# Unbalanced wheatstone bridge

Suppose I have an unbalanced Wheatstone bridge (resistances $R_1, R_2, R_3, R_4, R_g$), to which I apply some voltage $V$. I need to find the voltage across the the galvanometer. Solving the system of equations obtained by Kirchoff's Law would yield the answer, but it will definitely be enormous. I recall seeing somewhere it could be written as the determinant of a somewhat elegant matrix, but I can't see how I would find it (Cramer's rule applied to the system is way too ugly, and solving first, then backtracking to finding the determinant is unpractical). Does anybody know how to approach this problem?

• This is a circuits question. Flagged for migration to EE.SE – Alfred Centauri Dec 27 '13 at 14:47
• Being on-topic elsewhere does not necessarily render the questions off-topic here. – dmckee --- ex-moderator kitten Dec 27 '13 at 16:30
• If you are solving it for the sake of some practical use, I would prefer SPICE. – Torsten Hĕrculĕ Cärlemän Dec 27 '13 at 17:20
• Unless the bridge resistances are very high, the galvanometer resistance will have little effect. Ohms law will provide the voltage to each leg of the meter. The optimum operating point is in balance only. – Optionparty Dec 27 '13 at 23:19

Consider the given circuit as follows:

The first thing to realize that it is hopeless to work with currents as our independent variable.In the end there are going to be three independent currents which while solving can make the calculations messier.

So,we will use potentials of points as our independent variables.

Let,

$V_A=V$(Voltage of the source)

$V_D=0$

$V_B=V_1$

$V_C=V_2$

Also instead of working with resistances,lets work with conductance instead as the equations are comparatively cleaner.So let the resistances $R_1,R_2,R_3,R_4$and$R_g$ correspond to the conductances $G_1,G_2,G_3,G_4$and$G$.

Writing Kirchhoff's current rule at the node B,we get

$(V_1-V)G_1=(0-V_1)G_3+(V_2-V_1)G$

which then simplifies to,

$$V_1(G_1+G_3+G)-V_2(G)=VG_1$$

Similarly for node D,we get,

$$-V_1(G)+V_2(G_2+G_4+G)=VG_2$$

Using Cramer's rule,we can easily solve the two equations for $V_1$ and $V_2$ as,

$V_1=\frac{G_1G_2+G_1G_4+GG_1+GG_2}{\Delta}$

$V_2=\frac{G_1G_2+G_2G_3+GG_2+GG_1}{\Delta}$

where \begin{align}\Delta&=&(G_1+G_3+G)(G_2+G_4+G)-G^2 \\ &=&\begin{vmatrix} G_1+G_3 & 0 & -G_2-G_4 \\ 1 & 1 & 1 \\ \frac{G_1+G_3}{2} & -G & \frac{G_2+G_4}{2} \end{vmatrix} \end{align}

Using all of this,we get the potential difference between the terminals of the Galvanometer($\Delta V=V_1-V_2$) as:

\begin{align} \Delta V&=& \frac{G_1G_4-G_2G_3}{\begin{vmatrix} G_1+G_3 & 0 & -G_2-G_4 \\ 1 & 1 & 1 \\ \frac{G_1+G_3}{2} & -G & \frac{G_2+G_4}{2} \end{vmatrix}} \\ &=& \frac{\begin{vmatrix} G_1 & G_3 \\ G_2 & G_4 \end{vmatrix}}{\begin{vmatrix} G_1+G_3 & 0 & -G_2-G_4 \\ 1 & 1 & 1 \\ \frac{G_1+G_3}{2} & -G & \frac{G_2+G_4}{2} \end{vmatrix}}\end{align}

Note that the determinant in the numerator of the expression when zero implies that the balancing condition of the bridge.

$$\Delta V=\frac{\begin{vmatrix} G_1 & G_3 \\ G_2 & G_4 \end{vmatrix}}{\begin{vmatrix} G_1+G_3 & 0 & -G_2-G_4 \\ 1 & 1 & 1 \\ \frac{G_1+G_3}{2} & -G & \frac{G_2+G_4}{2} \end{vmatrix}}$$