VBQs Electricity Class 10 Science with solutions has been provided below for standard students. We have provided chapter wise VBQ for Class 10 Science with solutions. The following Electricity Class 10 Science value based questions with answers will come in your exams. Students should understand the concepts and learn the solved cased based VBQs provided below. This will help you to get better marks in class 10 examinations.

**Electricity VBQs Class 10 Science**

**Question. An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it? **

(a) 1 A

(b) 2 A

(c) 4 A

(d) 5 A

**Answer**

D

**Question. The maximum resistance which can be made using four resistor each of resistance 1/2 Ω is **

(a) 2 Ω

(b) 1 Ω

(c) 2.5 Ω

(d) 8 Ω

**Answer**

A

**Question. What is the minimum resistance which can be made using five resistors each of 1/5 Ω? **

(a) 1/5 Ω

(b) 1/25 Ω

(c) 1/10 Ω

(d) 25 Ω

**Answer**

B

**Question. When a 4 V battery is connected across an unknown resistor there is a current of 100 mA in the circuit. ****The value of the resistance of the resistor is:**

(a) 4 Ω

(b) 40 Ω

(c) 400 Ω

(d) 0.4 Ω

**Answer**

B

**Question. A cell, a resistor, a key, and ammeter are arranged as shown in the circuit diagrams given below. The current recorded in the ammeter will be: 5 **

(a) Maximum in (i).

(b) Maximum in (ii).

(c) Maximum in (iii).

(d) The same in all the cases.

**Answer**

D

**Question. In an electrical circuit three incandescent bulbs A, B, and C of rating 40 W, 60 W, and 100 W, respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness? **

(a) Brightness of all the bulbs will be the same.

(b) Brightness of bulb A will be the maximum.

(c) Brightness of bulb B will be more than that of A.

(d) Brightness of bulb C will be less than that of B.

**Answer**

C

**Question. Two bulbs of 100 W and 40 W are connected in series. The current through the 100 W bulb is 1 A. ****The current through the 40 W bulb will be:**

(a) 0.4 A

(b) 0.6 A

(c) 0.8 A

(d) 1 A

**Answer**

D

**Question. Unit of electric power may also be expressed as: **

(a) Volt-ampere

(b) Kilowatt-hour

(c) Watt-second

(d) Joule-second

**Answer**

A

**Question. What is the maximum resistance which can be made using five resistors each of 1/5 Ω? **

(a) 1/5 Ω

(b) 10 Ω

(c) 5 Ω

(d) 1 Ω

**Answer**

D

**Question. The resistivity does not change if : **

(a) the material is changed.

(b) the temperature is changed.

(c) the shape of the resistor is changed.

(d) both material and temperature are changed.

**Answer**

C

**Question. Electrical resistivity of a given metallic wire depends upon **

(a) its length.

(b) its thickness.

(c) its shape.

(d) nature of the material.

**Answer**

D

**Question. A cylindrical conductor of length ‘l’ and uniform area of cross section ‘A’ has resistance ‘R’. The area of cross section of another conductor of same material and same resistance but of length ‘2l’ is **

(a) A/2

(b) 3A/2

(c) 2 A

(d) 3 A

**Answer**

C

**Question. If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be **

(a) 100%

(b) 200%

(c) 300%

(d) 400%

**Answer**

C

**Assertion and Reason Based MCQs :**

**Directions : In the following questions, A statement of Assertion (a) is followed by a statement of Reason (R).****(a) Both A and R are true and R is the correct explanation of A.****(b) Both A and R are true but R is NOT the correct explanation of A.****(c) A is true but R is false.****(d) A is false and R is true.**

**Question. Assertion (A): Two resistance having value R each. ****Their equivalent resistance is R/2 .****Reason (R): Resistances are connected in parallel.**

**Answer**

A

**Question. Assertion (A): Silver is not used to make electric wires. ****Reason (R): Silver is a bad conductor of electricity.**

**Answer**

C

**Question. Assertion (A): Alloys are commonly used in electrical heating devices like electric iron and heater. ****Reason (R): Resistivity of an alloy is generally higher than that of its constituent metals but the alloys have low melting points than their constituent metals.**

