Wheatstone Bridge and Metre Bridge

Wheatstone Bridge

The Wheatstone bridge is a circuit used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component. It was invented by Samuel Hunter Christie in 1833 and improved by Sir Charles Wheatstone, who popularized it. The bridge consists of four resistors (P, Q, R, S) arranged in a diamond shape. A galvanometer is connected between the junction of P and Q, and the junction of R and S. A voltage source (like a battery) is connected across the top and bottom junctions.

Principle of Wheatstone Bridge

The bridge is said to be balanced when there is no current flowing through the galvanometer. This occurs when the potential difference between the two mid-points (where the galvanometer is connected) is zero.

Let the four resistors be P, Q, R, and S. Let the galvanometer be connected between points B and D in the circuit diagram. Let the voltage source be connected between points A and C.

When the bridge is balanced, the ratio of the resistances in the two arms is equal:

P/Q = R/S

This condition ensures that the potential at point B is equal to the potential at point D, hence no current flows through the galvanometer.

Derivation of the Balance Condition

Consider the bridge circuit. Let the current flowing from the source be I. Let the current through P be I₁ and the current through R be I₂. The voltage source has an EMF of V.

Using Kirchhoff's laws:

• In loop ABDA: V_A - I₁P - I_gG - I₂R = V_A (where V_A is the potential at A, and V_C=0 for simplicity at C) So, I₁P + I_gG = I₂R (Equation 1)
• In loop BCDA: (V_B - V_C) = I₃Q + I₄S = V_C (Assuming V_C=0) So, (V_B - V_C) = I₃Q + I₄S. If we consider potentials relative to C. Let's use a more direct approach with potentials. Let V_A be the potential at A, and V_C be the potential at C.
• Potential at B, V_B = V_A - I₁P
• Potential at D, V_D = V_A - I₂R

For the galvanometer to show no deflection, V_B = V_D.

So, VA - I1P = VA - I2R This implies I1P = I2R (Equation 2)

Also, consider the current division at point B. The total current entering B is I₁. Part of it goes through the galvanometer (I_g) and the rest goes through Q (I₃). So, I1 = Ig + I3.

Similarly, at point D, the total current entering D is I₂. Part of it comes from the galvanometer (I₄) and the rest comes from S. So, I2 + I4 = I. And I1 + I4 = I. From current continuity at B and D: If I_g = 0, then I₁ = I₃ and I₂ = I₄.

Now consider the potentials at the points connected to the galvanometer (B and D). Let the voltage source be connected between A and C. Let P and Q be in one branch, and R and S be in the other. Potential at B: VB = VA * (Q / (P + Q)) (if no galvanometer current) Potential at D: VD = VA * (S / (R + S)) (if no galvanometer current)

For balance, V_B = V_D.

VA * (Q / (P + Q)) = VA * (S / (R + S)) Q / (P + Q) = S / (R + S) Q(R + S) = S(P + Q) QR + QS = SP + SQ QR = SP P/S = R/Q This is slightly different from the initial statement. Let's redraw and re-label.

Standard Wheatstone bridge diagram: A --- P --- B --- Q --- C | | | R G S | | | A --- R --- D --- S --- C (Voltage source between A and C)

The condition for balance is when the potential at B equals the potential at D. Let I₁ be the current through P and R. Let I₂ be the current through Q and S. Potential at B: VB = VA - I1P Potential at D: VD = VA - I2R For balance, VB = VD => I1P = I2R.

Also, current through Q is I1 - Ig and current through S is I2 + Ig. If Ig = 0, then I1 = I3 (current through Q) and I2 = I4 (current through S).

Using voltage division from A to C: VB = VA - I1P VD = VA - I2R Currents I1 and I2 depend on the resistances.

