61. Which of the following test employed the Wheatstone bridge to detect the location of the fault in underground cables?
Murray loop test
Varley Loop test
Both 1 & 2
None of the above
Answer.3. Both 1 & 2
Explanation:-
Loop Tests For Location of Faults in Underground Cables
There are several methods for locating the faults in underground cables. However, two popular methods known as loop tests are
Murray loop test
Varley loop test
These simple tests can be used to locate the earth fault or short-circuit fault in underground cables provided that a sound cable runs along the faulty cable. Both these tests employ the principle of the Wheatstone bridge for fault location.
62. One of the simplest applications of a Wheatstone bridge is _______
63. 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.
The errors of measurement of the two students are the same.
Errors of measurement do depend on the accuracy with which R2 and R1 can be measured.
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.
Wheatstone bridge is a very accurate instrument and has no error in measurement.
Answer.2 & 3
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.
64. 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.
6.24 mA
5.18 A
5.18 mA
51.8 mA
Answer.3. 5.18 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
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
65. 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
40 Ω
50 Ω
20 Ω
10 Ω
Answer.1. 40 Ω
Explanation:-
The voltage drop across the arm BC is given by voltage divder rule
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
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
66. 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.
20 Ω
40 Ω
50 Ω
25 Ω
Answer.2. 40 Ω
Explanation:-
For Wheatstone bridge balanced condition
P/Q = R/S
S = (Q × R)/P
S = (100 × 20)/50
S = 40Ω
67. 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.
20.4 mm
204 mm
2.04 mm
0.204 mm
Answer.1. 20.4 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
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
= Sensitivity × Current = 10 mm/µA × 2.04 µA = 20.4 mm
68. What is the value of x when Wheatstone’s network is balanced?
P = 500 Ω, Q = 800 Ω, R = x + 400, S = 1000 Ω
200Ω
250Ω
150Ω
225Ω
Answer.4. 225Ω
Explanation:-
For Wheatstone bridge balanced condition
P/Q = R/S
500/800 = (x + 400)/1000
x + 400 = (500 × 1000)/800
x + 400 = 0.625 × 1000
x = 225 Ω
69. 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.
500Ω to 400Ω
4Ω to 500Ω
500Ω t0 4Ω
50Ω to 40Ω
Answer.3. 500Ω t0 4Ω
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
70. 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.
510.58Ω
504.97Ω
525.24Ω
425.22Ω
Answer.2. 504.97Ω
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
71. 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
5Ω
10Ω
1Ω
3Ω
Answer.4. 3Ω
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 3Ω
72. 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