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Advanced Physics Topics like quantum mechanics and relativity have revolutionized our understanding of the universe.
How is Wheatstone bridge calculated? What is the Wheatstone bridge used for?
Description: Four resistances P, Q, S, R form the four arms AB, BC, CD and DA respectively of a quadrilateral
ABCD [Fig.]. A battery E is connected between A and C and a galvanometer of resistance G is connected between B and D. This complete arrangement is called a Wheatstone bridge circuit. In this circuit the first arm AB has resistance P, the second arm BC has resistance Q, the third arm AD has resistance R and the fourth arm DC has resistance S.
Working principle: The resistances P, Q, R, S for the four arms of the bridge are so selected that galvanometer deflection is zero i.e., galvanometer current ‘G = o . This is called null condition or balanced condition of the bridge. In the null condition, the relation of the resistances is:
\(\frac{P}{Q}\) = \(\frac{R}{S}\) …… (1)
Of the four resistances, if three are known, the fourth resistance can be determined from the above relation. Usually, the unknown resistance is placed in the fourth arm i.e., in place of S.
Proof of the relation P/Q = R/S: In the null condition, IG = 0. So, the potential difference between the two ends of the galvanometer,
VB – VD = IG ᐧ G = 0 [G = resistance of galvanometer]
or, VB = VD
For IG = o, suppose current in resistance P = current in resistance Q = I1 and current in resistance R = current in resistance S = I2.
∴ VA – VB = I1P ; VA – VD = I2R
and VB – VC = I1Q ; VD – VC = I2S
Now, since VB = VD,
∴ I1P = I2R …. (2)
and I2Q = I2S …. (3)
Now, dividing equation (2) by equation (3) we have,
\(\frac{P}{Q}\) = \(\frac{R}{S}\)
On the basis of this relation the arms P and Q are called ratio arms.
Discussion:
i) Wheatstone bridge is used for measurement of ordinary resistances (from 1 Ω to about 1000 Ω).
ii) If a low resistance (less than 1 Ω) is placed in the position of S, the resistances of the connecting wires become nearly equal to S. So the value of S becomes erroneous due to the resistance of the connecting wires. Therefore, for measurement of low resistances Wheatstone bridge is not used.
iii) If a high resistance (higher than 1000 Ω) is placed in the position of S, a very small current will pass through it. So in the null condition the value of I1 is much greater than I2 ie., almost the entire current passes through P and Q. In
that case, even when the bridge is unbalanced, no appreciable current flows through the galvanometer, and the galvanometer deflection stays at zero. Then, it becomes difficult to identify the actual null condition. In other words, the sensitivity of the bridge decreases. So, Wheatstone bridge is not used for measurement of high resistances.
iv) If the bridge is sensitive enough, even in slightly off-balance condition the galvanorneter deflection is visible distinctly.
The conditions of sensitiveness of this bridge are:
- The galvanometer must be very sensitive. Even for the passage of a very feeble current through the galvanometer, the deflection of the pointer should be distinct.
- The four resistances P, Q, R, S as well as the resistance of the galvanometer G should be of almost equal value i.e., P ≈ Q ≈ R ≈ S ≈ G.
It is to be noted that, null condition does not depend on galvanometer resistance G, but sensitivity of the bridge rely on G. Current can flow through galvanometer easily if G has lesser value. On the other hand, if the number of turns of the coil of galvanometer is increased pointer of the galvanometer deflects more and G also increases. So a galvanometer is designed for optimum sensitivity. Various types of galvanometer are avialable based on which circuit it is connected. In Wheatstone bridge, if galvanometer resistances are nearly equal to the resistances of four arms i.e, P ≈ Q ≈ R ≈ S ≈ G bridge becomes most sensitive.
v) Galvanorneter is a very sensitive instrument br detection and measurement of current. While making adjustments for the balanced condition of the bridge, there is possibility of high current passing through the galvanometer. Thus the galvanometer may get damaged. For this reason, in the beginning, it is judicious—
- to connect a high resistance in series with the galvanometer, or
- to connect a shunt in parallel with the galvanometer.
