20.9 WHEATSTONE BRIDGE
20.9.1 Wheatstone Bridge
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| Figure 20.57 A Wheatstone bridge consisting of four resistors P, Q, R, and S connected in a bridge arrangement, with an additional resistor (or galvanometer) connected across points B and D. |
1 The circuit
(a) The Wheatstone bridge circuit consists of four resistors, P, Q, R and S connected in the manner as shown in Figure 20.57. Another resistor is connected across points B and D.
(b) Instead of a resistor, a galvanometer G may be connected across BD because G itself has resistance.
(c) A battery is connected across AC so that it can supply current to all the resistors.
(d) Normally one of the resistors, like S, is a variable resistor. The purpose of having a variable resistor in the circuit is to adjust its resistance so that a ‘balanced’ circuit can be obtained.
2 Condition for achieving a ‘balanced’ circuit
(a) In general, current flows through all the five resistors, i.e., P, Q, R, S and G. If we adjust the resistance of S, the current in each resistor will change.
(b) We could keep on adjusting the resistance of S so that the current flowing through the galvanometer G could be reduced to zero. When this condition (current through G = 0) is achieved, the circuit is said to be ‘balanced’.
3 Equation for balanced circuit
Suppose that no current flows through galvanometer G. Since no current flows through G, we would have the following conditions:
(a) The electric potential at point B must be equal to the electric potential at point D so that the p.d. across BD is zero. We get
p.d. across AB = p.d. across AD
p.d. across CB = p.d. across CD
(b) The current flowing through P must be equal to that flowing through Q. Let it be I1. Similarly, the current flowing through R must be equal to that flowing through S. Let it be I2.
Based on these two conditions, we get
$I_1P = I_2R$
$I_1Q = I_2S$
$\boxed{\frac{P}{Q} =\frac{R}{S}}$
We may also write the expression as
$\boxed{\frac{P}{R} =\frac{Q}{S}}$
20.9.2 The Sliding Wire Bridge
1. Structure
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| Figure 20.59A sliding wire bridge setup, which is a form of Wheatstone bridge, consisting of a uniform wire, resistors, and a galvanometer used to measure unknown resistance. |
(a) Figure 20.59 shows the structure of a sliding wire bridge which is actually a Wheatstone bridge.
(b) Resistors or conductors having resistances R1 and R2 are connected across the air gaps.
(c) A bare uniform wire is connected across AB. The wire normally has a length of 1.0 m and resistance of several ohms.
(d) A galvanometer G is connected to the circuit in the manner as shown. The main function of G is to indicate whether there is any current flowing through it.
(e) A jockey is connected to G. It is made to slide on wire AB.
(f) The wire, resistors and galvanometer are connected to metal (normally copper) plates whose resistances are very low and may be neglected.
2. Principle
Suppose that when the jockey touches wire AB at point C, the galvanometer G indicates that there is no current flowing through it. This means that the circuit is balanced. Let the lengths of AC and CB be l1 and l2 respectively. Then we have
$\frac{R_1}{R_2}= \frac{resistance \ of \ AC}{resistance \ of \ CB}$
But, resistance of $AC = rl_1$
resistance of $CB = rl_2$
where r is the resistance per unit cm of wire AB and so is a constant. Hence we have
$\frac{R_1}{R_2} = \frac{l_1}{l_2}$
If l1 and l2 are measured, and, say, R1 is known, then R2 can be determined. Hence, we can use this circuit to measure an unknown resistance whose value is not too low.
Note: This circuit is not suitable for measuring resistance whose value is 1 ohm or less.
Solved Examples of Wheatstone Bridge and Meter Bridge Circuits
EXAMPLE 20.19
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| Figure 20.58 A Wheatstone bridge used to determine the unknown resistance Q by adjusting resistors until the circuit is balanced under different conditions. |
Refer to the Wheatstone bridge circuit shown in Figure 20.58. The circuit is balanced when R = 100 Ω and S = 200 Ω. The resistance of Q is then changed to 250 Ω. The circuit is balanced again only when R has resistance 161 Ω. Determine the initial resistance of Q.
Answer
Since the circuit is balanced, we get
$\frac{Q}{P} = \frac{100}{200}$
$\frac{250}{P} = \frac{161}{200}$
$P=310.6 \ \Omega$
EXAMPLE 20.20
A bridge circuit is balanced when resistor X is connected in series with a 9 Ω resistor, or when resistor Y is connected in parallel with a 12 Ω resistor. Determine the values of X and Y.
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| Figure 20.59 Balanced bridge circuit used to determine the values of X and Y. |
Answer
For a balanced Wheatstone bridge, the ratio of resistances in one branch equals the ratio in the other branch:
$\frac{R_1}{R_2} = \frac{R_3}{R_4}$
From the problem, the circuit can be redrawn as follows.
