# 1. The measurement of an unknown resistance R is to be carried out using the Wheatstone bridge. Two students perform an experiment in two ways. The first student takes R2 = 10Ω and R1 = 5Ω. The other student takes R2 = 1000Ω and R1= 500Ω. In the standard arm, both take R = 5Ω. Both find R = (R2/R1)R3 = 10Ω within errors.

1. The errors of measurement of the two students are the same.
2. Errors of measurement do depend on the accuracy with which R2 and R1 can be measured.
3. If the student uses large values of R2 and R1 the currents through the arms will be feeble. This will make the determination of null point accurately more difficult.
4. Wheatstone bridge is a very accurate instrument and has no error in measurement.

Explanation:-

Explanation: As the ratio of R2/R3 and standard resistance are the same, so the value of unknown resistance for both students is 10.

So, we can say the Wheatstone Bridge is most sensitive. The results of both students depend on the accuracy of resistances used. So, option (2) is verified.

When R1 and R2 are larger, the current through the galvanometer becomes weak. It will make it difficult to find out the null point more accurately. So, option (3) is verified.

# 2. Four arms of a Wheatstone bridge are as follows: AB = 100 Ω, BC = 10 Ω, CD = 4 Ω, DA = 50 Ω. A galvanometer with an internal resistance of 20 Ω is connected between BD, while a battery of 10-V dc is connected between AC. Find the current through the galvanometer.

1. 6.24 mA
2. 5.18 A
3. 5.18 mA
4. 51.8 mA

Explanation:-

To find out the current through the galvanometer, it is required to find out Thevenin equivalent voltage across nodes BD and also the Thevenin equivalent resistance between terminals BD.

To find out Thevenin’s equivalent voltage across BD, the galvanometer is open-circuited, and the circuit then looks like the figure given below.

At this condition, the voltage drop across the arm BC is given by voltage divder rule

[katex]\begin{array}{l} {V_{BC}} = 10 \times \frac{{10}}{{100 + 10}}\\ \\ = 0.91V \end{array}[/katex]

Voltage drop across the arm DC is given by:

[katex]\begin{array}{l} {V_{BC}} = 10 \times \frac{4}{{50 + 4}}\\ \\ = 0.74V \end{array}[/katex]

Hence, voltage difference between the nodes B and D, or the Thevenin equivalent voltage between nodes B and D is

VTH = VBD = VB – VD = VBC – VDC

= 0.91 – 0.74 = 0.17 V

To obtain the Thevenin equivalent resistance between nodes B and D, the 10 V source need to be shorted, and the circuit looks like the figure given below.

The Thevenin equivalent resistance between the nodes B and D is thus

[katex]\begin{array}{l} {R_{TH}} = \frac{{100 \times 10}}{{100 + 10}} + \frac{{50 \times 4}}{{50 + 4}}\\ \\ {R_{TH}} = 12.79\Omega \end{array}[/katex]

Hence, current through galvanometer is

[katex]\begin{array}{l} {I_D} = \frac{{{V_{TH}}}}{{{R_D} + {R_{TH}}}}\\ \\ {I_D} = \frac{{0.17}}{{20 + 12.79}} = 5.18mA \end{array}[/katex]

# 3. Four arms of a Wheatstone bridge are as follows: AB = 100 Ω, BC = 10 Ω, CD = 4 Ω, DA = 50 Ω. A galvanometer with an internal resistance of 20 Ω is connected between BD, while a battery of 10-V dc is connected between AC. Find the value of the resistance to be put on the arm DA so that the bridge is balanced

1. 40 Ω
2. 50 Ω
3. 20 Ω
4. 10 Ω

Explanation:-

The voltage drop across the arm BC is given by voltage divder rule

[katex]\begin{array}{l} {V_{BC}} = 10 \times \frac{{10}}{{100 + 10}}\\ \\ = 0.91V \end{array}[/katex]

Voltage drop across the arm DC is given by:

[katex]\begin{array}{l} {V_{BC}} = 10 \times \frac{4}{{50 + 4}}\\ \\ = 0.74V \end{array}[/katex]

Hence, voltage difference between the nodes B and D, or the Thevenin equivalent voltage between nodes B and D is

VTH = VBD = VB – VD = VBC – VDC

= 0.91 – 0.74 = 0.17 V

To obtain the Thevenin equivalent resistance between nodes B and D, the 10 V source need to be shorted, and the circuit looks like the figure given below.

The Thevenin equivalent resistance between the nodes B and D is thus

[katex]\begin{array}{l} {R_{TH}} = \frac{{100 \times 10}}{{100 + 10}} + \frac{{50 \times 4}}{{50 + 4}}\\ \\ {R_{TH}} = 12.79\Omega \end{array}[/katex]

Hence, current through galvanometer is

[katex]\begin{array}{l} {I_D} = \frac{{{V_{TH}}}}{{{R_D} + {R_{TH}}}}\\ \\ {I_D} = \frac{{0.17}}{{20 + 12.79}} = 5.18mA \end{array}[/katex]

In order to balance the bridge, there should be no current through the galvanometer, or in other words, nodes B and D must be at the same potential.

