electric-circuitselectric-currentelectrical-resistance
Let,
P/Q = R/S in the usual sense of a Wheatstone bridge where PQRS are resistances.
Then we know that the potential between midpoints of PQ and RS are equal and there's no potential difference between the midpoints i.e. $V_b = 0$.
Now if it was
P/Q > R/S
It is said that potential between PQ is less than RS. And the bridge is unbalanced. How do you prove this inequality. I know the absense of proper images make it hard to grasp my question, but i tried my best to explain it with words.
A Wheatstone bridge is only a very sensitive way of finding a match between two resistances. You have got to have a nice, well calibrated variable resistance on the opposite side to the unknown resistance. The accuracy with which you can judge the unknown is limited by the calibration accuracy of the variable resistance.
Using Ohm's law to find an unknown resistance is a great way to do it. You just match the voltage you apply to the unknown based on the range of resistance you expect to encounter. Suppose you had a short piece of heavy copper buss bar you wanted to measure the resistance of. I have a nice AC welder that can put out 400 amps for brief periods. I will short the output with the buss bar, put 400 amps to it and while the current is flowing, I will measure the voltage difference between each end of the bar. With such a small resistance in the unknown resistor, it takes a lot of amps to generate a nice measurable voltage.
Conversely, the mega high resistance of thick layers of insulation can be measured by putting a couple of tens of kilovolts to it and reading the milliamperage. I use neon sign transformers and large resistor banks for stepping down and for amp measurements.
I have a series of calibrated shunts that are useful. They run a millivolt per amp. You read .5 volts of potential from one end to the other of the shunt - that is 500 amps. I have a big brass shunt on my car dash that gives me a reading of starting amps. I ran old welding cable to bring the starter amps into the passenger compartment. I have a giant switch on the dash, custom made of 1/4" x 2" copper bussbar in several pieces up to a foot long. No auto thief would ever have the balls to close that switch. I have skulls and red warnings. High voltage, etc.
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,
\begin{equation}V_1(G_1+G_3+G)-V_2(G)=VG_1\end{equation}
Similarly for node D,we get,
\begin{equation}-V_1(G)+V_2(G_2+G_4+G)=VG_2\end{equation}
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.
Thus,the final answer comes to,
$$\Delta V=\frac{\begin{vmatrix} G_1 & G_3 \\ G_2 & G_4 \end{vmatrix}}{\begin{vmatrix} G_1+G_3 & 0 & -G_2-G_4 \\ 1 & 1 & 1 \\ \frac{G_1+G_3}{2} & -G & \frac{G_2+G_4}{2} \end{vmatrix}}$$
Best Answer
Consider the left branch of the bridge. The total resistance is $P + Q$, so the current is:
$$ I_{left} = \frac{V}{P+Q} $$
The voltage drop across $Q$ is just $V = IR$, so the voltage at the $PQ$ midpoint is:
$$ V_{PQ} = V_{in}\frac{Q}{P+Q} $$
We argue in the same way for the right hand branch to get the corresponding equation:
$$ V_{RS} = V_{in}\frac{S}{R+S} $$
So the bridge voltage $V_b$ is given by:
$$ V_b = V_{PQ} - V_{RS} = V_{in}\left(\frac{Q}{P+Q} - \frac{S}{R+S}\right) \tag{1} $$
When the bridge is balanced $V_b = 0$. If we set $V_b = 0$ in equation (1) we get:
$$ \frac{Q}{P+Q} = \frac{S}{R+S} $$
and rearranging this gives:
$$ \frac{P}{Q} = \frac{R}{S} $$
Now suppose $V_b \lt 0$. If we put this into equation (1) we get:
$$ V_{in}\left(\frac{Q}{P+Q} - \frac{S}{R+S}\right) \lt 0 $$
or:
$$ \frac{Q}{P+Q} \lt \frac{S}{R+S} $$
and rearranging this gives:
$$ \frac{R}{S} \lt \frac{P}{Q} $$