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| Figure 20.60 Wheatstone bridge in balance condition showing the relationship between resistors X and Y. |
Case 1: X in series with 9 Ω
The equivalent resistance becomes:
$X + 9$
Applying the balance condition:
$\frac{2}{3} = \frac{Y}{X + 9}$
Cross-multiplying:
$2(X + 9) = 3Y$
$2X + 18 = 3Y$ ...(1)
Case 2: Y in parallel with 12 Ω
The equivalent resistance of Y in parallel with 12 Ω is:
$\frac{12Y}{12 + Y}$
Applying the balance condition:
$\frac{2}{3} = \frac{\frac{12Y}{12 + Y}}{X}$
Rearranging:
$2X = \frac{(3)(12Y)}{12 + Y}$ ...(2)
Solving the Equations
Substitute equation (2) into equation (1):
$\frac{(3)(12Y)}{12 + Y} = 3Y - 18$
Divide both sides by 3:
$\frac{12Y}{12 + Y} = Y - 6$
Multiply both sides by (12 + Y):
$12Y = (Y - 6)(12 + Y)$
Expand:
$12Y = Y^2 + 6Y - 72$
Rearrange:
$Y^2 - 6Y - 72 = 0$
Factor:
$(Y - 12)(Y + 6) = 0$
Thus:
$Y = 12 \ \Omega$ (valid solution)
Substitute into equation (1):
$2X + 18 = 3(12)$
$2X + 18 = 36$
$2X = 18$
$X = 9 \ \Omega$
EXAMPLE 20.21
The figure shows a Wheatstone bridge circuit with a homogeneous wire AB of length 1 m. The sliding contact C is positioned so that the bridge is balanced when CB = 60 cm (the galvanometer G shows zero reading).
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Figure 20.61 A Wheatstone bridge circuit using a 1 m uniform wire AB with a sliding contact C. The bridge is balanced when CB = 60 cm, and after adding a 10 Ω resistor in series with X, a new balance is achieved when the contact is shifted by 30 cm. |
If a resistor of 10 Ω is connected in series with resistor X, the bridge becomes balanced again when the contact C is moved 30 cm from its original balanced position.
Determine the values of X and Y.
Answer
$AB = 1 \ m = 100 \ cm$.
At the initial balanced position, $CB = 60 \ cm$, so:
$AC = 100 \ cm − 60 \ cm = 40 \ cm$
From the balance condition of the bridge:
$X(CB) = Y(AC)$
$X(60 \ cm) = Y(40 \ cm)$
$Y = (3/2)X$ ...(1)
Now, a 10 Ω resistor is added in series with X, so the total resistance becomes $(X + 10) \Omega$. The bridge becomes balanced again when contact C is shifted 30 cm to the right.
Thus:
$AC = (40 + 30) \ cm = 70 \ cm$
$CB = (100 − 70) \ cm = 30 \ cm$
Using the balance condition again:
$(X + 10)(CB) = Y(AC)$
$(X + 10)(30 \ cm) = Y(70 \ cm)$
$3X + 30 = 7Y$ ...(2)
Substitute equation (1) into equation (2):
$3X + 30 = 7(3X/2)$
$2(3X + 30) = 21X$
$6X + 60 = 21X$
$−15X = −60$
$X = 4 Ω$
Substitute into equation (1):
$Y = (3)(4)/2 = 6 \ \Omega$
EXAMPLE 20.22
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| Figure 20.62 A Wheatstone bridge circuit using a uniform wire AB of length 1.5 m, where the bridge is balanced at point J (AJ = 90 cm) with resistor Y = 3.0 Ω, resulting in zero galvanometer deflection. |
A practical Wheatstone bridge consists of a uniform wire AB with a total length of 1.5 m. The bridge is balanced when a known resistor Y = 3.0 Ω is used, and the jockey is placed at point J such that AJ = 90 cm. At this position, the galvanometer shows zero deflection. The resistance per unit length of the wire AB is 0.02 Ω/cm. Determine:
(a) The value of the unknown resistance X.
(b) The current supplied by a cell of emf 6.0 V, assuming negligible internal resistance.
Wheatstone bridge is an experimental setup which is used to measure an unknown resistance. It has four arms which act as resistors, and the ratio of two of the resistors is kept fixed. From balancing the other two arms of the bridge we can calculate the unknown resistance. We will be using a suitable formula in relation to wheatstone bridge to find the unknown resistance and the current drawn from the battery.
Complete Step-By-Step answer:
We have been given that,
AB = 2 m; $AB =1.5 \ m$
Y = 2 Ω; $Y=3 \ \Omega$
AJ = 1.20 m; $AJ=90 \ cm=0.9 \ m$
Emf = 4 V; $Emf = 4 \ V$$
The Resistance offered by wire $AB = 0.02 \ \Omega /cm$
We will start by looking into the diagram,
We can see that,
P = Resistance of wire AJ
$P = (90 \ cm) \times (0.02 \ \Omega /cm)$
$P = 1.80 \ \Omega$
Q = Resistance of wire BJ
$Q = [(1.50 − 0.09) × 100] cm \times (0.01 \ \Omega /cm)$
$Q = 1.41 \ \Omega$
We have the resistances of P and Q, now using this we will have to find the unknown resistance of resistor X.
