Wheatstone Bridge Important Questions

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Important Questions on Wheatstone Bridge

Ques. What is a Wheatstone bridge? (1 Mark)

Ans. The Wheatstone bridge is a circuit used to measure an unknown resistance with the help of three known resistances. It is also known as the resistance bridge.

Ques. What is the principle of Wheatstone Bridge? (2 Marks)

Ans. Wheatstone bridge works on the principle of zero deflection or null defection. This means that the ratio of resistances is equal and there is no current flowing through the electrical circuit.

Ques. Is the Wheatstone bridge a DC or AC bridge? (1 Mark)

Ans. The Wheatstone bridge is mainly designed for DC (direct current) applications.

Ques. What is the condition for a balanced Wheatstone bridge? (2 Marks)

Ans. Wheatstone bridge is said to be balanced when there is no current flow through the galvanometer. This condition occurs when the variable and known resistance are adjusted. 

Ques. What is the major limitation of a Wheatstone bridge? (2 Marks)

Ans. Wheatstone bridge is used to measure only medium resistance values. At lower resistance, the Wheatstone bridge measures the resistance of the leads and the contacts as a significant value thereby causing an error. 

Ques. Derive the conditions for the balance condition in the given Wheatstone bridge. (4 Marks)

Ans. Let us evaluate the current at the junctions using Kirchoff’s Laws,

At B junction,

i₁= i_g + i₃

At D junction,

I₂ + i_g = i₄

Now, when the current through the galvanometer is 0, 

i_g = 0, and hence, i₁ = i₃ and i₂ = i₄

Now, applying Kirchoff Law on the loop ABDA, 

i₁P + i_gG = i₂R

Similarly, on the loop BCDB

i_3Q = i₄S = + i_gG,

However, when i_g = 0, i₁P = i₂R and i₃Q = i₄S

But we know that, i = i₃ and i₂ = I₄,

Thus, 

\(\frac{P}{Q}= \frac{R}{S}\)

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Ques. Consider a Wheatstone bridge comprising four resistances 200 Ω, 20 Ω, 400 Ω, and 40 Ω. When the bridge is connected to a 1.5 V battery, what will be the current through the individual resistors? (4 Marks)

Ans. Let

• The resistance of arm P = 200 Ω
• The resistance of arm Q = 20 Ω
• The resistance of arm R = 400 Ω
• The resistance of arm S = 40 Ω
• The potential difference, V = 1.5 V

Thus,

P/Q = 200/20 = 10

R/S = 400/40 = 10

The null condition is satisfied by the above ratios, 

Therefore, P/Q = R/S

∴ Current in arm with resistance P = V/(P + Q)

⇒ 1.5/ (200 + 20) = 0.0681 A

Again, current in arm R with resistance R = V/(R + S)

⇒ 1.5/ (400 + 40) = 0.0340 A

Ques. The resistance ratios of the arms of a Wheatstone bridge are 300 Ω and 30 Ω. The fourth arm is connected to an unknown resistor. What is the unknown resistance value when the third arm has a resistance of 250 Ω in a balanced condition? (3 Marks)

Ans. Let

• The resistance of arm P = 300 Ω
• The resistance of arm Q =30Ω
• The resistance of arm R =250Ω
• The resistance of arm S = X

Then the ratio of resistances in the balanced condition 

R/X = P/Q

Therefore,

X = (Q/P)R

Substituting the values,

X = (30/300)250 Ω

⇒ X = 25 Ω

Ques. When the galvanometer and the cell are interchanged at the balance point of the bridge, would the galvanometer show any current? (3 Marks)

Ans. The left side of the below figure is a Wheatstone bridge and the right side of the image shows that the galvanometer and the battery are interchanged.

We can see that both the circuits are symmetrical. 

The balanced condition for the figure on the left is

( R₁ / R₂ ) = (R₃ / R₄ ) – (1) 

The balanced condition for the figure on the right is

( R₁ / R₃ ) = (R₂ / R₄ ) – (2)

At the balanced condition, if (1) is true, then (2) is also true. It can be checked mathematically.

Hence, the deflection in the galvanometer will be null for both the circuit arrangements. 

Ques. Which instrument is used as a null detector in a Wheatstone bridge? (2 Marks)

Ans. The galvanometer measures current and determines the voltage between the two terminals in a circuit. Hence, it can detect even minor currents in the circuit. This makes the galvanometer a null detector in a Wheatstone bridge.

Ques. List down the limitations of the Wheatstone bridge. (4 Marks)

Ans. The limitations of the Wheatstone bridge are as follows:

• The Wheatstone bridge is highly sensitive which may cause the measurements to be less precise in an off-balance situation.
• It measures resistances ranging from a few ohms to kilo-ohms.

Thus, for high resistance measurements, the galvanometer is an unsuitable device.

• Due to the heating effect of the current through the resistance, the value of resistance fluctuates sometimes causing a permanent change in the resistance during extreme current situations.
• The sensitivity is reduced when the four resistances are not comparable.

Ques. What are the major applications of the Wheatstone bridge? (4 Marks)

Ans. Some of the applications of the Wheatstone bridge are:

• Measurement of low resistance

Ohm’s law cannot be used in these situations to accurately determine the value. In these cases, the Wheatstone bridge can be installed to get accurate values.

• Measurement of temperature

Thermistors are semiconductors whose resistances are temperature-dependent. Thus a very minor change in temperatures can be measured by using thermistors in a Wheatstone bridge. 

• Measurement of light intensity

Wheatstone bridge can be used to measure the light intensity by replacing the unknown resistor in the circuit with a photoresistor. The resistance of a photoresistor changes depending on the amount of light it receives.

• Measurement of strain and pressure

Wheatstone bridge is used to measure the strain and pressure.

Ques. In a bridge circuit, the resistances are R₁ = 50 Ω, R₂ = 10 Ω, R₃ = 20 Ω. Calculate the value of the unknown resistance R_x when the circuit is in a balanced condition. (3 Marks)

Ans. In balanced condition, we have

R₁ / R₂ = R₃ / R_x

⇒ R_x = R₂ R₃ / R₁

⇒ R_x = (10 x 20) / 50

⇒ R_x = 200 / 50

⇒ R_x = 40 Ω

Ques. What is the current through the galvanometer that is connected across points P and R having resistances of 10 Ω and a potential difference of 20 V. (5 Marks)

Ans. In loop PRQP

50 I₁ – 30 I₂ + 10 I_g = 0

 = 5 I₁ – 3 I₂ + 1 I_g = 0    ….(1)

In loop PSQP

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

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

= 100 I₁ - 150 I_g - 40 I₂ = 0

= 10 I₁ – 15 I_g - 4I₂ = 0    ….(2)

In loop RQSVR

30 I₂ + 40 (I₂ + I_g) = 20

= 30 I₂ + 40 I₂ + 40 I_g = 20

= 70 I₂ +40 I_g = 20

= 7 I₂ +4 I_g = 2    ….(3)

Multiply (1) by (2)

10 I₁ –6 I₂ + 2 I_g = 0

Now, by subtracting (1) from (2), we get

10 I₁ – 6 I₂ + 2 I_g = 0 

10 I₁ – 4 I₂ – 15 I_g = 0

- 2 I₂ + 17 I_g = 0

17 I_g = 2 I₂

I₂ = 8.5 Ig

Substituting the value in (3), we get

59.5 I_g + 4 I_g = 2

63.5 I_g = 2

I_g = 0.0315 A

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