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Physics Topics are often described using mathematical equations, making them precise and quantifiable.
What do you Mean by End Error of a Metre Bridge? How can this Error be Remedied?
It is another practical form of the Wheatstone bridge. It also helps in measuring ordinary resistances very easily.
From the null condition of Wheatstone bridge we have,
S = R ᐧ \(\frac{Q}{P}\) ….. (1)
In metre bridge, steps that are followed are mentioned below.
- The unknown resistance is placed in the fourth arm S.
- The magnitude of R, i.e., the resistance of the third arm is kept constant.
- By increasing or decreasing the ratio \(\frac{Q}{P}\), the bridge is brought to the balanced condition.
- Now by using equation (1) the unknown resistance S is calculated.
Description: The Fig.(a) shows a metre bridge circuit. It consists of a one metre long thin uniform wire AB made of manganin or German silver. The wire is stretched and fixed to the two points a and f of two copper strips ab and fe and is placed along a metre scale M over a wooden board. There is another copper strip cd. The gaps bc and de are called the left gap and the right gap, respectively. A jockey J is connected to the mid point of the copper strip cd through a galvanometer G and its sharp end touches the wire AB and can be moved along it when required. A resistance box R is inserted in the left gap (third arm of the bridge). The resistance S to be determined is introduced in the right gap (fourth arm of the bridge). A cell E is connected to the bridge through a commutator C.
Working principle: Fig.(b) shows an equivalent metre bridge circuit. Obviously, it is a Wheatstone bridge circuit.
The direction of current through the wire AB can be reversed with the help of the commutator C. The left portion of the wire with respect to the position of the jockey is the first arm of the bridge and the right portion of the wire is the second arm, i.e., resistance of the portion AJ of the wire = P and resistance of the portion JB of the wire = Q.
A suitable resistance R is introduced in the resistance box. Now the jockey is moved along the wire till a null point is reached. At this balanced condition, let the length of AJ = 1 cm.
∴ Length of JB = (100 – l) cm [∵ AB = 1 m = 100 cm]
If ρ be the resistance per unit length of the wire then,
P = ρl and Q = ρ(100 – l)
∴ \(\frac{Q}{P}\) = \(\frac{100-l}{l}\) and S = \(R \cdot \frac{Q}{P}\) = \(R \cdot \frac{100-l}{l}\) ….. (2)
Knowing l from the scale, S can be calculated from the equation (2). Taking different resistances from the box R the experiment is repeated.
Discussion:
i) Material of the wire: The wire of the metre bridge is generally made of alloys like manganin, German silver, etc. The resistance of a uniform manganin wire of length 1 m and diameter 0.5 mm is about 2 Ω. The bridge becomes sensitive enough for this magnitude of resistance of the wire. On the other hand, the resistance of a copper wire of the same length and diameter is about 0.1 Ω. The bridge is not at all sensitive if such a low resistance is used. For this reason copper wire is never used in a metre bridge.
ii) Thermoelectric effect: The copper strips at the ends A and B of the wire come in contact with manganin or German silver. Now, if current passes through the junctions of two dissimilar metals, heat is evolved at one junction and is absorbed at the other due to thermoelectric effect. So, the wire AB does not remain in thermal equilibrium and hence its resistance per unit length, ρ is not uniform. Therefore, when we apply equation (2) to calculate the unknown resistance, error occurs. To remove this defect it is necessary to reverse the direction of current through the wire AB. Then, heat will be absorbed at the end where it was evolved originally and vice versa. In this way the ther-mal equilibrium of the wire AB will be maintained. To reverse the direction of current in the wire AB, a commu-tator C is used [Fig.].
iii) End error:
- Though the resistance of the copper strips is very small, it is finitely non-zero.
- The bridge wire is soldered at the two ends A and B with the copper strips. So some resistance exists at these two places.
- The 0 and 100 cm marks of the metre scale (M) may not coincide accurately with the ends A and B of the bridge wire. Hence it is assumed that some additional resistance exists at the two ends of the bridge wire.
This is called end resistance. So when we measure an unknown resistance, we face some end errors.
End correction: The end resistance may be assumed to be equivalent to the resistance of a certain length of the bridge wire. Suppose, end resistance of end A = resistance of λ1 cm of the bridge wire
and end resistance of end B = resistance of λ2 cm of the bridge wire
i.e., resistance of l1 length of the bridge wire = resistance of (1 + λ1) length
and resistance of l2 length of the bridge wire = resistance of (l2 + λ2) length
In the Fig.(a) if the position of null point be l1 cm then we have from equation (2),
\(\frac{S}{R}\) = \(\frac{\left(100-l_1\right)+\lambda_2}{l_1+\lambda_1}\) …… (3)
Now, by interchanging the position of R and S, if the null point is obtained at l2 cm, then we have
\(\frac{R}{S}\) = \(\frac{\left(100-l_2\right)+\lambda_2}{l_2+\lambda_1}\) …… (4)
If we use two known resistances R and S, we can find out λ1 and λ2 by solving equations (3) and (4). Next, using the values of λ1 and λ2, the unknown resistance can be measured accurately.
iv) Position of the null point: The bridge becomes sensitive if the position of the null point lies in between 40 cm and 60 cm of the bridge wire. But it is better to avoid the null point at 50 cm, because in that case no change is observed in the reading of the null point on interchanging the positions of R and S.
Numerical Examples
Example 1.
In a metre bridge experiment, a null point is obtained at a length of 39.8 cm when a 2 Ω resistance is placed in the left gap and a 3 Ω resistance in the right gap. If the two resistances are interchanged, the null point is obtained at 60.8 cm. Calculate the end errors of the bridge.