Let's use the potential divider rule directly assuming the bridge is balanced (Ig = 0). VB = VA * (Q / (P + Q)) VD = VA * (S / (R + S)) For balance, VB = VD. VA * (Q / (P + Q)) = VA * (S / (R + S)) Q / (P + Q) = S / (R + S) Q(R + S) = S(P + Q) QR + QS = SP + SQ QR = SP Rearranging this gives the standard form: P/R = S/Q or P/S = R/Q. The common form is P/Q = R/S. Let's ensure the labels are consistent. If P and R are adjacent arms, and Q and S are adjacent arms, and the galvanometer is across the junction of P and Q, and R and S. A -- P -- B | | G R Q | | D -- S -- C (Source across A and C) Balance condition: P/R = Q/S. This is the correct form for this labelling.

Let's use another common labelling: A -- P -- B -- S -- C | | G | R Q | | | | A -- R -- D -- Q -- C (Source across A and C) Balance condition: P/R = S/Q. This is the most common form.

If the bridge is balanced, the ratio of resistances in the ratio arms (P and R) must be equal to the ratio of resistances in the other two arms (S and Q).

Sensitivity of Wheatstone Bridge

The sensitivity of a Wheatstone bridge is its ability to detect small changes in resistance. It is maximum when all four resistances are of comparable magnitude. A higher sensitivity means a larger galvanometer deflection for a given imbalance.

The sensitivity is influenced by the EMF of the battery and the resistance of the galvanometer. For maximum sensitivity, the external resistance (battery resistance) should be equal to the internal resistance of the bridge (sum of the four arms).

Applications of Wheatstone Bridge

The Wheatstone bridge is used to measure unknown resistances with high accuracy. By keeping three resistances known and adjusting one until the galvanometer shows no deflection, the unknown resistance can be calculated using the balance condition.

It is also used in various sensors where a physical quantity (like temperature, pressure, strain) causes a change in resistance. Examples include:

• Resistance thermometers (RTDs)
• Strain gauges
• Photoconductive cells

Metre Bridge (Slide Wire Bridge)

The metre bridge is a practical application of the Wheatstone bridge principle. It is used to determine the unknown resistance of a wire or a resistor. As the name suggests, it uses a wire of uniform cross-section, typically 1 metre (100 cm) long, stretched along a scale.

Construction and Working Principle

A metre bridge consists of:

• A uniform wire of length 1 metre (or 100 cm), usually made of manganin or constantan (due to their low temperature coefficient of resistance).
• A metre scale along the wire.
• Two thick copper strips to hold the wire at the ends.
• A gap (usually two gaps) to insert the unknown resistance (X) and a known resistance (R).
• A sliding contact (jockey) that can move along the wire.
• A galvanometer to detect null deflection.
• A battery (usually 2V or 4V) connected across the ends of the wire.

The circuit is set up as follows:

1. The known resistance (R) is placed in one gap.
2. The unknown resistance (X) is placed in the other gap.
3. The galvanometer is connected between the jockey and the junction point between R and X.
4. The battery is connected across the ends of the 1-metre wire.

The jockey is slid along the wire until the galvanometer shows no deflection (null point). Let this point be J. Let the length of the wire from one end (say, A) to the jockey (J) be 'l' cm. The remaining length of the wire will be (100 - l) cm.

At the null point, the bridge is balanced. The resistance of the wire is uniform, so resistance is directly proportional to length. We can consider the wire as two resistances: 'l' cm and (100 - l) cm.

