Tag: electric power

Questions Related to electric power

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

Units of power and voltage are represented by

  1. W, V

  2. A, V

  3. W, $\Omega$
  4. V, $\Omega$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In our every day lives we use electrical energy to do work for us. This electrical energy is provided either by our power company through the outlets in our wall, or we can buy stored electrical energy in the form of batteries. Different electrical appliances use electrical energy at different rates. Power is the rate at which work is done. The power rating of an appliance tells us the rate at which it uses electrical energy. For example, a 1500 watt hair dryer uses 1500 joules of electrical energy per second.
In any electrical appliance their is a simple relationship between the power output of a component, the potential difference across that component, and the current running through it. This is because charge times potential difference is energy, so (charge per unit time) times (potential difference) gives (energy per unit time).
Thus, we have $P=V I.$
Substituting $V=I R$ from ohm's law, the power relationship can be rewritten as $P={ I }^{ 2 }R$. From this formula, the resistance can be calculated.

hence correct option is A

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

Fill the blank.
Voltmeter is always connected in . . . .with the device.

  1. series

  2. parallel

  3. series and parallel both.

  4. the cell.

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A voltmeter is an instrument used for measuring electrical potential difference between two points in an electric circuit. A voltmeter is connected in parallel with a device to measure its voltage.
Hence, the statement is false as the voltmeter is connected in parallel to the device in a circuit.

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

The melting point of filament of a bulb is 

  1. very low.

  2. very high.

  3. normal.

  4. none of these.

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The electric bulb is the glass enclosure around the filament that often contains a vacuum or is filled with a low-pressure noble gas to prevent the filament from burning out due to evaporation at high temperature. The coiled filaments of incandescent lamps are made up of tungsten, a high resistance material that is drawn into a wire .It has both a high melting point (3382 degrees Celsius) and a low pressure which keep it from melting or evaporating too quickly. Supplying electricity through this coiled tungsten wire generates the light. Due to the resistance, it is heated until the wire becomes white-hot. The emission of light by heating the filament wire to white-hot level is known as incandescence. Here, the electrical energy is converted to heat energy by the resistance of the wire.A fused electric bulb has its filament cut, because at higher temperature for a longer period of time, the tungsten filament melts and the flow of electric current stops and the bulb loses its original utility. 
Hence, the statement is true.

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

A certain network consists of two ideal and indentical voltage sources in series and a large number of ideal resistor. The power consumed in one of the resistor is $4W$ when either of the two sources is active and other is replaced by a short circuit. The power consumed by same resistor when both sources are simultaneously active would be: 

  1. $0$ or $16W$
  2. $4W$ or $8W$
  3. $0$ or $8W$
  4. $8W$ or $16W$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ P=VI$

$ I= \cfrac{nE}{R+r}$ (r=0)
In this case 
$ I= \cfrac{nE}{NR} ; E _1 = E _2$
$ P=VI$
In first case 
$ 4 =E \times \cfrac{E}{NR} = \cfrac{E^2}{NR}$
In second case 
$ x =2E \times \cfrac{2E}{NR} = \cfrac{4E^2}{NR}$
$ \cfrac{x}{4} = \cfrac{4E^2}{NR} \times \cfrac{NR}{E^2}$
$ x= 0 \quad or \quad 16W$