Tag: electricity and magnetism

Questions Related to electricity and magnetism

State whether true or false.
Electricity used at homes is generated in powerhouses.

  1. True

  2. False


Correct Option: A
Explanation:

Electricity used in homes is made at a power plant by huge generators. Most power plants use coal, but some use natural gas, water or even wind.

State whether true or false.
Electricity used in homes is generated from storage cells.

  1. True

  2. False


Correct Option: B
Explanation:

Electricity used in homes is made at a power plant by huge generators. Most power plants use coal, but some use natural gas, water or even wind.

An electric fan marked 60 watt consumes 3 units for the duration:

  1. 50 hours

  2. 150 hours

  3. 10 hours

  4. 15 hours


Correct Option: A
Explanation:

Answer is A.

The total energy consumed is 3 units = 3 kWh = 3000 Wh.
The total enrgy dissipated is given as Power * Time.
Therefore, time = Energy dissipated / Power.
Time taken = 3000 Wh / 60 W = 50 hours.
Hence, an electric fan marked 60 watt consumes 3 units for the duration of 50 hours.

Electricity generated at power station is

  1. AC

  2. DC

  3. Both of these

  4. None of these


Correct Option: A
Explanation:

Electricity generated at power station is AC usually $11  KV$ to $22  KV$.

The cost of electricity for domestic use is Rs.$2.30$ per unit. This unit is:

  1. $1$ Ampere

  2. $1$ Volt

  3. $1$ Joule

  4. $1$ Kilo-Watt-hour


Correct Option: D
Explanation:

The unit of electricity for domestic use is 'KWh'(KiloWatt-hour)

A battery is charged at a potential of 15V for 8 hours when the current flowing is 10A. The battery on discharge supplies a current of 5A for 15 hours. The mean terminal voltage during discharges is 14 V. The "Watt hour" efficiency of the battery is :-

  1. 80%

  2. 90%

  3. 87.5%

  4. 82.5%


Correct Option: C
Explanation:

% Watt hour efficiency = $\frac{E _{out}} {E _{in}} \times 100$
= $\frac{\left ( 14 \right )\left ( 5 \right )\left ( 15 \right )} {\left ( 15 \right )\left ( 10 \right )\left ( 8 \right )} \times 100 = 87.5$%

Watt-hour meter measures

  1. Electric Energy

  2. Current

  3. Voltage

  4. Power


Correct Option: A

Power transmission is carried out at ........................ and  ..........................?

  1. high voltage, high current

  2. low voltage, high current

  3. high voltage, low current

  4. low voltage, low current


Correct Option: C
Explanation:

One of the key concerns in the transmission of electricity is the power loss in transmission lines, dissipated as heat due to the resistance of the conductors.
The power loss ${P} _{loss} = { I }^{ 2 }R={ (\dfrac { P }{ V } ) }^{ 2 }R(\because p=VI)$
Since $P $and $R $ are fixed conditions, less power will be lost if high voltages V are used.
The transmission line then uses a transformer to step down the voltage.
For a given amount of power, a higher voltage reduces the current and thus the resistive losses occurs in the conductor. For example, raising the voltage by a factor of 10 reduces the current by a corresponding factor of 10 and therefore the$ {I}^{2}{R}$ losses by a factor of 100, provided the same sized conductors are used in both cases. 

X bulbs each of 80 watts are used for 5 hours a day for 30 days and they consume an electrical energy of 60 kWh. Then X is

  1. 1

  2. 3

  3. 5

  4. 7


Correct Option: C

A town situated $20km$ away from a power house at $440V$, requires $600KW$ of electric power at $220V$. The resistance of transmission line carrying power is $0.4\Omega$ per km. The town gets power from the line through a $3000V-220V$ step down transformer at a substation in the town. Which of the following is /are correct

  1. The loss in the form of heat is $640KW$

  2. The loss in the form of heat is $1240KW$

  3. Plant should supply $1340kW$

  4. Plant should supply $640kW$


Correct Option: A
Explanation:

Total length of wire = $20 \times 2 = 40km$


Resistance of wire $R = 40 \times 0.4 = 16\Omega $ ,


${I _{\max }} = \dfrac{P}{V} = \dfrac{{600kW}}{{3000V}} = 200A$

Power loss in the form of heat $ = {I^2} _{rms} \times R = {\left( {200} \right)^2} \times 16 = 640kW$