The metre bridge circuit can be represented as a Wheatstone bridge where:

• Resistance P = Known resistance R
• Resistance Q = Resistance of wire of length (100 - l) cm
• Resistance R = Unknown resistance X
• Resistance S = Resistance of wire of length l cm

Applying the Wheatstone bridge balance condition (P/R = S/Q): R / X = (Resistance of wire of length l cm) / (Resistance of wire of length (100 - l) cm)

Since the wire is uniform, resistance is proportional to length. Let the resistance per unit length be 'r'. Resistance of length l = r * l Resistance of length (100 - l) = r * (100 - l)

So, R / X = (r * l) / (r * (100 - l)) R / X = l / (100 - l)

Therefore, the unknown resistance X can be calculated as: X = R * (100 - l) / l

Sources of Error and Precautions

Several factors can lead to errors in measurements using a metre bridge:

• End Resistance: The copper strips and connecting wires have some resistance, which is not accounted for in the length measurement. This is known as end resistance. To minimize this, the jockey should be placed near the centre of the wire, and the null point should ideally be between 40 cm and 60 cm.
• Non-uniformity of the Wire: The wire might not have a perfectly uniform cross-section, leading to variations in resistance per unit length.
• Inaccurate Length Measurement: Errors in reading the scale or parallax error can occur.
• Contact Resistance: Poor connections can introduce additional resistance.
• Galvanometer Sensitivity: If the galvanometer is not sensitive enough, it might be difficult to find the exact null point.

Precautions:

• Use thick copper strips and wires for connections to minimize resistance.
• Ensure all connections are tight and clean.
• Do not pass current through the bridge for a long time, as the wire can heat up, changing its resistance (temperature coefficient).
• The null point should be obtained as close to the middle of the wire as possible (between 40 cm and 60 cm) to minimize the effect of end resistances.
• Repeat the experiment by interchanging R and X to get a more accurate result and to account for end resistances.
• The battery should be connected with the correct polarity.

Improving Accuracy: Two-Gap Metre Bridge

To account for end resistances, a two-gap metre bridge is used. In this setup, the known and unknown resistances are placed in two separate gaps. The experiment is performed twice:

1. First, R is in the left gap and X is in the right gap. Let the null point be at length l₁. The balance equation is: R / X = l1 / (100 - l1)
2. Then, R and X are interchanged. R is now in the right gap and X in the left. Let the new null point be at length l₂. The balance equation is: X / R = l2 / (100 - l2)

Multiplying these two equations: (R/X) * (X/R) = [l1 / (100 - l1)] * [l2 / (100 - l2)] 1 = [l1 * l2] / [(100 - l1) * (100 - l2)]

This equation doesn't directly help in finding X. Instead, we use the fact that the effective resistances of the gaps are different.

Let the resistance of the left gap (including end resistance) be rL and the right gap be rR. In the first reading: R / X = (rL + rwire1) / (rR + rwire2) where rwire1 = l1*runit and rwire2 = (100-l1)*runit. R / X = (rL + l1*runit) / (rR + (100-l1)*runit)

In the second reading (R and X interchanged): X / R = (rL + l2*runit) / (rR + (100-l2)*runit)

Taking the reciprocal of the second equation: R / X = (rR + (100-l2)*runit) / (rL + l2*runit)

Equating the two expressions for R/X: (rL + l1*runit) / (rR + (100-l1)*runit) = (rR + (100-l2)*runit) / (rL + l2*runit)

This becomes complicated. A simpler way to use the two readings is to realize that the average resistance of the wire per unit length can be considered.

From the first reading: X = R * (100 - l1) / l1 From the second reading: X = R * l2 / (100 - l2) (Note: here R is in the right gap, so the formula is X = R_known * (length_opposite_X) / (length_opposite_R_known)) Let's be careful with the formula in the second case. If X is in the left gap and R is in the right gap, and null point is l₂ from the left end (where X is), then X/R = l₂/(100-l₂). So X = R * l₂ / (100-l₂).

If we assume end resistances are negligible or equal, then both readings should give approximately the same X.

A better way to handle end errors: Let the resistance of the wire from the left end to the null point be rl and the resistance from the right end to the null point be rr. In the first case: R / X = rl1 / rr1, where rl1 = l1 * runit and rr1 = (100-l1) * runit. R / X = l1 / (100-l1). In the second case (R and X interchanged), let the null point be l2 from the end where X is now placed. X / R = l2 / (100-l2).

