge employing a standard variable mutual inductance are shown in figure 1. The primary of the mutual inductometer is inserted in the supply circuit and the secondary having self-inductance L_{2} and resistance R_{2} is put in arm 2 of the bridge. The unknown inductive impedance having self inductance of L_{1} and resistance R_{1} is placed in arm 1. The other two arms have pure resistances of R_{3} and R_{4}.

Balance is obtained by varying mutual inductance M and resistance R_{3} and R_{4}.

For balance, **I _{3}R_{3} = I_{2}R_{4}……………………..(i)**

**I _{1}(R_{1} + jωL_{1}) = I_{2}(R_{2} + jωL_{2}) + jωMI………..(ii)**

Since **I = I _{1} + I_{2}**, hence putting the value of I in equation (ii), we get

**I _{1}[R_{1} + jω(L_{1}-M)] = I_{2}[R_{2} + jω(L_{2} + M)]……….(iii)**

Dividing equation (iii) by (i), we have

**R _{3}[R_{2} + jω(L_{2} + M)] = R_{4}[R_{1} + jω(L_{1}-M)]**

Equation the reals and imaginaries, we have **R _{2}R_{3} = R_{1}R_{4}…………(iv)**

Also, **R _{3}(L_{2} + M) = R_{4}(L_{1}-M).**

If **R _{3} = R_{4}, then L_{2} + M = (L_{1} - M)**

Therefore, **L _{1 }- L_{2} = 2M……………..(v)**

This bridge, as modified by Campbell, is shown in figure 2. Here **R _{3} = R_{4}.** A balancing coil or a test coil of self-inductance equal to the self-inductance L

_{2}of the secondary of the inductometer and of resistance slightly greater then R

_{2}is connected in series with the unknown inductive impedance (R

_{1}and L

_{1}) in arm 1. A non-inductive resistance box along with a constant-inductance rheostat is also introduced in arm 2 as shown.

Balance is obtained by varying M and r. Two reading are taken; one when Z_{1} is in circuit and second when Z_{1} is removed or short-circuited across its terminals.

With unknown impedance Z_{1} still in circuit, suppose for balance the values of mutual inductance and r are M_{1} and r_{1}. With Z_{1} short-circuited, let these values be M_{2} and r_{2}. Then

**L _{1} = 2(M_{1 }– M_{2}) and R_{1} = r_{1} – r_{2}.**

By this method, the self – inductance and resistance of the leads are eliminated.