Question

Obtain the condition for balancing of Wheatstone’s bridge. The resistance of 4 arms of
the Wheatstone’s bridge are 10 ῼ,20 ῼ, Xῼ and 40 ῼ respectively. What is the value
of X.?

Answer 1

Answer:

P/Q=R/S

10/20=X/40SO X=20

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Explanation:

Answer 2

Answer:

please mark thank mark brilliant for this much long detailed answer

Explanation:

Method 1

Method one for finding the unknown resistance

The bridge is said to be balanced when there is no current flowing through the galvanometer. This means that the potential difference or voltage between the points Q and S is zero. In this case the current flowing through the fixed resistors R1 and R2 is the same and is considered as I1. The current flowing through the variable resistor R3 and the unknown resistor Rx will be the same and is I2. As the potential at Q and S is the same, the voltage drop from the point P to Q is equal to the voltage drop at from point P to point S.

Wheatstone Bridge

Wheatstone Bridge

Here I1 R1 = I2 R3 – equation 1. Now the voltage drop from point Q to point R is equal to the voltage drop from point S to R. So I1 R2 = I2 RX – equation 2. Now dividing the equation 1 by 2 we get

I1 R1 = I2 R3 / I1 R2 = I2 RX. Thus R1 / R2 = R3 / Rx.

R1 Rx = R2 R3

Thus Rx = R2 R3 / R1. The unknown resistance is determined in terms of other known resistors in the bridge.

Problem: Consider a bridge circuit where R1 = 50 Ω, R2 = 10 Ω, R3 = 20 Ω. Now find the value of unknown resistance Rx

Solution:

R1 / R2 = R3 / Rx.

Rx = R2 R3 / R1

= 10 * 20 / 50

= 200 / 50

= 40 Ω

Method 2

Method two for finding the unknown resistance

The value of the unknown resistance can be found in other ways too. Consider the same figure given above. When Ig, the current through the galvanometer is zero the bridge is said to be balanced. Apply Kirchhoff’s loop rule to the closed loops PSQP and RQSR. Kirchhoff’s loop rule or voltage law states that the sum of voltage drops is equal to the sum of voltage rise. The algebraic sum of all voltages across the circuit will be zero.

Wheatstone Bridge circuit

Wheatstone Bridge circuit

Consider the first loop PSQP.

-I2R3 + 0 + I1R1 = 0

From the loop RQSR, We get

I1R2 + 0 – I2Rx = 0

From the first loop we get, I2 / I1 = R1 / R3. Also from the second loop we get I2 / I1 = R2 / Rx.

Thus R1 / R3 = R2 / Rx, Rx = R2 R3 / R1.

Problem: Calculate the current through the galvanometer connected across P and R of 10 Ω resistance with a potential difference of 20 V

Solve the circuit

Solve the circuit

Solution:

Consider the mesh PRQP

50 I1 – 30 I2 + 10 Ig = 0

= 5 I1 – 3 I2 + 1 Ig = 0 – equation 1

Consider the mesh PSQP

100 (I1 – Ig) - 40 (I2 + Ig) - 10 Ig = 0

= 100 I1 - 100 Ig - 40 I2 - 40 Ig - 10 Ig = 0

= 100 I1 - 150 Ig - 40 I2 = 0

= 10 I1 – 15 Ig - 4I2 = 0 - equation 2

Consider the next mesh RQSVR

30 I2 + 40 (I2 + Ig) = 20

= 30 I2 + 40 I2 + 40 Ig = 20

= 70 I2 +40 Ig = 20

= 7 I2 +4 Ig = 2 – equation 3

Multiply the equation 1 by 2

10 I1 –6 I2 + 2 Ig = 0

Equation 1- 2

10I1 –6 I2 + 2 Ig = 0 -

10 I1 – 4 I2 – 15 Ig = 0

- 2 I2 + 17 Ig = 0

17 Ig = 2 I2

I2 = 8.5 Ig

Substituting the value in equation 3

59.5 Ig + 4 Ig = 2

63.5 Ig = 2

Ig =0.0315 A

Method 3

Method three for finding the unknown resistance

Consider the same figure. By adjusting the variable resistor the bridge is balanced as usual. Also the voltage at points Q and S should be equal for the bridge to be balanced. This method is applying the ohms’ law.

Ohm’s law states that V = I R. So I = V / R

Here I2 = V / Rx + R3. Substitute the values for V = I R

The voltage at point S = V (Rx / Rx + R3)

The voltage at point Q = V (R2 / R1 + R2)

Thus VSQ = V (Rx / Rx + R3) - V (R2 / R1 + R2)

When VSQ = 0

V (Rx / Rx + R3) = V (R2 / R1 + R2)

Rx R1 + Rx R2 = R2 R3 + R2 Rx

Rx R1 = R2 R3

Rx = R2 R3 / R1

Problem: Calculate the voltage across the points P and R. Also find the value of resistor R4 to balance the bridge.

Solve the Wheatstone Bridge

Solve the Wheatstone Bridge

Solution:

Consider the arm P1 Q1 P2

VQ1 = R3 / (R1 + R3) * Vs

= 40 / 90 * 100

= 0.44 * 100

= 44 V

Consider the next arm P1 Q2 P2

V Q2 = R4 / R2 + R4 * VS

= 50 / (100+ 50) * 100

= 50 / 150 * 100

= 0 .33 * 100

= 33 V

Thus Vout = 44 – 33 = 11 V.

The value of resistor R4 for balancing the bridge = R2 R3 / R1

= 140 / 50

= 2.8 Ω

when an external force is applied. When a metal conductor is stretched, or compressed, the resistance of the material changes as the length, diameter and the resistivity of the material changes. The resistance changes to a very small value for a particular strain. So, to measure such small resistance changes and for better accuracy we used the Wheatstone bridge. In this case, we replace the unknown resistance with a strain gauge.

Consider the bridge circuit given below. Here two resistors R1 and R3 have equal resistance. R2 is the variable resistor. R1 and R3 is one voltage divider circuit and R2 and RS makes the other voltage divider circuit. We know that the series circuit act as the voltage divider circuit. This means that the voltage across each resistor in a series connection will be different. When no force is applied to the strain gauge, the variable resistor is varied to obtain the zero deflection in the voltmeter. This is the balanced circuit.