Wheatstone Bridge - Definition, Working Principle, Formula, Application, FAQs

Wheatstone's Bridge

A Wheatstone bridge is an arrangement of four resistances used to determine the value of an unknown resistance. In this arrangement, one resistance is adjustable, two are known, and one is unknown.

Let $\mathbf{P}$ and $\mathbf{Q}$ be known resistances, $\mathbf{R}$ be an adjustable resistance, and $\mathbf{S}$ be the unknown resistance. To find the value of $\mathbf{S}$, the keys $\mathbf{K}_1$ and $\mathbf{K}_2$ are closed. The resistance $\mathbf{R}$ is then adjusted until the galvanometer shows no deflection.

In this condition, the bridge is said to be balanced.
Balanced condition:

$
\frac{P}{Q}=\frac{R}{S} \quad \text { and } \quad V_B=V_D
$

In the balanced condition, no current flows through the galvanometer.
If the bridge is not balanced, a current flows through the galvanometer. This is called the unbalanced condition. The direction of current depends on the potential difference between the points.

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Wheatstone bridge derivation

A Wheatstone bridge consists of four resistances $R_1, R_2, R_3$ and $R_x$. A battery is connected across points $\mathbf{A}$ and $\mathbf{D}$, and a galvanometer is connected between points $\mathbf{B}$ and $\mathbf{C}$.

When the galvanometer shows no deflection, the bridge is said to be balanced. In this condition, the potentials at points $\mathbf{B}$ and $\mathbf{C}$ are equal:

$
V_B=V_C
$

Applying Ohm's law in the two branches, we get:

$
\begin{aligned}
& I_1 R_1=I_2 R_3 \\
& I_1 R_2=I_2 R_x
\end{aligned}
$

Dividing the above equations:

$
\frac{R_1}{R_2}=\frac{R_3}{R_x}
$

Hence,

$
R_x=\frac{R_2 \times R_3}{R_1}
$

Thus, the unknown resistance can be calculated using the balanced condition of the Wheatstone bridge.

Application of Wheatstone bridge-

1. An ideal ammeter should have zero resistance, and an ideal voltmeter should have infinite resistance, however this is impossible to achieve in practice. As a result, this circuit is unable to provide precise readings. In such instances, a Wheatstone bridge circuit can be used to get extremely exact measurements. The Wheatstone bridge circuit has been adapted in two ways for experimental purposes:

• Box at the post office
• Bridge with meter

2. Some materials (such as semiconductors) have temperature-dependent resistance. When compared to typical resistors, the variances are rather considerable. Thermistors are what they're called. The Wheatstone bridge system wheatstone bridge uses thermistors to detect small temperature changes.
3. The unknown resistor in a Wheatstone bridge circuit can be replaced with a photoresistor to measure changes in light intensity. A photoresistor's resistance is proportional to the amount of light it receives.
4. Strain and pressure can also be measured using the Wheatstone bridge.

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Limitations of Wheatstone Bridge

1. The Wheatstone Bridge is a delicate device. In an off-balance situation, measurements may not be precise.
2. Wheatstone bridges are commonly used to measure resistances of a few ohms to a few kilo-ohms.
3. If the four resistances are not equivalent, the circuit's sensitivity is reduced.

Some important points-

• Samuel Hunter Christie devised the Wheatstone bridge circuit, which was later refined by Charles Wheatstone.
• In AC circuits, many variations of the Wheatstone bridge can be used to measure impedance, inductance, and capacitance.
• Maxwell improved the Maxwell bridge circuit, which is used in AC circuits.

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Solved Examples Based on Wheatstone's bridge

Example 1: In a Wheatstone’s bridge, three resistances P, Q and R, are connected in the three arms and the fourth arm is formed by two resistances S₁ and S₂ connected in parallel. The condition for the bridge to be balanced will be

(a) $\frac{P}{Q}=\frac{R}{S_1+S_2}$
(b) $\frac{P}{Q}=\frac{2 R}{S_1+S_2}$
(c) $\frac{P}{Q}=\frac{R\left(S_1+S_2\right)}{S_1 S_2}$
(d) $\frac{P}{Q}=\frac{R\left(S_1+S_2\right)}{2 S_1 S_2}$.