Later when the bridge comes close to the balanced condition,
- the resistance can be removed or a shunt can he added to it, or
- the shunt which is connected with the galvanometer may be removed.
vi) The balanced condition does not depend on the emf of the battery, its internal resistance or any resistance connected with the battery. However, generally a cell of low emf (e.g., Leclanche cell) is used. Otherwise, due to flow of large currents, the resistances of P, Q, R, S may increase due to heating effect.
vii) If the position of the battery and the galvanometer are interchanged, the battery exists in between B and D and the galvanometer in between A and C. Then Q, S, P and R become first, second, third and fourth arms respectively of the bridge. So in this case the null condition is \(\frac{Q}{S}\) = \(\frac{P}{R}\).
Obviously this condition is identical with the condition \(\frac{P}{Q}\) = \(\frac{R}{S}\). So it can be said that the balanced condition of the Wheatstone bridge remains unchanged in spite of the interchange of the positions of the battery and the galvanometer. For this reason the diagonals AC and BD of the bridge are called mutually conjugate.
viii) The sensitivity of the bridge Is not the same in the two mutually conjugate positions. The bridge becomes more sensitive if the galvanometer is connected between the junction of the two smaller resistances and the junction of the two larger resistances.
ix) When the equivalent resistance of the bridge is calculated in balanced condition, the resistance of the galvanometer i.e., the resistance of the arm BD should not be taken into consideration.
Practical applications of Wheatstone bridge: For measurement of various electrical quantifies in the laboratory, Wheatstone bridge and some of its modified forms are extensively used. Two familiar forms of Wheatstone bridge for measurement of resistances in the laboratory are:
- post office box and
- metre bridge. (Discussion on post office box is beyond our present syllabus.)
Numerical Examples
Example 1.
Resistances of the ratio arms of a Wheatstone bridge are 100Ω and 10Ω respectively. An unknown resistance is placed in fourth arm and the galvanometer current becomes zero when 153Ω resistance is placed in third arm. What is value of unknown resistance?
Solution:
Here, P = 100Ω, Q = 10Ω and R = 153Ω
Since, \(\frac{P}{Q}\) = \(\frac{R}{S}\) ∴ S = \(\frac{R Q}{P}\) = \(\frac{153 \times 10}{100}\) = 15.3Ω
Example 2.
Five resistances are connected according to the Fig. What ¡s the effective resistance between the points A and B?
Solution:
Here, \(\frac{2 \Omega}{3 \Omega}\) = \(\frac{4 \Omega}{6 \Omega}\)
So, the circuit between the points A and B is a balanced Wheat-stone bridge. So no current flows through the resistance of 7 Ω.
In this condition to calculate the effective resistance, the resistance of 7 Ω will be ignored.
Equivalent resistance of 2 Ω and 3 Ω = 2 + 3 = 5 Ω
Equivalent resistance of 4 Ω and 6 Ω = 4 + 6 = 10 Ω
Now at the points A and B this 5 Ω and 10 Ω resistances are connected in parallel.
So, the effective resistance = \(\frac{5 \times 10}{5+10}\) = \(\frac{50}{15}\) = \(\frac{10}{3}\)Ω
Example 3.
In a Wheatstone bridge, a 15Ω resistance and an unknown resistance are placed in third arm and fourth arm respectively. Current through galvanometer becomes zero when ratio of resistances of first and second arms is 3 : 2. Find the value of unknown resistance.
Solution:
Here, R = 15Ω and \(\frac{P}{Q}\) = \(\frac{3}{2}\)
Since, \(\frac{P}{Q}\) = \(\frac{R}{S}\)
∴ S = R ᐧ \(\frac{Q}{P}\) = 15 × \(\frac{2}{3}\) = 10Ω
Example 4.
The resistances of the four arms of a Wheatstone bridge are 100Ω, 10Ω, 300 Ω and 30 Ω respectively. A battery of emf 1.5 V and negligible internal resistance is connected to the bridge. Calculate the current flowing through each resistance.
Solution:
Here, P = 100 Ω,
Q = 10Ω, R = 300Ω, S = 30Ω and E = 1.5V [Fig.].
Since, \(\frac{100}{10}\) = \(\frac{300}{30}\)
ie., \(\frac{P}{Q}\) = \(\frac{R}{S}\), the bridge is in a balanced condition.
