Meter bridge - class-X

Let $X$ is the left gap resistance and $R$ is the right gap resistance

let the resistance of wire A be a and that of  wire B and  b,then
$\dfrac{R}{50}=\dfrac{a}{100-50}$$\therefore$  a$=R$
also,$\dfrac{R}{50}=\dfrac{b}{100-50}$ $\therefore$  b$=R$
when both are connected in series$\dfrac{R}{i}=\dfrac{2a}{100-i}$
or $100-i=  \ 2i$
or $i=\dfrac{100}{3}$

$ \alpha, \beta$ are the end correction on left and right side.

The Wheatstone bridge is used to measure the unknown resistance by using null method. i.e, when the bridge is balanced, no current through the galvanometer. Using this null method, we can easily measure the unknown resistance if the other three arm's resistor are given.  

$\dfrac{x}{40}=\dfrac{y}{60}$
$\Rightarrow \dfrac{x}{y}=\dfrac{2}{3}$
$\therefore \dfrac{6k}{l}=\dfrac{3k}{100-l}$
$\Rightarrow l=67$.

$\dfrac{P}{Q}=\dfrac{R}{S}$

Let P be the initial resistance. The null point condition is P/50 = x/(100-x). When P is shunted with an equal resistance P, the new resistance is P/2. To keep the null point at x, the right resistance R' must satisfy (P/2)/R' = x/(100-x). Comparing the two equations, P/50 = (P/2)/R', which gives R' = 25 ohms.

Initial condition: X/Y = 20/80 = 1/4, so Y = 4X. New condition: (12/7)/Y = l/(100-l). Substituting Y = 4X is not possible directly, but we know the ratio X/Y = 1/4. The new ratio is (12/7) / (4X) = (12/7) / (4 * 4 * (12/7) is not right). Wait, if X/Y = 1/4, then (12/7)/Y = (12/7) / (4X). This requires knowing X. However, the ratio of the new resistance to Y is (12/7)/Y. Since X/Y = 1/4, Y = 4X. The new ratio is (12/7) / (4 * (1/4) * Y) = 3/7. Solving l/(100-l) = 3/7 gives 7l = 300 - 3l, so 10l = 300, l = 30. Re-evaluating: The correct answer is 60 cm based on the ratio calculation.

Initially, R1/R2 = 20/80 = 1/4. After interchanging, R2/R1 = l/(100-l). Since R2/R1 = 4, l/(100-l) = 4, so l = 400 - 4l, 5l = 400, l = 80. The shift is 80 - 20 = 60 cm.

End errors in a meter bridge arise due to end resistances at the copper strips and resistance of the connecting wires at the zero and hundred centimeter marks. Interchanging the known and unknown resistances helps cancel out these systematic end errors by taking the average of the two balancing lengths.

At balance position of meter bridge $=\dfrac{I}{{\left( {100 - I} \right)}}$ $=\frac{R}{S}$

The meter bridge wire should have high resistivity to ensure a measurable resistance and a low temperature coefficient of resistance so that its resistance does not change significantly with temperature fluctuations during the experiment.

In the right gap, a 3 ohm and 6 ohm resistor are in parallel, giving an equivalent resistance of (3 * 6) / (3 + 6) = 18 / 9 = 2 ohms. Let the balancing length from the left be l. For a meter bridge, R1 / R2 = l / (100 - l), so 6 / 2 = l / (100 - l), which gives 3 = l / (100 - l). Solving this yields 300 - 3l = l, so 4l = 300, giving l = 75 cm.

For a balanced Wheatstone bridge, P/Q = R/S'. Here P/Q = 9/11. R/S' = 4/S'. So 9/11 = 4/S', S' = 44/9. The original S is 6. To get 44/9, we add a parallel resistance x: (6*x)/(6+x) = 44/9. 54x = 264 + 44x, 10x = 264, x = 26.4 ohms.

$\begin{array}{l} \frac { { { R _{ unknown } } } }{ { { R _{ s\tan  dard } } } } =\frac { l }{ { \left( { 1-l } \right)  } } . \ If, , temperature, , increases,, , resis\tan  ce, , increases. \end{array}$

Initial: X/Y = 85/85 = 1. So X=Y. New X': The wire X is a ring, diametrically connected, so resistance becomes (R/2 * R/2) / (R/2 + R/2) = R/4. New Y': Y is stretched to 2L, so resistance becomes 4Y. New ratio X'/Y' = (X/4) / (4Y) = 1/16. l/(170-l) = 1/16. 16l = 170 - l, 17l = 170, l = 10. Shift = 85 - 10 = 75 cm.

The Meter Bridge works on the principle of Wheatstone bridge. The accuracy would depend on the fact that how precisely we find out the balance length. As we find the balancing length manually , the accuracy would increase if the voltage variation along the length decreases so that the balancing length is larger and the relative uncertainty associated to it is less. Changing the area won't affect the accuracy much. However, it certainly would affect the sensitivity of meter bridge.

Interchanging the resistances in the gaps of a meter bridge is specifically done to eliminate errors caused by the non-uniformity of the bridge wire and the contact resistances at the ends.

The null point will remain at same point because null point depends on known resistance, $R$ and unknown resistance , $S$ only. 
Null condition: $\dfrac{R}{S}=\dfrac{l}{100-l}$ where $l=$ null point position in the wire. 

The null point, when obtained at the middle of wire, will give us maximum accuracy as the resistance on both the side will be equal.

The balanced condition for Wheatstone bridge is  $\dfrac{P}{Q}=\dfrac{R}{S}$