Household Electricity and Power Transmission

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.

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.

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 AC usually $11  KV$ to $22  KV$.

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

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

A watt-hour meter is an electrical instrument that measures the amount of electric energy used by an electric load over a period of time, typically in kilowatt-hours.

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. 

Total energy = (Number of bulbs * Power per bulb * Hours per day * Days) / 1000. 60 kWh = (X * 80 * 5 * 30) / 1000. 60 = X * 12000 / 1000. 60 = 12X. X = 5.

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

Long distance power transmission is done at high voltage.

The electric meter used in home measures electric energy consumed in units of $kWh$ that is $kilowatt$ $hours$.

All options A, B, and C are correct advantages of a parallel circuit in a domestic setting. Since the question asks for a 'not correct' advantage and none of the listed options are incorrect, 'none' is the correct choice.

The higher the voltage V, the lower the current I and therefore much lower power loss.Transmitting electricity at high voltage reduces the fraction of energy lost to resistance

Power supplied is equal to to the voltage multiplied by the current flowing through the wire.

When power is transmitted from power station to different different substation then one important factor is kept in mind that power loss in pathway during transmission over large distances should be minimum.