∴ Current in the resistance P = current in the resistance Q
= \(\frac{V_{A B}}{P+Q}\) = \(\frac{1.5}{100+10}\) = \(\frac{1.5}{110}\) = 0.0136 A
Again, current in the resistance R = current in the resistance S
= \(\frac{V_{A B}}{R+S}\) = \(\frac{1.5}{300+30}\) = \(\frac{1.5}{330}\) = 0.0045 A
Example 5.
In Fig. (a), every resistance is of magnitude r. What is the equivalent resistance between A and B?
Solution:
In Fig. (b) the Wheatstone equivalent circuit of Fig.(a).
Here, P = Q = R = S = r
∴ \(\frac{P}{Q}\) = \(\frac{R}{S}\)
i.e., the circuit is in a balanced condition. So for calculation of equivalent resistance, the value of CD is of no use.
Now, resistance along the path ADB = r+ r = 2r and resistance along the path ACB = r + r = 2r
Since ADB and ACB are connected in parallel, the equivalent resistance between the points A and B is
R’ = \(\frac{2 r \times 2 r}{2 r+2 r}\) = \(\frac{4 r^2}{4 r}\) = r
Example 6.
Each resistance In the given circuit in Fig.(a) is of value R. Calculate the equivalent resistance of the circuit with respect to the points A and B.
(Karnataka CET ‘02)
Solution:
The equivalent circuit has been shown in Fig.(b).
Now proceeding in the same way as done in example 1, it can be shown that equivalent resistance = R.
Example 7.
A coil of wire is kept in melting ice its resistance measured by a Wheatstone bridge is 5 Ω. If the coil is heated to 100°C and another wire of resistance 100 Ω is connected in parallel to it, the balanced condition of the bridge remains unchanged. Determine the temperature coefficient of resistance of the wire of the coil.
Solution:
If the resistance of the coil at 100°C be R, then the equivalent resistance of the parallel combination of R and 100 Ω is obviously 5 Ω.
∴ \(\frac{R \times 100}{R+100}\) = 5 or, 95R = 500 or, R = \(\frac{100}{19} \Omega\)
Again, if α be the temperature coefficient of resistance of the wire of the coil, then
Example 8.
An electrical circuit is shown in Fig.(a). Calculate the potential difference across the resistance 400 Ω as will be measured by the voltmeter V of resistance 400 Ω.
Solution:
The equivalent circuit is shown in Fig.(b).
The equivalent resistance between B and C = \(\frac{400 \times 400}{400+400}\)
= 200 Ω
This is a Wheatstone bridge circuit, where
So, the bridge is in balanced condition, i.e., no current flows in the arm BD. Under this condition the reading of the voltmeter C is
VB – VC = \(\frac{V_A-V_C}{P+Q}\) × Q
= \(\frac{10}{100+200}\) × 200 = \(\frac{20}{3}\) = 6.67 V
Example 9.
ABCÐ is a Wheatstone bridge in which the resistance of the arms AB, BC, CD and DA are respectively 2 Ω, 4 Ω, 6 Ω and 8 Ω. The points A and C are connected to the terminals of a cell of emf 2 V and negligible internal resistance. The points B and D are connected to a galvanometer of resIstance 50 Ω. Using Kirchhoff’s laws find the current flowing through the galvanometer. [WBCHSE Sample Question]
Solution:
In the circuit of Fig., let VC = 0; then VA = 2 V; again, if VB = V1 and VB – VD = V, then VD = V1 – V
Now, applying Kirchhoff’s 1st law at the junction B, we get
Doing (1) × 7 – (2) × 3 we get,
575V = 250 or, V = \(\frac{250}{575}\)V
∴ IG = \(\frac{V}{50}\) = \(\frac{250}{575 \times 50}\) = \(\frac{1}{115}\) = 0.0087 A = 8.7 mA
Example 10.
We determine the value of the fourth resistance to be 8 Ω with the help of a Wheatstone bridge with three known resistances 100 Ω, 10 Ω, 80 Ω respectively. If the emf of the cell and Its internal resistance be 2V and 1.1 Ω respectively, find the current passing through the cell. [HS ‘06]
Solution:
The bridge is in balanced condition, because \(\frac{100}{10}\) = \(\frac{80}{8}\), i.e., galvanometer current is zero. In this case the effective circuit diagram is shown in Fig. Total resistance of the circuit,
∴ Current through the cell, i.e., main current of the circuit,
I = \(\frac{2}{50}\) = 0.04 A