This set of Electrical Measurements Objective Questions & Answers focuses 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) R_{x} + R_{v}

b) \(\frac{R_x^2}{R_x + R_v}\)

c) \(\frac{R_v^2}{R_x + R_v}\)

d) \(\frac{R_x R_v}{R_x + R_v}\)

View Answer

Explanation: Measured resistance R

_{m}= \(\frac{V_x}{I_A} = \frac{V_x}{I_v + I_R}\)

\(\frac{I_v}{V_x} = \frac{1}{R_v}\) And \(\frac{I_R}{V_x} = \frac{1}{R_X}\)

So, R

_{m}= \(\frac{R_x R_v}{R_x+R_v}\).

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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: In Wheatstone bridge, balance condition is

R

_{1}R

_{3}= R

_{2}R

_{4}

Here, R

_{1}= 5, R

_{2}= 10, R

_{3}= 16, R

_{4}= 8

And when the Wheatstone bridge is balanced then, at V

_{out}voltage will be Zero.

**Sanfoundry Global Education & Learning Series – Electrical Measurements.**

To practice all objective questions on Electrical Measurements, __here is complete set of 1000+ Multiple Choice Questions and Answers__.

**If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]**

**Related Posts:**