**Answer**

C

**Question. Assertion (A): The resistivity of conductor increases with the increasing of temperature. ****Reason (R): The resistivity is the reciprocal of the conductivity.**

**Answer**

B

**Question. Assertion (A): Electric appliances with metallic body have three connections, whereas an electric bulb has a two pin connection. ****Reason (R): Three pin connections reduce heating of connecting wires.**

**Answer**

C

**Question. Assertion (A): Copper is used to make electric wires. ****Reason (R): Copper has very low electrical resistance.**

**Answer**

A

**Question. Assertion (A): Bending a wire does not affect electrical resistance. ****Reason (R): Resistance of wire is proportional to resistivity of material.**

**Answer**

A

**Very Short Answer Type Questions :**

**Question. Write SI unit of resistivity.****Answer:** Ohm metre (ohm m).

**Question. Two unequal resistances are connected in parallel.****If you are not provided with any other parameters (eg. numerical values of I and R), what can be said about the voltage drop across the two resistors?****Answer:** Voltage-drop is same across both.

**Question. What is meant by the statement. “The resistance of a conductor is one ohm” ?****Answer:** The resistance of a conductor is said to be 1 ohm if a current of 1 ampere flows through it when the potential difference across it is 1 volt.

**Question. Why are the heating elements of electric toasters and electric irons made of an alloy rather than a pure metal ? ****Answer:** Due to high resistivity of alloys rather than its constituting metals.

**Question. Write the mathematical expression for Joule’s law of heating.****Answer:** Mathematical expression of Joule’s law of heating is : H = I2Rt

Where, H = Produced Heat

I = Current flowing through the device

t = Time of current flow

R = Resistance of the appliance

**Question. What does the cord of an electric oven not glow while its heating element does ?****Answer:** Cord is made up of copper wire whereas heating element is made up of alloy.

**Question. Should the resistance of a voltmeter be low or high? Give reason.****Answer:** High. In parallel connection, less current passes through high resistance.

**Short Answer Type Questions :**

**Question. A student has two resistors- 2 Ω and 3 Ω. She has to put one of them in place of R2 as shown in the circuit. The current that she needs in the entire circuit is exactly 9A. Show by calculation which of the two resistors she should choose **

**Answer:** The overall current needed = 9A

The voltage is 12V

Hence by Ohm’s Law V=IR

The resistance for the entire circuit = 12/9 = 4/3 Ω

= R

R1 and R2 are in parallel. Hence, R=(R1 R2)/(R1 + R2)

or, 4R2/(4+R2) = 4/3

∴ R2 = 2Ω

**Question. Compute the heat generated while transferring 96,000 coulomb of charge in two hours through a potential difference of 40 V. A****Answer:** Given, Charge (Q)= 96000 C, Time (t) = 2 h,

Potential difference (V) = 40 V

Heat generated, H = VIt (where I = Q/t)

H = d

H = V × Q

H = 40 ×96000

H = 384000 J

**Question. Define electric power. Write an expression relating electric power, potential difference and resistance.****Answer:** Electric power : It is the amount of electric energy consumed in a circuit per unit time.

Expression : P = V2/R

**Question. How many 132 Ω resistors in parallel are required to carry 5 A on a 220 V line ?****Answer:** Given V = 220 V, I = 5 A

V = IR

Or R = V/I

In parallel combination, let the no. of resistors = x

132/x = 220/5

or, 132/x = 44

or, x = 132/44

∴ x = 3

The number of resistors = 3

**Question. A V-I graph for a nichrome wire is given below. **

**What do you infer from this graph? Draw a labelled circuit diagram to obtain such a graph.****Answer:** Graph between V and I is a straight line.

So, this infers that the flow of current (I) in the conductor is directly proportional to the potential difference (V) established across it. This is ohm’s law.

Resistance of the wire can be calculated as :

R = V/I 0.8/0.2 = 4 ohm

This means nichrome wire has a constant value of the resistance 4 ohm.

Circuit diagram:

**Question. (a) State the relation correlating the electric current flowing in a conductor and the voltage applied across it. Also, draw a graph to show this relationship.****(b) Find the resistance of a conductor if the electric current flowing through it is 0.35 A when the potential difference across it is 1.4 V.****Answer:** (a) The flow of current (I) in the conductor is directly proportional to the potential difference (V) established across it provided the physical

conditions remain same.

