Previous Year Questions- Measurement of Resistance and Potentiometers | Electrical and Electronic Measurements - Electrical Engineering (EE) PDF Download

Q1: An unbalanced DC Wheatstone bridge is shown in the figure. At what value of p will the magnitude of V_o be maximum? (SET-1(2015))
(a)
(b) (1 + x)
(c)
(d)
Ans: (a)
Sol: For maximum V₀,
dV₀/dP = 0  
⇒ [1 + x + P(2+P+x)] = P(2P+x+2)
⇒ 1 + x + 2P + P² + Px = 2p2 + Px + 2P
⇒ 1 + x = P²
⇒ P = 
Q2: Suppose that resistors R₁ and R₂ are connected in parallel to give an equivalent resistor R. If resistors R₁ and R₂ have tolerance of 1% each, the equivalent resistor R for resistors R₁ = 300Ω and R₂ = 200Ω will have tolerance of   (SET-2 (2014))
(a) 0.50%
(b) 1%
(c) 1.20%
(d) 2%
Ans: (b)
Sol: Given, R₁=300Ω,  R₂=200Ω
Equivalent resistance of parallel combination of R₁ and R₂

= 120Ω = R_T 
Given tolerance of R₁ = 1%
∴ R₁ = 300 ± 3
R₁ = 303Ω or 297Ω
Tolerance of R₂ = 1%
∴ R₂ = 200 ± 2
R₂ = 202Ω or 198Ω
Equivalent resistance of the parallel combination of R₁ and  R₂ in high range is:
∴ Tolerance for equivalent resistance R in high range of R₁ and R₂ is  
Note: In low range of R₁ and R₂,
∴ Tolerance of equivalent resistance, R in low range is,  
= -1%

Q3: A strain gauge forms one arm of the bridge shown in the figure below and has a nominal resistance without any load as R_s = 300Ω. Other bridge resistances are R₁ = R₂ = R₃ = 300Ω. The maximum permissible current through the strain gauge is 20 mA. During certain measurement when the bridge is excited by maximum permissible voltage and the strain gauge resistance is increased by 1% over the nominal value, the output voltage V₀ in mV is (2013)
(a) 56.02
(b) 40.83
(c) 29.85
(d) 10.02
Ans: (c)
Sol: V_ab = V_a − V_b
Using voltage division rule,
= 29.85mV

Q4: R₁ and R₄ are the opposite arms of a Wheatstone bridge as are R₃ and R₂. The source voltage is applied across R₁ and R₃. Under balanced conditions which one of the following is true  (2006)
(a) R₁ = R₃R₄/R₂
(b) $R1=R2R3/R4$R₁ = R₂R₃/R₄
(c) $R1=R2R4/R3$R₁ = R₂R₄/R₃
(d) R₁ = R₂ + R₃ + R₄ 
Ans: (b)
Sol: Adjustment are made in various arms of the bridge so that the voltage across the detector is zero and hence no current slows through it, when no current flows through detector the bridge is said to be balanced.
Under condition of balance = R₁ = R₂(R₃/R₄)

Q5: The set-up in the figure is used to measure resistance R. The ammeter and voltmeter resistances are 0.01 Ω and 2000 Ω, respectively. Their readings are 2A and 180V, respectively, giving a measured resistances of 90 Ω. The percentage error in the measurement is (2005)
(a) 2.25%
(b) 2.35%
(c) 4.50%
(d) 4.71%
Ans: (c)
Sol: Measured value of resistance = 90Ω
Resistance of voltmeter, R_v = 2000Ω
Voltage across voltmeter,  V = 180
Current through voltmeter = V/R_v = 180/2000 = 0.09 A
Current through resistance,
(R) = I_R = 2 − I_v = 2 − 0.09
⇒ I_R = 1.91A
True value of resistance,

Q6: A dc potentiometer is designed to measure up to about 2 V with a slide wire of 300 mm. A standard cell of emf 1.18 V obtains balance at 600 mm. A test cell is seen to obtain balance at 680 mm. The emf of the test cell is (2004)
(a) 1.00 V
(b) 1.34 V
(c) 1.50 V
(d) 1.70 V
Ans: (b)
Sol: E₁= standerd call of emf 1.18
l₁ = 600 mm
E₂ = emf of the test cell
i₂ = 680 mm
The voltage of any point along the slide wire is proportional to length of slide wire.
E ∝ l
= (680/600) x 1.18 = 1.34 V