Power

Rate Of Doing Work

In the previous section, we discussed work and energy, focusing on the amount of work done or the amount of energy transferred. However, the rate at which work is done or energy is transferred is also a crucial concept in physics. This rate is called power.

Imagine two machines doing the same amount of work, say lifting a heavy object to a certain height. If one machine takes 10 seconds to do the work, and the other takes 5 seconds, the second machine is more powerful because it does the same work in less time.

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Definition of Power

Power ($P$) is defined as the rate at which work is done or the rate at which energy is transferred.

If an agent (like a person or a machine) does work $W$ in time $t$, the average power ($P_{avg}$) delivered by the agent is given by:

$ P_{avg} = \frac{\text{Work Done}}{\text{Time Taken}} = \frac{W}{t} $

Similarly, if energy $\Delta E$ is transferred or converted in time $\Delta t$, the average power is $P_{avg} = \frac{\Delta E}{\Delta t}$.

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Instantaneous Power

If the rate of doing work is not constant, we can define instantaneous power ($P$) as the limiting value of the average power as the time interval approaches zero:

$ P = \lim_{\Delta t \to 0} \frac{\Delta W}{\Delta t} = \frac{dW}{dt} $

For an object moving under the influence of a force $\vec{F}$, experiencing a small displacement $d\vec{r}$ in a small time interval $dt$, the work done is $dW = \vec{F} \cdot d\vec{r}$.

The instantaneous power can then be written as:

$ P = \frac{dW}{dt} = \frac{\vec{F} \cdot d\vec{r}}{dt} = \vec{F} \cdot \left(\frac{d\vec{r}}{dt}\right) $

Since $\frac{d\vec{r}}{dt}$ is the instantaneous velocity $\vec{v}$ of the object, the instantaneous power delivered by the force $\vec{F}$ on the object is:

$ P = \vec{F} \cdot \vec{v} $

This formula $P = \vec{F} \cdot \vec{v}$ is very useful. It tells us that the instantaneous power is the dot product of the force applied and the instantaneous velocity of the object. If the force and velocity are in the same direction, $P = Fv\cos(0^\circ) = Fv$. If they are in opposite directions, $P = Fv\cos(180^\circ) = -Fv$ (power is being absorbed or removed). If they are perpendicular, $P = Fv\cos(90^\circ) = 0$.

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Units of Power

The SI unit of power is the watt (W). It is named after James Watt, the Scottish engineer who improved the steam engine.

One watt is defined as the power delivered when one joule of work is done in one second.

$ 1 \text{ Watt (W)} = \frac{1 \text{ Joule (J)}}{1 \text{ Second (s)}} = 1 \text{ J/s} $

For larger amounts of power, we use kilowatts (kW), megawatts (MW), etc.

• $1 \text{ kW} = 10^3 \text{ W}$
• $1 \text{ MW} = 10^6 \text{ W}$

Another historical unit of power, still sometimes used (especially for motors in agriculture or older vehicles in India), is the horsepower (HP) or (PS - Pferdestärke, metric horsepower).

$ 1 \text{ HP} \approx 746 \text{ W} $ (British/US horsepower)

$ 1 \text{ PS} \approx 735.5 \text{ W} $ (Metric horsepower)

Answer:

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Commercial Unit Of Energy

While the joule (J) is the standard SI unit of energy and work, it is a relatively small unit for practical purposes, especially when dealing with large amounts of energy consumed in households and industries.

Electricity meters in homes and businesses measure the total electrical energy consumed over a period (usually a month). This energy is billed in a unit called the kilowatt-hour (kWh).

The kilowatt-hour is not a unit of power, but a unit of energy.

It is defined based on the relationship between power, energy, and time: Energy = Power $\times$ Time.

One kilowatt-hour is the energy consumed when an appliance with a power of 1 kilowatt (kW) runs for 1 hour (h).

$ 1 \text{ kWh} = 1 \text{ kilowatt} \times 1 \text{ hour} $

Let's convert kWh to the standard SI unit (Joules):

$ 1 \text{ kilowatt} = 1000 \text{ Watts} = 1000 \text{ J/s} $

$ 1 \text{ hour} = 60 \text{ minutes} \times 60 \text{ seconds/minute} = 3600 \text{ seconds} $

So,

$ 1 \text{ kWh} = (1000 \text{ J/s}) \times (3600 \text{ s}) $

$ 1 \text{ kWh} = 3,600,000 \text{ J} $

$ 1 \text{ kWh} = 3.6 \times 10^6 \text{ J} $

This unit is convenient for calculating electricity bills. For example, if your electricity consumption for a month is 100 kWh, and the cost per kWh is, say, ₹8, then your energy cost would be $100 \times ₹8 = ₹800$ (excluding other charges and taxes).

The term "Unit" in electricity bills in India often refers to 1 kWh. So, consuming "100 Units" means consuming 100 kWh of electrical energy.