From the first equation: X = R * (100-l1) / l1 From the second equation: X = R * l2 / (100-l2)

The actual resistance of the wire segments are affected by end resistances. Let a be the resistance per unit length and e1 and e2 be the end resistances of the two gaps. In the first setup: R / X = (e1 + a*l1) / (e2 + a*(100-l1)) In the second setup: X / R = (e1 + a*l2) / (e2 + a*(100-l2))

Taking the reciprocal of the second equation: R / X = (e2 + a*(100-l2)) / (e1 + a*l2) Equating the two expressions for R/X: (e1 + a*l1) / (e2 + a*(100-l1)) = (e2 + a*(100-l2)) / (e1 + a*l2) This is complex.

The standard method to handle end errors is to assume that the resistance of the wire segments are l and 100-l and the balance condition is R/X = l/(100-l). If end resistances are significant and unequal (e_1 and e_2), the equations become: Case 1: R/X = (e_1 + l_1 r) / (e_2 + (100-l_1) r) Case 2: X/R = (e_1 + l_2 r) / (e_2 + (100-l_2) r) From Case 1: R(e_2 + (100-l_1)r) = X(e_1 + l_1 r) From Case 2: X(e_2 + (100-l_2)r) = R(e_1 + l_2 r)

To find X, we can rearrange the equations. A common simplification for exam purposes is to assume that e_1 = e_2 = e. Then: Case 1: R/X = (e + l_1 r) / (e + (100-l_1) r) Case 2: X/R = (e + l_2 r) / (e + (100-l_2) r)

A more direct method using the two readings: From Case 1: X = R * (e_2 + (100-l_1)r) / (e_1 + l_1 r) From Case 2: X = R * (e_1 + l_2 r) / (e_2 + (100-l_2) r)

If we multiply the apparent values of X obtained from the simple formula X = R * (100-l)/l: Apparent X₁ = R * (100-l1)/l1 Apparent X₂ = R * (100-l2)/l2 (This is actually X = R * l₂/(100-l₂) if X is in left gap and R in right) Let's correct the second equation: If R is in the left gap and X in the right, null point is l₁. R/X = l1/(100-l1) => X = R * (100-l1)/l1. If X is in the left gap and R in the right, null point is l₂. X/R = l2/(100-l2) => X = R * l2/(100-l2).

The true value of X is the geometric mean of the two apparent values of X obtained from the two settings: Xtrue = sqrt( Apparent X1 * Apparent X2 ) Xtrue = sqrt( [R * (100-l1)/l1] * [R * l2/(100-l2)] ) Xtrue = R * sqrt( [(100-l1)/l1] * [l2/(100-l2)] ) This formula helps to cancel out the effect of unequal end resistances.

Applications of Metre Bridge

The metre bridge is primarily used to:

• Determine the unknown resistance of a wire or component.
• Compare two unknown resistances.
• Determine the specific resistance of a wire.

It is a simple and effective instrument for resistance measurement in educational laboratories.

Comparison: Wheatstone Bridge vs. Metre Bridge

While both are based on the same principle, they differ in application and complexity:

Feature	Wheatstone Bridge	Metre Bridge
Principle	Balance of four resistances	Application of Wheatstone bridge principle
Components	Four fixed resistors, galvanometer, battery	Uniform wire (1m), jockey, known/unknown resistors, galvanometer, battery
Measurement	Measures unknown resistance with high accuracy, especially small resistances. Can be used for sensors.	Measures unknown resistance using a null point on a wire. Primarily used for medium resistances.
Accuracy	Can be very accurate, especially with sensitive galvanometers and precise known resistors.	Accuracy limited by wire uniformity, end resistances, and reading precision. Generally less accurate than a well-set-up Wheatstone bridge for small resistances.
Practicality	Can be complex to set up for precise measurements, especially for sensors.	Simple to construct and use in labs for educational purposes.