# Measurement of DC Power MCQ || DC Power measurement Questions and Answers

1. The DC power can be measured by using ______

1. Ammeter Only
2. Voltmeter Only
3. Both Ammeter and Voltmeter
4. None of the above

Explanation:

The DC power is the product of voltage across the load and current through the load. Therefore, the power can be determined by using a voltmeter and ammeter. Instruments can be connected in two different ways. The first option is to connect the ammeter in front of the voltmeter and the second option is to connect the ammeter behind the voltmeter.

2. What is the power loss in the 10 Ω resistor in the Network shown in the figure?

1. 15.31 W
2. 14.9 W
3. 12.3 W
4. 13.2 W

Explanation:

The given circuit diagram can be redrawn as shown below

By nodal analysis,

KCL at V1:

${\frac{{{V_1}}}{3} + \frac{{{V_1} – 15}}{8} + \frac{{{V_1} – {V_2}}}{5} = 0}$

= 8V2 − 2V1 = 15

By solving above two equations, V1 = 3.69 V, V2 = 2.79 V

Voltage across 10 Ω resistor = 15 – V2 = 15 – 2.79 = 12.21 V

Power loss in the 10 Ω resistor,

P = V2/R = 12.212/10 = 14.9 W

3. The error Caused by the Ammeter and Voltmeter connection in DC power measurement is

1. Insertion Error
2. Gross Error
3. Temperature Error
4. All of the above

Explanation:

Ammeter and Voltmeter connection in DC power measurement introduces a systematic error, generally referred to as insertion error.

Insertion errors:- These are errors that result from the insertion of the instrument into the position to measure a quantity affecting its value. For example, inserting an ammeter into a circuit to measure the current will change the value of the current due to the ammeter’s own resistance.

In the Ammeter in front of the voltmeter connection as shown in the fig. the ammeter measures the current flowing into the voltmeter, IL, as well as that flowing into the load, IV.

In ammeter Behind the voltmeter connection, the voltmeter measures the voltage drop across the load, VL, in addition to that dropping across the ammeter, VA. So both connections give an excess of measured power, representing the power absorbed by the instrument connected closer to the load.

4. Estimate the relative error that we will have in the wattmeter reading in % of true load power being consumed in the presence of the wattmeter as shown in fig.

1. 5%
2. 15%
3. 10%
4. 30%

Explanation:

As the ammeter is connected on the load side, the error is due to the ammeter and the error is equal to the power dissipated across the ammeter.

Measured value:

VL = E

IL = E/100

Measured power = (E)(E)/100 = E2/100

True value:

VL = E – voltage drop across ammeter

E − 10E/100

IL = E/110

True power

$= \left( {E – \frac{{10E}}{{110}}} \right)\left( {\frac{E}{{110}}} \right)$

Error

$= \left( {E – \frac{{10E}}{{110}}} \right)\left( {\frac{E}{{110}}} \right$

5. The DC power can be measured by a wattmeter called as ______

1. Power Meter
2. Ammeter
3. Micrometer
4. Dynamometer

Explanation:

In a dc circuit, the electrical power can also be measured by a wattmeter. The instrument most commonly used is the dynamometer.

An electrodynamometer or simply Dynamometer wattmeter is an instrument that is universally used for the measurement of DC as well as AC electric power. It works on the principle of dynamometer i.e. a mechanical force acts between two current-carrying conductors.

As the dynamometer instrument has a Square Law response so can be used on both the DC as well as on AC

6. An ammeter of range (0-5A) is connected in a series with a resistive load reads 3 A. A voltmeter of range (0 – 600 V) reads 300 V. The power measured by voltmeter and ammeter is 2 kW. The current flowing through a resistive load is 1.2 A. The power loss in the voltmeter in kW is ______.

1. 1.64 kW
2. 3.28 kW
3. 5.46 kW
4. 3.20 kW

Explanation:

As shown in the figure

IL + IV = 3 A

Pm = 2 kW

IL = 1.2 A

VL = 300 V

Now IV = 3 – IL

IV = 3 – 1.2 = 1.8 A

And Pt = VL × IL = 300 × 1.2 = 360 W = 0.36 kW

Pm = Voltmeter × Ammeter = 2 kW

P$= {P_t} + \frac{{V_L^2}}{{{R_V}}}$

Where

V2L/RV is power loss in the voltmeter.

∴ 2 = 0.36 + V2L/RV

V2L/RV = 1.64 kW

7.  In DC power Measurement when the load resistance is much higher than the ammeter impedance then the voltmeter Ammeter connection chosen is

1. Ammeter in front of Voltmeter
2. Ammeter Behind the Voltmeter
3. Both 1 and 2
4. Wattmeter

Explanation:

When the load resistance is much higher than the ammeter impedance Ammeter Behind the Voltmeter connection is used.

In the ammeter Behind the voltmeter connection, the voltmeter measures the voltage drop across the load, VL, in addition to that dropping across the ammeter, VA. So both connections give an excess of measured power, representing the power absorbed by the instrument connected closer to the load.

Ammeter Behind the Voltmeter is maximally efficient when the load resistance is much greater than the source (loss) resistance. The useful power delivered to the load resistance is much larger than the power loss in Ammeter.

8. A resistance is measured by the voltmeter-ammeter method. The voltmeter reading is 100 V on 150 V scale and the ammeter reading 40 mA on 80 mA scale. If both the meters are guaranteed for accuracy within 3% of full scale what is the limit within which resistance can be measured

1. 200 ohms
2. 300 Ohms
3. 100 Ohms
4. 150 Ohms

Explanation:

Resistance R = 100/(40 × 10−3) = 2.5 kΩ

Full scale Accuracy δR/R = ±( 3 + 3)% = ±6%

The limit within which resistance can be measured

δR = (2.5 × 103 × 6)/100

δR = 100 Ohms

9. In DC Power Measurement which of the following instrument has High sensitivity?

1. Wattmeter
2. Electrodynamometer
3. Ammeter and Voltmeter
4. All of the above

Explanation:

Since the voltmeter and ammeter are more sensitive than a wattmeter, the measured value is more accurate than that obtained by a wattmeter. Therefore, DC power can be calculated by the reading obtained by the meters.

10. In the power measurement by voltmeter ammeter method, if the voltmeter is connected across supply voltage, then the load power will be

1. The power consumed by the load
2. The sum of power consumed by the load and ammeter.
3. The sum of power consumed by the load and voltmeter.
4. The sum if power consumed by the load, ammeter and voltmeter