Meter Bridge: Definition, Working Principle & Examples

Meter Bridge, the slide wire bridge, is an instrument that works on the principle of a Wheatstone bridge. It is one of the systems under Current electricity Mechanism. Meter Bridge has a uniform cross sectional area stretched tightly and clamped between two thick metallic strips which bent at right angles. It is basically used to measure the resistance of wires, coils or any other material.

Key Takeaways: Capacitance, Electric Charge, Electric Potential, Current Electricity, Meter bridge, Wire bridge, Electric current, Flow rate, Angle

Current Electricity 

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• The flow of electric charge is known as electric current. It is calculated as a charge flow rate. It refers to the volume of charge flowing over time.
• Consider a charge of 20Coulomb flowing at a rate of one per second.
• Then it becomes 20Coulomb/sec.
• I stand for electric current.
• It's written as I=dQ/dt, where dQ is the rate of charge shift over time.
• Ampere is a unit of measurement (A).

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Current Electricity Detailed Video Explanation:

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What is Meter Bridge?

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A Meter connect is a sort of electrical circuit utilized in estimating an electrical obstruction, which is obscure by adjusting its two legs of the scaffold circuit, where one of the legs incorporates an obscure segment. The Wheatstone Bridge can in any case be utilized in estimating light estimations of protections around the scope of milli-Ohms.

Read More: Electric Charges and Fields Notes

How does Meter Bridge use the Unknown Resistance?

A meter connect is a device used in finding the obscure opposition of a curl. Underneath figure 12 is the graph of a valuable meter connect instrument.

In the above figure, R is called the Resistance, P is the Resistance running over AB, S is the Unknown Resistance, Q is the Resistance between the joints BD. 

AC is the long wire estimating 1m long and it is made of constantan or manganin having a uniform territory of the cross-area Such that L₁ + L₂ = 100 

Expecting that L₁ = L => L₂ = 100 – L 

Connection acquires the obscure obstruction 'X' of the given wire: 

X = RL₂/L₁ = R(100 – L)/L 

Read More: Unit of Voltage

What's more, the particular obstruction of the material for a given wire is gotten by the connection = (3.14) r2X/l 

where r = the sweep of the link and l = length of the wire.

Also Read: Verify the laws of parallel combination of resistances using a metre bridge experiment

How to Find Unknown Resistance Using Meter Bridge?

To find the unknown resistance using meter bridge, given below step-by-step procedure must be followed:

• Gather the instruments and get ready associations as demonstrated in the above figure. 
• Take some appropriate sort of obstruction 'R' from the opposition box. 
• Contact jockey at point A; look that there exists an avoidance in the galvanometer on one of the sides, at that point contact the rider on point C of wire, at that point the diversion in the galvanometer must be on another side. 
• Discover the situation of the invalid point having redirection in the galvanometer that gets zero.
• Proceed with the above strategy for various estimations of the 'R'. Note probably around 5 readings. 
• Consider where the galvanometer shows a 0 avoidance; this is known as the equilibrium point. 
• Presently, Measure the length of a given wire by the utilization of standard scale and range of the wire by the usage of a screw check, (Take in any event five readings). 
• Compute Mean Resistance of Single Unknown Resistance = Total Sum of protections of Unknown opposition from the over five readings)/5.

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Sample Questions

Ques. What is a Meter Bridge? For what reason is it purported? [1 mark]

Ans. Meter Bridge is an instrument for looking at protections. It consists of a scaffold wire of length 1m. It has a uniform cross sectional area stretched tightly and clamped between two thick metallic strips which bent at right angles. 

Ques. What is the rule of Meter Bridge? [1 mark]

Ans. At the point when the extension is adjusted P/Q = R/S 

Ques. Would you be able to discover exceptionally high protections precisely with the assistance of a meter connect? [1 mark]

Ans. No, high protections precisely with the assistance of a meter connect cannot be discovered.

Ques. What is the material of the scaffold: What is the measure for determinations?[1 mark]

Ans. Manganin wire of one meter is used as a scaffold wire. Measure for determination is ‘Low-temperature coefficient’.

Ques. Why are copper strips on the scaffold thick? [1 mark]

Ans. Copper strips on the scaffold thick are placed to limit the obstruction in the meter bridge.

Ques. a) Write the principle of working of a meter bridge. [2 marks]

(b) In a meter bridge, the balance point is found at a distance  I₁ with resistances R and S as shown in the figure.

An unknown resistance X is now connected in parallel to the resistance S and the balance point is found at a distance I₂ . Obtain a formula for X in terms of I₂ , I₂ and S. [Outside Delhi, 2017, 3 marks]

Ans.

