Electrical Measurements Objective Questions & Answers on “Advanced Problems on Measurement of Low Medium and High Resistance”.
1. Circuit shows the method of Measurement of low resistance by Ammeter-Voltmeter method. The measured resistance Rm for the given Circuit is _________
a) Rx + Rv
b) (frac{R_x^2}{R_x + R_v})
c) (frac{R_v^2}{R_x + R_v})
d) (frac{R_x R_v}{R_x + R_v})
Answer: d
Clarification: Measured resistance Rm = (frac{V_x}{I_A} = frac{V_x}{I_v + I_R})
(frac{I_v}{V_x} = frac{1}{R_v}) And (frac{I_R}{V_x} = frac{1}{R_X})
So, Rm = (frac{R_x R_v}{R_x+R_v}).
2. Circuit shows the method of Measurement of low resistance by Ammeter-Voltmeter method. What is the percentage error?
a) Zero
b) (frac{R_x}{R_x + R_v}) × 100
c) (– frac{R_x}{R_x + R_v}) × 100
d) (– frac{R_v}{R_x + R_v}) × 100
Answer: c
Clarification: Percentage Error = (frac{R_m – R_x}{R_x}) × 100
(= frac{R_x R_v – R_x(R_x-R_v)}{R_x (R_x+R_v)}) × 100
∴ Percentage Error = (– frac{R_x}{R_x + R_v}) × 100.
3. The readings of polar type potentiometer are I = 12.4∠27.5°, V = 31.5∠38.4°. Then, reactance of the coil will be ________
a) 2.51 Ω
b) 2.56 Ω
c) 2.54 Ω
d) 2.59 Ω
Answer: c
Clarification: Here, V = 31.5∠38.4°
I = 12.4∠27.5°
Z = (frac{31.5∠38.4°}{12.4∠27.5°}) = 2.54∠10.9°
But Z = R + jX = 2.49 + j0.48
∴ Reactance X= 2.54 Ω.
4. The voltage drop across a standard resistor of 0.2 Ω is balanced at 83 cm. Find the magnitude of the current, if the standard cell emf of 1.53 V is balanced at 42 m.
a) 13.04 A
b) 10 A
c) 14.95 A
d) 12.56 A
Answer: c
Clarification: Voltage drop per unit length = (frac{1.53}{42}) = 0.036 V/cm
Voltage drop across 83 cm length = 0.036 × 83 = 2.99 V
∴ Current through resistor, I = (frac{2.99}{0.2}) = 14.95 A.
5. A resistance R is measured using the connection shown in the below figure.
The current measured is 10 A on ranges 100A and the voltage measured is 125 V on 150 V range. The scales of the ammeter and voltmeter are uniform. The total number of scale divisions of the ammeter is 100 and that of the voltmeter is 150. The scale division can be distinguished. The constructional error of the ammeter is ± 0.3% and that of voltmeter±0.4%. The resistance of the ammeter is 0.25 Ω.
The value of R is?
a) 12.75 Ω
b) 12.0 Ω
c) 12.25 Ω
d) 12.5 Ω
Answer: c
Clarification: Percentage error in ammeter = (± frac{1}{10×100} × 100) = ± 0.1%
Percentage error in voltmeter= (± frac{1}{10×150} × 100) = ± 0.067%
So, δI = ± 0.3 ± 0.1 = ± 0.4%
δV = ± 0.4 ± 0.067 = ± 0.467%
R = (frac{V}{I})
So, error = ± δV ± δI = ± 0.867
Measured value of resistance = (R_m = frac{125}{10}) = 12.5
∴ True value = (R_m(1-frac{Ra}{R_m})) = 12.25 Ω.
6. A resistance R is measured using the connection shown in the below figure.
The current measured is 10 A on ranges 100A and the voltage measured is 125 V on 150 V range. The scales of the ammeter and voltmeter are uniform. The total number of scale divisions of the ammeter is 100 and that of the voltmeter is 150. The scale division can be distinguished. The constructional error of the ammeter is ± 0.3% and that of voltmeter±0.4%. The resistance of the ammeter is 0.25 Ω.
The possible error in the measurement of R is?
a) ±0.11 Ω
b) ±0.15 Ω
c) ±0.867 Ω
d) ±0.625 Ω
Answer: a
Clarification: Possible error is ± 0.867, so,
12.25 ± 0.867%
Or, 12.25 ± 0.11 Ω.
7. Low resistance is measured by ___________
a) De-Sauty’s bridge
b) Maxwell’s bridge
c) Kelvin double bridge
d) Wein’s bridge
Answer: c
Clarification: De-Sauty’s bridge is used for measurement of Capacitance; Maxwell’s bridge is used for measurement of Inductance and Wein Bridge for Frequency. Kelvin double bridge is used for measurement of Low resistance.
8. The resistance can be measured most accurately by _________
a) Voltmeter-Ammeter method
b) Bridge method
c) Multimeter
d) Megger
Answer: b
Clarification: Bridge method applies the concept of null point or bridge balance condition. Multimeter and Megger are used for measuring very high resistances and Voltmeter-Ammeter method is used for Low resistances. A null type instrument has higher accuracy as compared to a deflection type instrument.
9. A slide wire potentiometer has 10 wires of 2 m each. With the help of a standard voltage source of 1.045 V, it is standardized by keeping the jockey at 104.5 cm. If the resistance of potentiometer wires is 2000 Ω, then the value of working current is?
a) 1 mA
b) 10 mA
c) 0.1 mA
d) 0.5 mA
Answer: b
Clarification: Total length of the slide wire = 10 × 200
Total resistance of slide wire = 2000 Ω
∴ Resistance per cm = 1 Ω
Resistance of 104.5 cm = 104.5 Ω
This corresponds to a voltage of 1.045 V
∴ Current = (frac{1.045}{104.5}) = 10 mA.
10. Which of the following method is used for the measurement of Medium Resistance?
a) Kelvin’s double bridge method
b) Carey-Foster bridge method
c) Anderson Bridge
d) Direct-Deflection method
Answer: b
Clarification: Kelvin’s double bridge method is used for measurement of Low Resistance, Anderson Bridge is not used for measurement of Resistance and Direct-Deflection method is used for Measurement of High Resistance.
11. In the Wheatstone bridge shown below, if the resistance in each arm is increased by 0.05%, then the value of Vout will be ________
a) 50 mV
b) Zero
c) 5mV
d) 0.1mV
Answer: b
Clarification: In Wheatstone bridge, balance condition is
R1R3 = R2R4
Here, R1 = 5, R2 = 10, R3 = 16, R4 = 8
And when the Wheatstone bridge is balanced then, at Vout voltage will be Zero.
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