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What is wheatstone bridge?????

reddevil_2009

A Wheatstone bridge is a measuring instrument invented by Samuel Hunter Christie in 1833 and improved and popularized by Sir Charles Wheatstone in 1843. It is used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component. Its operation is similar to the original potentiometer except that in potentiometer circuits the meter used is a sensitive galvanometer.

Wheatstone's bridge circuit diagram.

In the circuit on the right, $R x$ is the unknown resistance to be measured; $R 1$ , $R 2$ and $R 3$ are resistors of known resistance and the resistance of $R 2$ is adjustable. If the ratio of the two resistances in the known leg $(R 2 / R 1)$ is equal to the ratio of the two in the unknown leg $(R x / R 3)$ , then the voltage between the two midpoints (B and D) will be zero and no current will flow through the galvanometer $V g$ . $R 2$ is varied until this condition is reached. The current direction indicates whether $R 2$ is too high or too low.

Detecting zero current can be done to extremely high accuracy (see galvanometer). Therefore, if $R 1$ , $R 2$ and $R 3$ are known to high precision, then $R x$ can be measured to high precision. Very small changes in $R x$ disrupt the balance and are readily detected.

At the point of balance, the ratio of $R 2 / R 1 = R x / R 3$

Therefore, $R_x = (R_2 / R_1) \cdot R_3$

Alternatively, if $R 1$ , $R 2$ , and $R 3$ are known, but $R 2$ is not adjustable, the voltage or current flow through the meter can be used to calculate the value of $R x$ , using Kirchhoff's circuit laws (also known as Kirchhoff's rules). This setup is frequently used in strain gauge and Resistance Temperature Detector measurements, as it is usually faster to read a voltage level off a meter than to adjust a resistance to zero the voltage.

is used to find the currents in junctions B and D:

$I_3 \ - I_x \ + I_g = 0$

$I_1 \ - I_g \ - I_2 = 0$

Then, Kirchhoff's second rule is used for finding the voltage in the loops ABD and BCD:

$(I_3 \cdot R_3) - (I_g \cdot R_g) - (I_1 \cdot R_1) = 0$

$(I_x \cdot R_x) - (I_2 \cdot R_2) + (I_g \cdot R_g) = 0$

The bridge is balanced and $I g = 0$ , so the second set of equations can be rewritten as:

$I_3 \cdot R_3 = I_1 \cdot R_1$

$I_x \cdot R_x = I_2 \cdot R_2$

Then, the equations are divided and rearranged, giving:

$R_x = {{R_2 \cdot I_2 \cdot I_3 \cdot R_3}\over{R_1 \cdot I_1 \cdot I_x}}$

From the first rule, $I 3 = I x$ and $I 1 = I 2$ . The desired value of $R x$ is now known to be given as:

$R_x = {{R_3 \cdot R_2}\over{R_1}}$

If all four resistor values and the supply voltage ( $V s$ ) are known, the voltage across the bridge ( $V$ ) can be found by working out the voltage from each potential divider and subtracting one from the other. The equation for this is:

$V = {{R_x}\over{R_3 + R_x}}V_s - {{R_2}\over{R_1 + R_2}}V_s$

This can be simplified to:

$V = \left({{R_x}\over{R_3 + R_x}} - {{R_2}\over{R_1 + R_2}}\right)V_s$

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