Power Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. An electric fan, with an effective area of cross-section A, accelerates air of density d to a speed v. What is the power needed for this process?

1. dA
2. 1/2 dAv
3. 1/2 dAv²
4. 1/2 dAv³

Ans. The correct answer is d. 1/2 dAv³

Explanation: The mass of the air passes the cross-section A in time t is given by

m = density (d) x volume (V)

⇒ m = dAvt

Kinetic energy, E = 1/2 mv² = 1/2 x dAvt x v² = 1/2 x dAv³t

Power = E/t = (1/2 x dAv³t)/t = 1/2 dAv³

Ques. One horsepower is equivalent to approximately

1. 645.7 watts
2. 845.7 watts
3. 745.7 watts
4. 545.7 watts

Ans. The correct answer is c. 745.7 watts

Explanation: Horsepower, which is approximately equivalent to 745.7 watts, is used to measure the power delivered by a machine.

Ques. Which among the following formulas is used to find power?

1. Power = Force x Velocity
2. Power = 2(Force – Velocity)
3. Power = Force + Velocity
4. Power = Force / Velocity

Ans. The correct answer is a. Power = Force x Velocity

Explanation: Power is defined as the rate of change of doing work i.e.

Power = Work/time

⇒ Power = (Force x displacement) / time

But displacement/time = velocity

⇒ Power = Force x Velocity

Ques. Can the power of a motor be represented using horsepower?

1. True
2. False

Ans. The correct answer is a. True

Explanation: The term horsepower refers to the amount of power delivered by a machine or motor.

Ques. One Gigawatt =

1. 1 x 10³ W
2. 1 x 10⁶ W
3. 1 x 10⁹ W
4. 1 x 10¹² W

Ans. The correct answer is c. 1 x 10⁹ W

Explanation: The metric unit prefix Giga denotes a factor of a short-scale billion or long-scale milliard (10⁹ or 1000000000). It is denoted by the letter G.

Therefore. One Gigawatt = 1 x 10⁹ W

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Short Answers Questions [2 Marks Questions]

Ques. Define power.

Ans. The quantity of energy transferred or converted per unit of time is referred to as power. It is a scalar quantity. The watt is the unit of power in the International System of Units, equivalent to one joule per second. Power is also referred to as activity in older works.

Ques. Define Average power.

Ans. The definition of average power (P_avg) is the ratio of total work done to total time taken. The watt is the SI unit of measurement for average power. A wattmeter is used to measure average power.

Ques. Define Instantaneous power.

Ans. The power used at a certain point in time is referred to as instantaneous power. In electronics, the instantaneous power is an important measure. It allows the designer to project the components' capacity to tolerate power consumption.

Instantaneous power is defined as the limiting value of the average power as the time interval Δt approaches zero.

Ques. Define one watt.

Ans. The rate of work done is one watt when an object's velocity is maintained constant at one meter per second against a constant opposing force of one Newton.

In electromagnetism, One watt is the rate at which electrical work is done when a current of one ampere (A) travels across an electrical potential difference of one volt (V).

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Long Answers Questions [3 Marks Questions]

Ques. To produce 106 Joule of heat in 20 seconds, how much voltage should be applied to 2000-ohm resistance?

Ans. Given

• The heat energy to be produced by the resistance, H = 10⁶ J
• Time required to produce heat, t = 20 seconds
• Resistance, R = 2000 ohm

Heat energy is given by, 

H = V²t/R

⇒ V² = HR/t = (10⁶ x 2000)/20 = 10⁸

⇒ V = 10⁴ V

Hence the 10⁴ V voltage should be applied to 2000 ohm resistance to produce 10⁶ Joule of heat in 20 seconds.

Ques. An electric bulb is connected to a 220 V power supply line. If the bulb draws a current of 0.5 A, calculate the power of the bulb.

Ans. Given

• The voltage applied to the electric bulb, V = 220 V
• The current drawn by the bulb, I = 0.5 A

The formula to calculate power in terms of voltage and current is given by

P = V x I

On substituting the values, the power of the electric bulb is given by

⇒ P = 220 x 0.5 = 110 watts

Ques. When a constant force is applied to a body moving with constant acceleration, is the power of the force constant? If not, how would force have to vary with speed for power to be constant?