Solution:
Suppose, resistance per unit length of the bridge wire = ρ Ω ᐧ cm<sup.-1. End resistance of the left end of the bridge = resistance of λ1 cm of the wire. End resistance of the right end of the bridge = resistance of λ2 cm of the wire. These two are the end errors of the bridge.
Therefore, if l be the position of the null point, then according to the relation = we have,
Solving (1) and (2) we get,
λ1 = 2.2 and λ2 = 2.8
So, the left end resistance and the right end resistance of the bridge are equal to the resistances of 2.2 cm and 2.8 cm of the bridge wire respectively.
Example 2.
In the left gap of a metre bridge there is a coil of copper and in the right gap there is a fixed resistance. If the coil of copper is dipped in ice the balance point is obtained at 41.2 cm of the bridge wire. Next the coil is taken off from ice and placed in a container of hot water. Now the balance point is shifted by a distance 8.1 cm towards right. What is the temperature of hot water? (Temperature coefficient of resistance of copper = 42.5 × 10-4 °C-1.)
Solution:
Suppose, fixed resistance = It, temperature coefficient of resistance of copper = α, resistance of the coil at 0°C = R0, position of the balance point = l0, resistance of the coil at t°C = Rt and position of the balance point = l.
∴ Rt = R0(1 + αt)
According to the principle of the metre bridge,
In case of 0°C, \(\frac{R_0}{R}\) = \(\frac{l_0}{100-l_0}\) …. (1)
In case of t °C, \(\frac{R_0(1+\alpha t)}{R}\) = \(\frac{l}{100-l}\) ….(2)
Dividing (2) by (1) we have,
Example 3.
In a meter bridge the balance point is found to be at 40 cm from one end when the resistor at the end is 15 Ω. Find the resistance at the other side. [WBCHSE Sample Question]
Solution:
If ρ be the resistance per 1 cm length of the metre wire, then in the present case,
P = 40ρΩ, Q = (100 – 40)ρ = 60ρΩ, R = 15Ω,
S = ?
In balance condition, \(\frac{P}{Q}\) = \(\frac{R}{S}\)
or, S = R\(\frac{Q}{P}\) = 15 × \(\frac{60 \rho}{40 \rho}\) = 22.5Ω
Example 4.
In the circuit given, E1 = 6 V, E2 = 2V, E3 = 3V, C’ = 5µF, R1 = 2R2 = 6Ω, R3 = 2R4 = 4Ω. Find the current in R3 and the energy stored in the capacitor.
Solution:
Let the distribution of currents in the various branches be as shown below.
In steady state a capacitor offers infinite resistance and the branch containing the capacitor acts as an open path. So no cur-rent flows through C’.
Considering the closed loop CDEHC,
I2R3 = E1 or, 4I2 = 6 or, I2 = 1.5 A
∴ Current through the resistance R3 = 1.5 A
Considering the closed loop AHEFGA,
(I1 – I2) × R2 – I2R3 + (I1 – I2)R4 = -E2 – E3
or, (I1 – 1.5) × 3 – 1.5 × 4 + (I1 – 1.5) × 2 = – 2 – 3
∴ I1 = 1.7 A
Let VA and VC be the potentials at the points A and C respec-tively. As the current flows from the point A to C along the path AHC,
VA = VC + potential drop across R2 + E2
or, VA – VC = (I1 – I2)R2 + E2 = (1.7- 1.5) × 3 + 2
∴ VA – VC = 2.6 V
Through branch ABC current = 0. Hence the potential differ-ence across the capacitor is equal to potential difference between the points A and C.
∴ Potential difference across the capacitor,
VA – VC = 2.6 V
∴ Energy stored in the capacitor,
U = \(\frac{1}{2} C V^2\) = \(\frac{1}{2}\) × 5 × 10-6 × (2.6)2 = 1.69 × 10-5J
Example 5.
If the resistance X and Y (X < Y) be placed in the two gaps of a metre bridge, null point is obtained at a length of 20 cm. Keeping Y unchanged if a resistance 4X is placed in place of X what will be the position of the null point? [AIEEE 04]
Solution:
In the first case, \(\frac{X}{Y}\) = \(\frac{20}{100-20}\)
or, Y = 4X
So, if resistance 4X be placed in place of X, resistances of the two gaps will be equal, since Y = 4X.
Therefore, the null point will be at the middle of the bridge wire i. e., at 50 cm.
Example 6.
The distance between the positions of two null points obtained in a metre bridge wire of length 100 cm, by interchanging a known resistance of 2.5 Ω and an unknown resistance in the two gaps, is 28.6 cm. Find the value of the unknown resistance. [HS ’11]
Solution:
In first case, if the unknown resistance r is kept at the left gap and resistance of 2.5Ω at the right gap, the position of null point is l1.
∴ \(\frac{r}{2.5}\) = \(\frac{l_1}{100-l_1}\) …. (1)
In second case, after interchanging the resistances, the position of null point is l2.
∴ \(\frac{2.5}{r}\) = \(\frac{l_2}{100-l_2}\) …… (2)
According to the question,
l1 – l2 = 28.6 ….. (3)
By (1) × (2) we get,
l = \(\frac{l_1 l_2}{\left(100-l_1\right)\left(100-l_2\right)}\)
Solving equation (3) and (4),
∴ l1 = 64.3
and l2 = 100 – l1 = 35.7
∴ The value of the unknown resistance
r = 2.5 × \(\frac{l_1}{100-l_1}\) = 2.5 × \(\frac{64.3}{100-64.3}\) = 4.5 Ω