Tag: commercial unit of energy

Questions Related to commercial unit of energy

Smallest commercial unit of energy is

  1. Kilowatt hour

  2. Watt second

  3. Watt hour

  4. Watt minutes


Correct Option: C
Explanation:

Smallest commercial unit of energy is called Watt hour. It is defined as the amount of electric energy, which flows through a conductor in one hour, at a power of one watt.

Convert $28.8  khJ$ into kilo watt hour.

  1. $8 \times {10}^{-3} kwh$

  2. $8 kwh$

  3. $8 \times {10}^{3} kwh$

  4. None of these


Correct Option: A
Explanation:

$E = \dfrac{28.8 \times {10}^{3}}{3.6 \times {10}^{6}}$
   $= 8 \times {10}^{-3}  kwh$

Biggest commercial unit of energy.

  1. $kwh$

  2. $BOT$

  3. $kVAh$

  4. All


Correct Option: D
Explanation:

The biggest commercial unit of energy is kilowatt hour $\left(kwh\right)$. It is also known as kilo-volt-ampere hour $\left(kVAh\right)$ and board of trade unit (BOT) or simply electric unit.

Unit used in selling electrical energy to consumer.

  1. Volt-Amper

  2. Kilowatt-hour

  3. Volt/second

  4. None


Correct Option: B
Explanation:

Commercial unit of electrical energy is  Kilowatt-hour  i.e  $kWh$ and 1 kWh is equal to 1 unit of energy.

Watt hour meter means

  1. Electric energy

  2. Current

  3. Voltage

  4. Power


Correct Option: A
Explanation:

As watt hour is the unit of energy, thus watt hour meter measures the amount of energy consumed for a given time period.

$1  kwh$ is equal to

  1. $3.6 \times {10}^{6} J$

  2. $100 J$

  3. $1000 J$

  4. $3.6 \times {10}^{3} J$


Correct Option: A
Explanation:

$1  kwh = 1  kw \times 1  h$
     $= 1000  w \times 1 \times 60 \times 60$
     $= 3.6 \times {10}^{6}  J$

A lamp rated 20w and an electric iron rated 50w are used for 2 hour everyday. Calculate the total energy consumed in 20 days.

  1. 14kwh

  2. 2.8kwh

  3. 40kwh

  4. All


Correct Option: B
Explanation:

Energy consumed by lamp in 2 hour     $E _l = 0.02\times 2 = 0.04$ kWh per day

Energy consumed by iron in 2 hour     $E _i = 0.05\times 2 = 0.1$ kWh per day
$\therefore$ Total energy consumed by both appliance     $E _T = (E _l+E _i)\times 20 = (0.04+0.1)\times 20 = 2.8$ kWh

One kilowatt hour is equal to

  1. $\displaystyle 36\times { 10 }^{ 5 }$ joules

  2. $\displaystyle 36\times { 10 }^{ 3 }$ joules

  3. $\displaystyle { 10 }^{ 3 }$ joules

  4. $\displaystyle { 10 }^{ 5 }$ joules


Correct Option: A
Explanation:

$1$ kW  $ = 1000$ $\dfrac{J}{s}$            

$1$ h $ = 3600$ s
$\therefore$  $1$ kWh $ = 1000\dfrac{J}{s}\times 3600$ $s  =36\times 10^5$  $J$

Number of KWh in 1Joule.

  1. $\displaystyle 3.6\times { 10 }^{ 6 }KWh$

  2. $\displaystyle 2.77\times { 10 }^{ -7 }KWh$

  3. $\displaystyle 600KWh$

  4. $\displaystyle 1.6\times { 10 }^{ -19 }KWh$


Correct Option: B
Explanation:

We know  $1$ kWh $ = 3.6\times 10^6$ $J$

$\therefore$  $1$ $J = \dfrac{1}{3.6\times 10^6} = 2.77\times 10^{-7}$ kWh

Calculate the number of Joules in 1KWh.

  1. $\displaystyle 6\times { 10 }^{ -19 }J$

  2. $\displaystyle 3.6\times { 10 }^{ 6 }J$

  3. $\displaystyle 60J$

  4. $\displaystyle 59J$


Correct Option: B
Explanation:

$1$ kW $ = 1000$ $\dfrac{J}{s}$

$1$ h $=3600$ s 
$\therefore$  $1$ kWh $ = 1000\dfrac{J}{s}\times 3600$ $s  =3.6\times 10^6$  $J$