To find the resistance of X we will use the formula,
$\frac{P}{Q} = \frac{X}{Y}$
$X = Y \times \left(\frac{P}{Q}\right)$
$X = 3 \times \left(\frac{1.8}{1.41}\right)$
$X = 3.83 \ \Omega$
Hence the resistance of the resistor X is $3.83 \ \Omega$.
Now, let us find the total resistance of X and Y when they are connected in series,
$R_1= X + Y$
$R_1 = 3.82 + 3$
$R_1 = 6.82 \ \Omega$
Now, finding the total resistance when P and Q are connected in series (or of wire AB),
$R_2 = 1.5 \times 100 \times 0.02$
$R_2 = 3 \ \Omega$
Now we will find the resistance when R1 and R2 are connected in parallel, meaning the effective or total resistance between the terminals A and B of Wheatstone bridge will be,
$R_{AB}= \frac{R_1R_2}{R_1+R_2}$
$R_{AB}= \frac{6.82 \times 3}{6.82+3}=2.08 \ \Omega$
Now using the above values, we will find the current drawn from the battery using the formula,
$I = \frac{emf}{R_{AB}}$
$I=\frac{6}{2.08}$
$I = 2.88 \ A$
Hence the current drawn from the battery is 2.88 A.
Note:
The Wheatstone bridge is an instrument which can give very precise values of resistances. It is used to measure stress, strain, temperature, etc., along with the help of an operational amplifier.
The formula $\frac{P}{Q} = \frac{X}{Y}$
It is essential and students are recommended to remember it.
Applications of Wheatstone Bridge and Sliding Wire Bridge
The Wheatstone Bridge and Sliding Wire Bridge are important electrical circuits used to measure unknown resistances with high accuracy. These methods are widely applied in laboratories, industry, and electronic instrumentation.
Applications of Wheatstone Bridge
- Measurement of Unknown Resistance: Used to determine an unknown resistance by balancing the bridge circuit.
- Precision Instruments: Applied in devices such as galvanometers and sensitive measuring equipment.
- Strain Gauge Measurements: Used in engineering to measure strain, pressure, and force in materials.
- Temperature Measurement: Used with thermistors and resistance temperature detectors (RTDs).
- Fault Detection: Helps detect faults in electrical circuits and wiring systems.
Applications of Sliding Wire Bridge (Meter Bridge)
- Comparison of Resistances: Used to compare two resistances accurately using a uniform wire.
- Determination of Resistivity: Helps calculate the resistivity of a material.
- Verification of Ohm’s Law: Commonly used in school laboratories to verify Ohm’s law.
- Calibration of Instruments: Used to calibrate ammeters and voltmeters.
- Educational Experiments: Widely used in physics experiments for students to understand bridge principles.
Both Wheatstone Bridge and Sliding Wire Bridge are essential tools in electrical measurements. They provide accurate and reliable methods for determining resistance and are widely used in scientific, industrial, and educational applications.
Conclusion-Wheatstone Bridge & Sliding Wire Bridge
The Wheatstone Bridge and Sliding Wire Bridge are important electrical circuits used for precise measurement of resistance. Both operate based on the principle of balance, where no current flows through the galvanometer, ensuring high accuracy.
The Wheatstone Bridge is highly effective for comparing and determining unknown resistances using the ratio of known resistors, while the Sliding Wire Bridge provides a practical method by relating resistance to the length of a uniform wire.
These methods are widely used in laboratories, education, and instrumentation due to their sensitivity, reliability, and ability to detect very small changes in resistance.
Overall, both bridges play a crucial role in electrical measurements and form the foundation for many modern sensing and measurement techniques.
Frequently Asked Questions (FAQ) – Wheatstone Bridge & Sliding Wire Bridge
The main application is to measure unknown resistance accurately by balancing the bridge circuit.
It operates under a null condition where no current flows through the galvanometer, reducing measurement errors.
It is used in sensors such as strain gauges, thermistors, and LDRs in industrial and scientific applications.
It is used to measure unknown resistance by comparing lengths along a uniform wire.
It provides a simple method using proportional wire lengths, ideal for laboratory experiments.
Yes, both bridges are very sensitive and can detect small variations in resistance.
They provide accurate and reliable methods for teaching and experimental measurements.
It is simpler and directly relates resistance to measurable wire length.
Yes, especially the Wheatstone bridge in sensor circuits and measurement systems.
The bridge must be balanced so that no current flows through the galvanometer.
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