Balance equation is thus

100/10 = RDA/4

RDA = 40Ω

# 4. In a Wheatstone’s bridge P = 50 Ω, Q = 100 Ω and R = 20 Ω. If the galvanometer shows zero deflection, determine the value of S.

1. 20 Ω
2. 40 Ω
3. 50 Ω
4. 25 Ω

Explanation:-

P/Q = R/S

S = (Q × R)/P

S = (100 × 20)/50

S = 40Ω

# 5. 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. 204 mm
3. 2.04 mm
4. 0.204 mm

Explanation:-

To find out the current through the galvanometer, it is required to find out the Thevenin equivalent voltage across nodes AC and also the Thevenin equivalent resistance between terminals AC.

To find out Thevenin equivalent voltage across AC, the galvanometer is open-circuited. At this condition, the voltage drop across the arm AB is given by

[katex]\begin{array}{l} {V_{AB}} = 4 \times \frac{{100}}{{100 + 401}}\\ \\ = 0.798V \end{array}[/katex]

The voltage drop across the arm CB is given by

[katex]\begin{array}{l} {V_{BC}} = 4 \times \frac{{1000}}{{4000 + 1000}}\\ \\ = 0.8V \end{array}[/katex]

Hence, the voltage difference between the nodes A and C, or the Thevenin equivalent voltage between nodes A and C is

VTH = VAC = VA – VC = VAB – VCB

VTH = 0.798 – 0.8 = –0.002 V

To obtain the Thevenin equivalent resistance between nodes A and C, the 10 V source needs to be shorted. Under this condition, the Thevenin equivalent resistance between the nodes A and C is thus

[katex]\begin{array}{l} {R_{TH}} = \frac{{100 \times 401}}{{100 + 401}} + \frac{{1000 \times 4000}}{{1000 + 4000}}\\ \\ {R_{TH}} = 880.04\Omega \end{array}[/katex]

Hence, the current through the galvanometer is

[katex]\begin{array}{l} {I_D} = \frac{{{V_{TH}}}}{{{R_D} + {R_{TH}}}}\\ \\ {I_D} = \frac{{0.002}}{{100 + 880.04}} = 2.04\mu A \end{array}[/katex]

Deflection of the galvanometer

= Sensitivity × Current = 10 mm/µA × 2.04 µA = 20.4 mm

# P = 500 Ω, Q = 800 Ω, R = x + 400, S = 1000 Ω

1. 200Ω
2. 250Ω
3. 150Ω
4. 225Ω

Explanation:-

P/Q = R/S

500/800 = (x + 400)/1000

x + 400 = (500 × 1000)/800

x + 400 = 0.625 × 1000

x = 225 Ω

# 7. A Wheatstone bridge has P = 3.5 kΩ, Q = 7 kΩ, and galvanometer null is obtained when S = 5.51 kΩ. Determine the resistance measurement range for the bridge if S is adjustable from 1 kΩ to 8 kΩ.

1. 500Ω to 400Ω
2. 4Ω to 500Ω
3. 500Ω t0 4Ω
4. 50Ω to 40Ω

Explanation:-

Under Balanced condition

P/Q = R/S

When S = 1 KΩ

R = (1 KΩ × 3.5 KΩ)/7 KΩ

R = 500 Ω

When S = 8 KΩ

R = (8 KΩ × 3.5 KΩ)/7 KΩ

R = 4 Ω

Hence the measurement range is 500 Ω to 4 Ω

# 8. In Wheatstone’s bridge arrangement PQRS, the ratio arms P and Q are nearly equal. The bridge is balanced when R = 500Ω. On interchanging P and Q, the value of R for balancing is 510Ω. Find the value of S.

1. 510.58Ω
2. 504.97Ω
3. 525.24Ω
4. 425.22Ω

Explanation:-

For balanced Wheatstone’s bridge

P/Q = R/S

In the first case, R = 500Ω

P/Q = 500/S ——— (1)

In the second case when P and Q are interchanged, R = 510Ω

Q/P = 510/S ——— (2)

Multiplying the equation 1 by 2

1 = (500 × 510)/S2

S = 504.97 Ω

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

1. 10Ω

Explanation

Given: R1 = 4Ω, R2 = 8Ω, R4 = 6Ω

To find: Unknown resistance (X)

For balance wheatstone bridge

R3 = (R1 × R4)/R2

R3 = (4 × 6)/8

X = 3Ω

The unknown resistance is

10. Resistances in the branches of Wheatstone’s bridge are 30Ω, 60Ω, 15Ω and a series combination of X and 5Ω resistances. If the bridge is balanced then value of unknown resistance X is

1. 50Ω
2. 40Ω
3. 25Ω
4. 15Ω

Explanation:-

Given: R1 = 30Ω, R2 = 60Ω, R3 = 15Ω, R4 = (X + 5)Ω

For balanced Wheatstone’s bridge

R1/R2 = R3/R4

30/60 = 15/(X + 5)

X + 5 = (15 × 60)/30

X = 25Ω

Scroll to Top