Or V = IR

Graph: 2

(b) Given :

Potential Difference (V)= 1.4 V

Current (I) = 0.35 A

As per formula, V = IR

So, V/I = 1.4/0.35 = 4 ohm

**Question. The figure below shows three cylindrical copper conductors along with their face areas and lengths.****Compare the resistance and the resistivity of the three conductors. Justify your answer. **

**Answer:** Ra = L/A

Rb = ρ (3L/A/3) = 9 ρL/A = 9 Ra

Rc = ρ L/3/3 A = 1.9 ρL/A = 1/9 Ra

Hence Rb > Ra > Rc

ra = rb = rc because all the three conductors are of same material.

**Question. While studying the dependence of potential difference (V) across a resistor on the current (I) passing through it, in order to determine the resistance of the resistor, a student took 5 readings for different values of current and plotted a graph between V and I. He got a straight line graph passing through the origin. What does the straight line signify? Write the method of determining resistance of the resistor using this graph.****Answer:** The graph between V and I is a straight line and passes the origin, this verifies the Ohm’s law.

The slope gives the resistance of the resistor used in the circuit.

Slope = QM/MP = V2/I2 – V2/I2

Or R = Value of potential differnce at a point/Value of current at the same point

**Question. Show four different ways in which three resistors of ‘r ’ ohm each may be connected in a circuit.**

In which case is the equivalent resistance of the combination:

(i) Maximum

(ii) Minimum**Answer:**

(i) Circuit (a) has maximum resistance

(ii) Circuit (b) has minimum resistance

**Question. Consider the following circuit: **

**What would be the readings of the ammeter and the voltmeter when key is closed ? Give reason to justify your answer.****Answer:** R = R1 + R2 + R3

R = 5 Ω + 8 Ω + 12 Ω = 25 Ω 1

V = 6 V

V = IR

∴ I = V/I = 6 V/25Ω = 0.24 A

Hence, Current through 12 Ω resistance is 6/25 A = 0.24 A

V = IR = 6A x 12Ω/25 = 2.88 V

**Question. The current flowing through a resistor connected in a circuit and the potential difference developed across its ends are as shown in the diagram by milliammeter and voltmeter readings respectively.**

(a) What are the least counts of these meters ?

(b) What is the resistance of the resistor ?

**Answer:** (a) least count of ammeter = 10 mA

least count of voltmeter = 0.1 V

(b) 2. 4/0. 25 = 9.6 ohm (250 mA = 0.25 A)

**Question. State Ohm’s Law. Draw a circuit diagram to verify this law indicating the positive and negative terminals of the battery and the meters. Also show the direction of current in the circuit.****Answer:** Statement of Ohm’s Law, Circuit diagram with polarity of battery, ammeter and voltmeter Direction of current by arrow.

**Question. Calculate the resistance of a 1 km long copper wire of area of cross section 2 × 10–2 cm2. The resistivity of copper is 1.623 × 10–8 ohm-meter.****Answer:** R = ρ l/A

= 1 x 623 x 10-8 x 100/2 x 10-2 x 10-4 m2

= 0.81 × 10 Ω = 8.1 Ω.

**Question. Give reason for the following:**

(i) Why are copper and aluminium wires used as connecting wires ?

(ii) Why is tungsten used for filament of electric lamps?

(iii) Why is lead-tin alloy used for fuse wires ?**Answer:** (i) These are good conductors of electricity/low resistance, low resistivity.

(ii) Very high melting point and high resistivity.

(iii) Low melting point.

**Question. Calculate the total cost of running the following electrical devices in the month of September, if the rate of 1 unit of electricity is ` 6.00.**

(i) Electric heater of 1000 W for 5 hours daily.

(ii) Electric refrigerator of 400 W for 10 hours daily.**Answer:** P1 = 1000 W = 1000/1000 kW, t1 = 5h

P2 = 400 W = 400/1000 1000 kW, t2 = 10h

No. of days, n = 30

E1 =P1 × t1 × n

= 1 kW × 5h × 30 = 150 kWh

E2 = P2 × t2 × n

= 400/1000 kW × 10 h × 30

= 120 kWh

∴ Total energy = (150 + 120) kWh = 270 kWh ½

∴ Total cost = 270 × 6 = ` 1620

**Question. What is electrical resistivity ? Derive its SI unit.****In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 100 mA. If the length of the wire is doubled, how will the current in the circuit change ? Justify your answer.****Answer:** Electrical resistivity of the material of a conductor is the resistance offered by the conductor of length 1 m and area of cross-section 1 m2.

