4.1. Condition of balance of bridge according to Ohm’s law
We can make Ohm's law for the homogeneous arm AB:
,
or
.
(25)
The same for the arm AD
,
or
.
(26)
As at the moment of balance В=D, then the right parts of equations (25) and (26) are equal, so the left are equal too:
.
(27)
Same for arms BC and DC
.
(28)
At the moment of balance current doesn't flow through the microampermeter, so at the point B current I4 doesn't branch out, but completely flows through resistor R3, so I3=I4. At the point D current I1 also moves to the part DC, so I2=I1. Then we can write equation (28) as
(29)
If we divide the equation (27) on the (29), we will get
.
(30)
4.2. Condition of balance of bridge according to Kirchhoff rules
Formula (30) can be derived with help of Kirchhoff’s rules for the circuits ABDA and BCDB, and junctions B and D.
Before a compiling of the equations, it’s necessary arbitrarily to choose:
a) Directions of currents (see fig. 12); b) directions of path-tracing of contours АDС and АВС (clockwise on fig. 12).
In order to compile the equations with the help of first Kirchhoff’s rule it is necessary to take into account: we should take a current with the plus sign if it flows into the junction, we should take it with minus sign, if it outflows from the junction.
According the first Kirchhoff’s rule for the junction B:
.
(31)
According the first Kirchhoff’s rule for the junction D:
.
(32)
We will get two necessary equations with the help of second Kirchhoff’s rule. It is necessary to follow the sign rules:
a) We should take a voltage drop (product IR or Ir) with plus sign, if the current’s direction coincides with the direction of path-tracing, in other case – with minus sign; b) We should take an EMF with plus sign, if it enlarges the potential in the direction of path-tracing (we pass through the source from minus to plus), in other case – with minus sign.
It is necessary for us to obtain a relation between resistances of arms of the bridge, therefore for compilation of the equations we will choose the closed circuits consisting only from these arms (see fig. 12).
According the second Kirchhoff’s rule for the closed loop АВDA:
,
(33)
and for the closed loop ВСDВ:
.
(34)
As at the moment of balance I6=0, then from (33) and (34) we obtain:
(35)
From (31) and (32)
(36)
After the substitution (36) in (35), we obtain:
,
(37)
.
(38)
If we divide the equation (37) on the (38), we will get
.
(39)
From the same equations (30) or (39) we obtain condition of the bridge's balance:
.
(40)
Using the formula (40) the unknown resistor RX can be calculated.