.What is wheatsone bridge and derive the expression and please fast give answer please my aal frds
Answers
A Wheatstone bridge is an electrical circuit used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component. The primary benefit of the circuit is its ability to provide extremely accurate measurements (in contrast with something like a simple voltage divider).[1] Its operation is similar to the original potentiometer.
The Wheatstone bridge was invented by Samuel Hunter Christie in 1833 and improved and popularized by Sir Charles Wheatstone in 1843. One of the Wheatstone bridge's initial uses was for the purpose of soils analysis
First, Kirchhoff's first law is used to find the currents in junctions B and D:
{\displaystyle {\begin{aligned}I_{3}-I_{x}+I_{G}&=0\\I_{1}-I_{2}-I_{G}&=0\end{aligned}}} \begin{align}
I_3 - I_x + I_G &= 0 \\
I_1 - I_2 - I_G &= 0
\end{align}
Then, Kirchhoff's second law is used for finding the voltage in the loops ABD and BCD:
{\displaystyle {\begin{aligned}(I_{3}\cdot R_{3})-(I_{G}\cdot R_{G})-(I_{1}\cdot R_{1})&=0\\(I_{x}\cdot R_{x})-(I_{2}\cdot R_{2})+(I_{G}\cdot R_{G})&=0\end{aligned}}} \begin{align}
(I_3 \cdot R_3) - (I_G \cdot R_G) - (I_1 \cdot R_1) &= 0 \\
(I_x \cdot R_x) - (I_2 \cdot R_2) + (I_G \cdot R_G) &= 0
\end{align}
When the bridge is balanced, then IG = 0, so the second set of equations can be rewritten as:
{\displaystyle {\begin{aligned}I_{3}\cdot R_{3}&=I_{1}\cdot R_{1}-(1)\\I_{x}\cdot R_{x}&=I_{2}\cdot R_{2}-(2)\end{aligned}}} {\displaystyle {\begin{aligned}I_{3}\cdot R_{3}&=I_{1}\cdot R_{1}-(1)\\I_{x}\cdot R_{x}&=I_{2}\cdot R_{2}-(2)\end{aligned}}}
Then, the equations (1)&(2) are divided(=equation(1)/equation(2)) and rearranged, giving:
{\displaystyle R_{x}={{R_{2}\cdot I_{2}\cdot I_{3}\cdot R_{3}} \over {R_{1}\cdot I_{1}\cdot I_{x}}}} R_x = {{R_2 \cdot I_2 \cdot I_3 \cdot R_3}\over{R_1 \cdot I_1 \cdot I_x}}
From the first law, I3 = Ix and I1 = I2. The desired value of Rx is now known to be given as:
{\displaystyle R_{x}={{R_{3}\cdot R_{2}} \over {R_{1}}}} R_x = {{R_3 \cdot R_2}\over{R_1}}
If all four resistor values and the supply voltage (VS) are known, and the resistance of the galvanometer is high enough that IG is negligible, the voltage across the bridge (VG) can be found by working out the voltage from each potential divider and subtracting one from the other. The equation for this is:
{\displaystyle V_{G}=\left({R_{2} \over {R_{1}+R_{2}}}-{R_{x} \over {R_{x}+R_{3}}}\right)V_{s}} {\displaystyle V_{G}=\left({R_{2} \over {R_{1}+R_{2}}}-{R_{x} \over {R_{x}+R_{3}}}\right)V_{s}}
where VG is the voltage of node D relative to node B.