ρ = RA/l

Unit of ρ = ohm meter2 / metre = ohm meter

Resistance of wire is doubled if its length is doubled.

Hence current is reduced to half.

∴ Ammeter reading = 100 mA/2 = 50 mA

**Question. Show how would you join three resistors, each of resistance 9 Ω so that the equivalent resistance of the combination is (i) 13.5 Ω, (ii) 6 Ω ?****Answer:**

(i)

Two 9 ohm resistors in parallel connected to one 9 ohm resistor in series.

1/RP = 1/9 + 1/9 = 2/9

∴ Rp = 9/2 Ω

R = 9 Ω + 9/2 Ω = 13.5 Ω

(ii)

Two 9 ohm resistors in series connected to one 9 ohm resistor is parallel

Rs = 9 Ω + 9 Ω = 18 Ω

1/R = 1/18 + 1/9

= 3/18

∴ R = 6 Ω

**Question. Three resistors of 10 Ω, 15 Ω and 20 Ω are connected in series in a circuit. If the potential drop across the 15 Ω resistor is 3 V, find the current in the circuit and potential drop across the 10 Ω resistor.****Answer:** In series circuit same current flows through all the

resistors. Current through 15 Ω resistor,

I = V/R = 3 V/15Ω = 1/5

= 0.2 A

∴ Current in the circuit = 0.2 A

∴ Potential drop across 10 Ω resistor is

V = IR

= 0.2 A × 10 Ω

= 2 V

**Question. (a) Write Joule’s law of heating.****(b) Two lamps one rated 100 W 220 V, and the other 60 W 220 V, are connected in parallel to electric mains ****supply. Find the currents drawn by two bulbs from the line, if the supply voltage is 220 V.****Answer:** (a) Joule’s law of heating: Heat produced in a resistor is (i) directly proportional to the square of current for a given resistance, (ii) directly proportional to the resistance for a given current and (iii) directly proportional to the time for which the current flows through the resistor.

H = I2Rt where, H = Heat produced, I = current,

R = Resistance of the conductor and t = Time for

which the current flows through the resistor.

(b) Current in 1st bulb,

I1 = p1/V = 100/220 = 5/11 A or 0.45 A

Current in 2nd bulb,

I2 = p2/V = 60/220 = 3/11 A or 0.27 A

**Question. A bulb is rated at 200V – 40W. What is its resistance ? 5 such bulbs are lighted for 5 hours. Calculate the electrical energy consumed ? Find the cost if the rate is 5.10 per kWh.****Answer:** V = 200 V, P = 40 W

P = VI

I = V/I = 40/200 = 1/5 A

R = V/I = 200/1/5

= 200 × 5 = 1000 Ω

Total Power = 40 W × 5 = 200 W

Time = 5 hrs

Electrical energy = 200 W × 5 hrs.

= 1000 Wh

= 1 kWh.

Cost of 1 kWh = ₹ 5.10

∴ Total cost = ₹ 1 × 5.10 = ₹ 5.10

**Question. A circuit has a line of 5 A. How many lamps of rating 40 W, 200 V can simultaneously run on this line safely ?****Answer:** Given, V = 200 V, P = 40 W, I = 5A, n = ?

nP = VI

n = VI/P = 200 X 5/40

= 1000/4 = 25 lamps

**Question. (a) List the factors on which the resistance of a conductor in the shape of a wire depends.**

(b) Why are metals good conductors of electricity whereas glass is a bad conductor of electricity ? Give reason.

(c) Why are alloys commonly used in electrical heating devices ? Give reason.**Answer:** (a) Factors on which resistance of a conductor depends :

(i) Length of conductor [or R ∝ l]

(ii) Area of cross-section of the conductor

[or R ∝ 1/A]

(b) Metals are good conductor of electricity – as they have low resistivity/have free electrons. Glass is a bad conductor of electricity – as it has high resistivity/have no free electrons.