Solution

Wheat stone Bridge

It is an arrangement of four resistances which can be used to measure one of them in terms of rest

wherein

For balanced Wheatstone's bridge, $\mathrm{P} / \mathrm{Q}=\mathrm{R} / \mathrm{S}$

$
\begin{aligned}
S & =\frac{S_1 S_2}{S_1+S_2}\left(\because S_1 \text { and } S_2 \text { are in parallel }\right) \\
\frac{P}{Q} & =\frac{R\left(S_1+S_2\right)}{S_1 S_2} .
\end{aligned}
$

Example 2: The current I drew from the 5-volt source will be :

1) 0.17 A

2) 0.33 A

3) 0.5 A

4) 0.67 A

Solution:

The given circuit can be redrawn as which is a balanced Wheatstone's bridge and hence, no current flows in the centre resistor, so equivalent circuit would be as shown below.

So, $I=\frac{V}{R}=5 / 10=0.5 A$

Example 3: In a metre bridge experiment, the null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y . If X<Y, then where will be the new position (in cm) of the null point from the same end, if one decides to balance a resistance of 4X against Y?

1) 50

2) 80

3) 40

4) 70

Solution:

Meter bridge is an arrangement which works on Wheatstone's principle, so the balancing condition is

$
\frac{R}{S}=\frac{l_1}{l_2}
$

where, $\quad l_2=100-l_1$

$
\begin{array}{ll}
\text { Case I } & R=X, S=Y, l_1=20 \mathrm{~cm} \\
& l_2=100-20=80 \mathrm{~cm} \\
\therefore & \frac{X}{Y}=\frac{20}{80}----(i)
\end{array}
$

Case II Let the position of null point be obtained at a distance $l$ from same end.

$
\begin{array}{ll}
\therefore & R=4 X, S=Y, \\
& l_1=l, l_2=100-l
\end{array}
$

So, from Eq. (i), we get

$
\begin{aligned}
& \frac{4 X}{Y}=\frac{l}{100-l} \\
& \frac{X}{Y}=\frac{l}{4(100-l)}----(ii)
\end{aligned}
$

Therefore, from Eqs. (i) and (ii), we get

$
\frac{l}{4(100-l)}=\frac{20}{80} \Rightarrow \frac{l}{4(100-l)}=\frac{1}{4}
$

or

$
l=100-l
$

or

$
2 l=100
$

Hence,

$
l=50 \mathrm{~cm}
$

Example 4: Four resistances of 15Ω,12Ω,4Ω and 10Ω respectively in cyclic order to form Wheatstone's network. The resistance that is to be connected in parallel with the resistance of 10Ω to balance the network is Ω⋅

1) 10

2) 5

3) 15

4) 20

Solution:

Let the resistance to be connected is $x$.
For balanced wheatstone bridge,

$
\begin{aligned}
& 15 \times 4=12 \times \frac{10 x}{10+x} \\
& \Rightarrow x=10 \Omega
\end{aligned}
$

So the answer will be 10.

Example 5: The Wheatstone bridge shown in the figure here gets balanced when the carbon resistor used as R₁ has the colour code (Orange, Red, Brown) The resistors R₂ and R₄ are 80$\Omega$ and 40$\Omega$, respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as R₃ , would be :

1) Brown, Blue, Brown

2) Brown, Blue, Black

3) Red, Green, Brown

4) Grey, Black, Brown

Solution:

In balanced Wheatstone bridge condition,

$
\frac{R_1}{R_2}=\frac{R_3}{R_4}
$

Calculation:
The value of $\mathrm{R}_1$ (orange, red, brown) $=32 \times 10=320 \Omega$
Given, $\mathrm{R}_2=80 \Omega$ and $\mathrm{R}_4=40 \Omega$

$
\begin{aligned}
& \Rightarrow R_3=R_4 \times \frac{R_1}{R_2} \\
& R_3=\frac{40 \times 320}{80} \\
& \mathrm{R}_3=160 \Omega=16 \times 10^1
\end{aligned}
$

Comparing the value of $\mathrm{R}_3$ with the colours assigned for the carbon resistor, we get

$
\mathrm{R}_3=16 \times 10^1
$

Here, 1 represents brown
6 represents blue
In $10^1$, the power 1 represents brown
Therefore, the colour code for the carbon resistor used in $R_3$ would be Brown, Blue, Brown.

Summary

Wheatstone bridge method for measuring an unknown resistance has the following merits over the other methods for measuring the resistance. It has a null method therefore, the measurement of resistance made by this method is not affected by the internal resistance of the battery used. As the measurement of current or potential difference is involved, the measurements are not affected because of the fact that the ammeters and voltmeters are not ideal ones.

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