Question

Wheatstone bridge derivation​

Answer 1

Hello guys Namaste,

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}\quad {\text{(1)}}\\I_{x}\cdot R_{x}&=I_{2}\cdot R_{2}\quad {\text{(2)}}\end{aligned}}} {\displaystyle {\begin{aligned}I_{3}\cdot R_{3}&=I_{1}\cdot R_{1}\quad {\text{(1)}}\\I_{x}\cdot R_{x}&=I_{2}\cdot R_{2}\quad {\text{(2)}}\end{aligned}}}

Then, equation (1) is divided by equation (2) and the resulting equation is 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}}

Due to: I3 = Ix and I1 = I2 being proportional from Kirchhoff's First Law in the above equation I3 I2 over I1 Ix cancel out of the above equation. 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}}

On the other hand, if the resistance of the galvanometer is high enough that IG is negligible, it is possible to compute Rx from the three other resistor values and the supply voltage (VS), or the supply voltage from all four resistor values. To do so, one has to work out the voltage from each potential divider and subtract one from the other. The equations for this are:

{\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}}

{\displaystyle R_{x}={{R_{2}\cdot V_{s}-(R_{1}+R_{2})\cdot V_{G}} \over {R_{1}\cdot V_{s}+(R_{1}+R_{2})\cdot V_{G}}}R_{3}} {\displaystyle R_{x}={{R_{2}\cdot V_{s}-(R_{1}+R_{2})\cdot V_{G}} \over {R_{1}\cdot V_{s}+(R_{1}+R_{2})\cdot V_{G}}}R_{3}}

where VG is the voltage of node D relative to node B.