Wheatstone Bridge Questions

Wheatstone bridge questions 1

Four resistances, 4Ω, 8Ω, XΩ, and 6Ω, are connected in a series to form Wheatstone’s network. If the network is balanced, find the value of ‘X’.

1. 3 Ω
2. 5 Ω
3. 7 Ω
4. 10 Ω

Answer: A. 3Ω

Solution:

Given that,

R₁=4Ω, R₂=8Ω, R₃=XΩ, R₄=6Ω

at null condition,

R₁/R₃=R₂/R₄

4/X=8/6

⇒X=3Ω

Hence option A is the correct answer.

Wheatstone bridge questions 2

In Wheatstone’s bridge arrangement PQRS, the ratio arms P and Q are nearly equal. The bridge is balanced when R = 700Ω. On interchanging P and Q, the value of R for balancing is 720Ω. Find the value of S.

1. 700 Ω
2. 720 Ω
3. 709.92 Ω
4. 705.75 Ω

Answer: 709.92Ω

Solution:

Given that, the ratio arms P and Q are nearly equal.

 R = 700Ω

at null condition

P/Q=700/S ________(1)

P and Q are interchanged, and the bridge is balanced at R=510 Ω.
Q/P=720/S ________(2)

Multiply both equations to obtain the value of S.

1=700×720/S²

S=709.92 Ω

Hence option C is the correct answer.

Wheatstone bridge questions 3

In the Wheatstone bridge given below, what should be the values of R₁ and R₂ so that the bridge is balanced?

A) R₁ = 6 ohms; R₂ = any finite value
B) R₁ = 6 ohms; R₂ = 3 ohms
C) R₁ = 29.4; R₂ = 2 ohms
D) R₁ = any finite value; R₂ = 3 ohms

Answer: A. R₁ = 6 ohms; R₂ = any finite value

Solution:

In the given bridge circuit, if the outer bridge is balanced, then there is no current flow in the branch BD, and then R₂ can be any finite value.

Hence, when the bridge is balanced.

R_ABR_CD=R_BCR_AD

21×10=R₁×35

R₁=210/35=6 Ohms

Hence option A is the correct answer.

Wheatstone bridge questions 4

The four arms of a Wheatstone bridge are as follows: AB = 100 Ω, BC = 1000 Ω, CD = 4000 Ω, DA = 400 Ω. A galvanometer with an internal resistance of 100 Ω and a sensitivity of 10 mm/μA is connected between AC, while a battery of 4 V dc is connected between BD. Calculate the deflection if the resistance of arm DA is changed from 400 Ω to 401 Ω.

1. 20.4 mm
2. 40.8 mm
3. 2.04 mm
4. 4.08 mm

Answer: A. 20.4 mm

Solution:

Given the resistance of the arm, DA is changed to 401 Ω.

The given Wheatstone bridge can be drawn as shown below.

Deflection of galvanometer=(sensitivity)×(current flowing through Galvano meter)

We can use Thevenin’s theorem to find the current through the Galvano meter.

I_D=V_ThR_D+R_Th

 So to apply Thevenin’s theorem, remove the Galvanometer and assume that the branch AC is open-circuited.

V_Th=V_AC=V_A-V_C

V_A=4×100/(401+100)=0.798 V

V_C=4×1000/(4000+1000)=0.8 V

V_Th=V_AC=V_A-V_C=0.798-0.8=-0.002 V

While finding R_Th, the Voltage source is replaced with a short circuit.

R_Th=(401×100)/(401+100)+(4000×1000)/(4000+1000)=880.03

I_D=V_Th/(R_D+R_Th)=-0.002/(100+880.03)=-2.04 μA

The magnitude of I_D=2.04 μA

Deflection of galvanometer =2.04×10^-6×10×(10^-3/10^-6)=20.4 mm

Hence option A is the correct answer.

Wheatstone bridge questions 5

Which instrument is used as the null detector in the Wheatstone bridge?
a) Voltmeter
b) Ammeter
c) Galvanometer
d) Multimeter

Answer:

Option C

Explanation:

A galvanometer is used as a null detector in the Wheatstone bridge. When the bridge is balanced, the current through the galvanometer is zero. Due to its high sensitivity, the Galvano meter can sense the smallest values of currents.

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