Ans. The relationship between power (P), force (F), and velocity (v) is given by

P = Fv

• The rate of change in velocity is known as acceleration. Because the body exhibits constant acceleration, it can be said that its velocity is increasing.
• Power changes as velocity changes. As a result, the power of the force is not constant.
• From P = Fv, we get F = P/v
• If power is to be kept constant, then F ∝ 1/v

If the body experiences acceleration, its velocity changes, and therefore its power changes. Force should vary inversely to velocity in order for power to be constant. In other words, while velocity is increasing (due to acceleration), force should be decreased to maintain power constant.

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Very Long Answers Questions [5 Marks Questions]

Ques. An engine pumps 400 kg of water through a height of 10 m in 40 seconds. Find the power of the engine if its efficiency is 80%. Take the value of g as 10 m/s²

Ans. Given

• Mass of water pumps through the engine, m = 400 kg
• Height through which water pumps, h = 10 m
• Time taken by the engine to pump water = 40 seconds
• Efficiency of the engine, ɲ = 80%

Work done by the engine against gravity is given by

W = mgh

⇒ W = 400 x 10 x 10 = 40 x 10³ J

The power used by the engine is given by

P’ = W/t

⇒ P’ = (40 x 10³)/10 = 10³ watt

If P is the power of the engine, then

P x Efficiency of the engine(ɲ) = Power used by the engine(P’)

⇒ P = P’/ɲ

⇒ P = (10³/80) x 100 = 1.25 kW

Hence, the power of the engine is 1.25 kW.

Ques. An elevator that can carry a maximum load of 1500 kg (elevator and passengers) is moving up with a constant speed of 2 m/s. The frictional force opposing the motion is 3000 N. Find the minimum power delivered by the motor to the elevator in watts as well as in horsepower. (g = 10 m/s²)

Ans. Given

• Maximum mass can be carried by the elevator, m = 1500 kg
• Speed of the elevator, v = 2 m/s
• Frictional force opposing the motion of the elevator, F_f = 3000 N

The total downward force acting on the elevator is given by

Downward force (F) = Weight of the elevator (mg) + Frictional force (F_f)

⇒ F = (1500 x 10) + 3000

⇒ F = 18000 N

The motor of the elevator must apply a force equal to 18000 N in the opposite direction to balance this downward force. 

Hence, the minimum power delivered by the motor is given by

P = F x v

⇒ P = 18000 x 2 = 36000 watts

Also, we know 1 horsepower(hp) = 746 watts, therefore

P = 36000/746 hp = 48.3 hp

Therefore, the minimum power delivered by the motor is 36000 watts or 48.3 horsepower.

Ques. The windmill sweeps a circle of area A with its blades. 

• If the velocity of the wind is perpendicular to the circle, find the mass of the air passing through it in time t.
• What is the kinetic energy of the air? 
• Assume that the windmill converts 25% of the wind's energy into electrical energy, and v = 36 km/h, A = 30 m2, and the density of the air is 1.2 kg m-3. What is the electrical power produced?

Ans. Given

• The speed of the wind, v = 36 km/h = 10 m/s
• Area sweeps by the fan of the windmill, A = 30 m²
• The density of the air, ρ = 1.2 kg m^-3
• Efficiency of the windmill, ɲ = 25%

• Let the air travel distance d in time t with velocity v, then the distance covered by the air is given by

d = vt 

The volume of the air, V = Area (A) x distance (d) = Avt

Therefore, the mass of air passed from area A in time t can given by

m = density of air (ρ) x Volume of air (V)

⇒ m = ρAvt

On substituting the values, we get

m = 1.2 x 30 x 10 x t = 360t

• The kinetic energy of the air is given by

K.E = 1/2 mv²

⇒ K.E. = 1/2 x 360t x 10² = 18000t

• Power is defined as the work done (energy) per unit of time. Therefore power delivered by the kinetic energy of the air is given by

P_air = K.E/t = 18000t/t = 18000 watts

Since the efficiency of the windmill is 25%, therefore 25% of this power will be converted into electric power. Therefore

P_electric = 25% x P_air

⇒ P_electric = (25/100) x 18000

⇒ P_electric = 25 x 180 = 4500 watts = 4.5 kW

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