(c) Reason: Alloys have high resistivity/high melting point/alloys do not oxidize (or burn) readily at high temperatures.

**Long Answer Type Questions :**

**Question. (a) An electric bulb is rated at 200 V-100 W. What is its resistance ? ****(b) Calculate the energy consumed by 3 such bulbs if they glow continuously for 10 hours for complete month of November. ****(c) Calculate the total cost if the rate is Rs 6.50 per unit.****Answer:** (a) Given, V = 200 volts and P = 100 watt

As P = V2/R or R = V2/R = (200)2/100W = 40000/100Ω = 400 Ω

(b) Electrical energy consumed, E = number of units ×

Power of each unit × time × total days

Here, n = 3, P = 100 W, t = 10 hours, Days = 30

So, E = 3 × 100 W × 10 h × 30 = 90,000 Wh

= 90 kWh 2

(c) Total cost of electricity = Total unit of energy

consumed × Cost per unit

= 90 kWh × 6.50 = ₹ 585

**Question. An electric lamp of resistance 20 Ω and a conductor of resistance 4 Ω are connected to a 6 V battery as shown in the circuit. Calculate:**

(a) the total resistance of the circuit.

(b) the current through the circuit.

(c) the potential difference across the (i) electric lamp and (ii) conductor, and

(d) power of the lamp.**Answer: **

(b) Using Ohm’s law I = V/RTotal

= 6V/24Ω = 0.25 A

(c) Potential difference across

(i) Electric lamp, V1 = IR1

= 0.25 A × 20 Ω

= 5V

(ii) Conductor, V2 = IR2

= 0.25 A × 4 Ω

= 1V

(d) Power of the lamp = I2R

= (0.25)2 × 20 Ω

= 1.25 W

**Question. A bulb is rated 40W, 220V. Find the current drawn by it, when it is connected to a 220V supply. Also find its resistance. If the given bulb is replaced by a bulb of rating 25W, 220V,will be there be any change in the value of current and resistance ?****Justify your answer and determine the change.****Answer:** P = 40 W, V = 220V

P = VI

∴ I = P/V = 40/220 = 2/11 A = 0.18 A

From Ohm’s law,

V = IR

R = V/I = 220/2 = 1210 W

When replaced by 25 W, 220 V lamp:

I = P/V = 25/220 = 5/44 A = 0.113 A

R = V2/P = 2202/25 = 1936 Ω

Yes there is change in current and resistance.

Change in current = 0.18 – 0.1136 = 0.0664 A Change in resistance = 1936 – 1210 = 726 ohm.

Hence, from the above justification, we can see that current decreases and resistance increases when we use a 25 W bulb in place of a 40 W.

**Question. (i) Draw a labelled circuit diagram to study a relationship between potential difference (V) across the two ends of a conductor and the current (I) flowing through it. State the formula to show how I in a conductor varies when V across it is increased step wise. Show this relationship also on a schematic graph.****(ii) Calculate the resistance of a conductor if the current flowing through it is 0.25 A when the applied potential difference is 1.0 V.****Answer:** (i)

The formula states that the current passing through a conductor is directly proportional to the potential difference across its ends, provided the physical conditions like temperature, density, etc.

remain unchanged. This is Ohm‘s law.

**Question. (a) With the help of a suitable circuit diagram prove that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.****(b) In an electric circuit two resistors of 12Ω each are joined in parallel to a 6 V battery. Find the current drawn from the battery.****Answer: **

From figure:

**Question. (a) How will you infer with the help of an experiment that the same current flows through every part of a circuit containing three resistors in series connected to a battery ?**

(b) Consider the given circuit and find the current flowing in the circuit and potential difference

across the 15 Ω resistor when the circuit is closed.

**Answer:** (a) (i) Join the three resistors of different values in series.

(ii) Connect them with battery, an ammeter and plug key.

(iii) Plug the key and note the ammeter reading.

(iv) Change the position of ammeter to anywhere in between the resistors and note the ammeter reading each time.

(v) The ammeter reading will remain same every time. Therefore when resistors are connected in series same current flows through all resistors, when it is connected to a battery.

Note: If explained with the help of diagram give full credit.