(a) Meter bridge works on the principle of balanced Wheatstone bridge i.e., when the bridge is balanced

Ques. (a) State the working principle of a meter bridge used to measure an unknown resistance.
(b)Give reason.
(i) why the connections between the resistors in a meter bridge are made of thick copper strips.
(ii) why is it generally preferred to obtain the balance length near the mid-point of the bridge wire.
(c) Calculate the potential difference across the 4 ? resistor in the given electrical circuit, using Kirchhoff’s rules. [Outside Delhi, 2019, 5 marks]

Ans. Meter bridge is a practical apparatus which works on principle of Wheat-Stone bridge. It is used to measure unknown resistance experimentally.

(b) (i) Connection between resistors are made of thick copper strips so that it will have maximum resistance and location of point of balance (D) will be more accurate which results in correct measurement of unknown resistance.

(ii) It is preferred to obtain the balance length near the mid-point of the bridge wire because it increase the sensitivity of meter bridges.

(c) From KCL (Kirchhoff’s current law) at point D

Ques. Two resistances are connected in two gaps of Meter Bridge, and the balance is 20cm from the zero end. A resistance of 15 ohms is connected in series with the smaller of the two. The null point shifts to 40cm. Find the value of the bigger resistance. (3 marks)

Ans. Let P be the smaller resistance and Q be the bigger resistance.
First case → P/Q=2080=14
Second case → (P+15)/Q=4060=23
Comparing both → P/(P+15)=14×32=38
8P = 3P + 45 → 5P = 45 → P = 9 ohms
Therefore, substituting in P/Q = 1/4 → 9/Q = 1/4 → Q = 36 ohms.

Ques. When the radius of the wire is tripled what is the effect on null deflection of galvanometer? (2 marks)

Ans. For a balanced Meter Bridge P/Q = x/(100−x), we can understand that the null deflection of galvanometer does not depend on the radius of the wire. Therefore, even if the radius of the wire is tripled, the null deflection of the galvanometer undergoes no change.

Ques. A resistance of 5 ohms is connected across the gap of a Meter Bridge and an unknown resistance, whih is greater than 5 ohms, is connected across the other gap. When these resistances are interchanged, the balance point shifts by 50 cm. What is the unknown resistance? The length of the wire is 150 cm. (4 marks)

Ans. Let x be the unknown resistance.
First case → 5/x = l/(150–l)
750 – 5l = xl ……….. 1
Second case → x/5 = (l+50)/(100–l)
100 – xl = 5l + 250
xl = 100x – 5l – 250 ………… 2
Comparing 1 and 2
750 – 5l = 100x – 5l – 250
100x = 750 + 250
100x = 1000
x = 10 ohms
Therefore, the unknown resistance is 10 ohms.

Ques. State whether it is true or false: The sensitivity of the meter bridge is at the peak when all resistors have the same order. (2 marks)

Ans. The statement is true.
Explanation: Yes, when all resistors have the same order, the sensitivity of the meter bridge is at the peak. The sensitivity can be increased by keeping the current in the galvanometer high and this can be achieved by reducing the values of the resistors used in the Meter Bridge.

Ques. The four arms of a Wheatstone connect (Fig) have the accompanying protections: AB = 100ω, BC = 10ω, CD = 5ω, and DA = 60Ω.A galvanometer of 15ω obstruction is associated across BD. Compute the current through the galvanometer when a likely contrast of 10 V is kept up across AC.

Ans. Considering the cross-section BADB, we have 

100I₁ + 15Ig – 60I₂ = 0 

or on the other hand 20I₁ + 3Ig – 12I₂= 0 (a) 

Considering the lattice BCDB, we have 

10 (I₁ – Ig) – 15Ig – 5 (I₂ + Ig) = 0 

10I₁ – 30Ig – 5I₂ = 0 

2I₁ – 6Ig – I₂ = 0 (b) 

Considering the lattice ADCEA, 

60I₂ + 5 (I₂ + Ig) = 10 

65I₂ + 5Ig = 10 

13I₂ + Ig = 2 (c) 

Increasing Eq. () by 10 

20I₁ – 60Ig – 10I₂ = 0 (d) 

From Equations (d) and (a) we have 

63Ig – 2I₂ = 0 

I₂ = 31.5Ig [(e)] 

Subbing the estimation of I₂ into Eq. (c), we get 

13 (31.5Ig) + Ig = 2 

410.5 Ig = 2 

Ig = 4.87 mA.

Ques. In a meter connect (Fig.), the invalid point is found a good way off of 33.7 cm from A. If now opposition of 12ω is associated with S, the invalid point happens at 51.9 cm. Decide the estimations of R and S.

Ans. From the first balance point, we get

(R/S) = (33.7)/ (66.3) = (3.87) (i)

After S is connected with a resistance of 12Ω, the gap changes from S to Seq, where

Seq= (12S)/(S+12) and hence the new balance condition now gives

(51.9)/ (48.1) =(R/ Seq) =R(S+12)/ (12S)

Substituting the value of (R/S) from Eq. (i), we get

(51.9)/ (48.1) = ((S+12)/12) x (33.7)/ (66.3)which gives

S = 13.5Ω. Using the value of (R/S) above, we get

R = 6.86 Ω

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