(b) Total resistance of the circuit = 1

R = R1 + R2 + R3 = 5 + 10 + 15 = 30 ohm

Potential difference across the circuit / By

ohm’s law

V = IR or I =V/R = 30V/30Ω = 1A

Potential difference across 15 ohm Resistor = 1A × 15 Ω = 15 V

**Question. In the given circuit, A, B, C and D are four lamps connected with a battery of 60 V. **

Analyse the circuit to answer the following questions.

(i) What kind of combination are the lamps arranged in (series or parallel)?

(ii) Explain with reference to your above answer, what are the advantages (any two) of this combination of lamps?

(iii) Explain with proper calculations which lamp glows the brightest?

(iv) Find out the total resistance of the circuit**Answer:** (i) The lamps are in parallel.

(ii) Advantages: If one lamp is faulty, it will not affect the working of the other lamps. They will also be using the full potential of the battery as they are connected in parallel.

(iii) The lamp with the highest power will glow the

brightest.

P=VI In this case, all the bulbs have the same

voltage. But lamp C has the highest current. Hence, for Lamp

P = 5 × 60 Watt = 300 W. (the maximum).

(iv) The total current in the circuit = 3+4+5+3 A = 15A

Voltage = 60V

V= IR and hence R = V/I

= 60/15 A = 4A

**Question. Draw a labelled circuit diagram to study the relationship between the current (I) flowing through a conductor and the potential difference (V) applied across its two ends. State the formula co-relating the I in a conductor and the V across it. Also show their relationship by drawing a diagram.**

What would be the resistance of a resistor if the current flowing through it is 0.15 A when the potential difference across it is 1.05 V ?**Answer: **

**Question. Establish a relationship to determine the equivalent resistance R of a combination of three resistors having resistances R1, R2 and R3 connected in series. Calculate the equivalent resistance of the combination of three resistors of 2 Ω, 3 Ω and 6 Ω joined in parallel.****Answer:** An applied potential V produces current I in the resistors R1, R2 and R3, causing a potential drop V1, V2 and V3 respectively through each resistor.

Total Potential, V = V_{1} + V_{2} + V_{3}

By Ohm’s Law, V_{1} = IR_{1}

V_{2} = IR_{2}

V_{3} = IR_{3}

Thus, V = IR_{1} + IR_{2} + IR_{3}

= I(R_{1} + R_{2} + R_{3})

If R is the equivalent resistance, V = IR

Hence, IR = I(R_{1} + R_{2} + R_{3})

R = R_{1} + R_{2} + R_{3}

This proves that overall resistance increases when resistors are connected in series.

Three resistors 2 Ω, 3 Ω and 6 Ω, are connected in parallel combination.

Equivalent resistance,

1/R_{P }= 1/2 + 1/3 + 1/6

= 3+2+1/6 = 6/6

R_{P} = 1 Ω

**Question. (a) Define Power and state its SI unit.**

(b) A torch bulb is rated 5 V and 500 mA. Calculate:

(i) Power (ii) Resistances (iii) Energy consumed when it is lighted for 2½ hours.**Answer:** (a) Power: It is the amount of electric energy

consumed in a circuit per unit time.

P = W/t

Its S.I unit is Watt (W).

(b) V = 5V

I = 500 mA = 0.5 A 2

(i) P = V × I = 5 × 0.5 = 2.5 W

(ii) Resistance R = V/I = 5/0.5 = 10 ohms

(iii) Energy consumed = P × t

= 2.5 × 2.5

= 6.25 Wh

**Question. (i) Consider a conductor of resistance ‘R’, length ‘L’, thickness ‘d’ and resistivity ‘r’.**

Now this conductor is cut into four equal parts.

What will be the new resistivity of each of these parts? Why?

(ii) Find the resistance if all of these parts are connected in:

(a) Parallel

(b) Series

(iii) Out of the combinations of resistors mentioned above in the previous part, for a given voltage which combination will consume more power and why?**Answer:** (i) Resistivity will not change as it depends on the material of the conductor.

(ii) The length of each part become L/4 . ρ A constant

R = ρL/A

Resistance of each part = Rpart = ρL/A /A = R/4.

If Reqv is less, power consumed will be more.

In the given case, Reqv is lesser in the parallel and thus power consumed will be more.