Tag: electric current

Questions Related to electric current

Calculate the energy consumed by $2\ kW$ heater used for $1\ hr$ every day in a period of $30\ \text{days}$.

physics electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $60\ \text{units}$

  2. $120\ \text{units}$

  3. $15\ \text{units}$

  4. $30\ \text{units}$

Show answer
Correct Option: A
Explanation:

$E= Pt = 2 kW\times (30 \times 1 hr)=60 kW h=60\ \text{units}$

You have the following appliances each of $500\ W$ running on  $220\ V$ a.c.: 

(1) Electric iron.
(2) Electric lamp. 
(3) Electric room heater. 
The electric resistance is:

physics electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. maximum for the heater.

  2. maximum for the electric lamp.

  3. maximum for the electric iron.

  4. same in all the three cases.

Show answer
Correct Option: D
Explanation:

Resistance$,\ R= \dfrac{V^2}{P} =\dfrac{(220)^2}{P}$. Since $P$ of each appliance is the same, hence $R$ is same for all the three appliances.

An electric lamp is marked $60\ W,\ 220\ V$. The cost of kilo watt hour of electricity is $Rs.\ 1.25$. The cost of using thing lamp on $220\ V$ for $8\ hrs$ is:

physics electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $Rs.\ 0.25$

  2. $Rs.\ 0.60$

  3. $Rs.\ 1.2$

  4. $Rs.\ 4.00$

Show answer
Correct Option: B
Explanation:

Given:

Power, $P=60\ W$
Time, $t=8\ hrs$

Energy consumed per day, $E=Pt=60\times 8=480\ Whr=\dfrac{480}{1000}=0.48\ kW hr$
Hence total cost, $C=0.48\times 1.25=Rs. 0.60 $

An electric bulb of $60\ W$ is used for $6\ hrs/day$. Calculate the units of energy consumed in one day by the bulb.

physics electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $0.18\ \text{units}$

  2. $0.36\ \text{units}$

  3. $0.54\ \text{units}$

  4. $0.72\ \text{units}$

Show answer
Correct Option: B
Explanation:

Given :    $t = 6\ hrs$             

                $P = 0.06\ kW$
Thus energy consumed per day is:

$E=  Pt = 0.06\times 6  = 0.36\ kWh$$ = 0.36\ \text{units}$

An electric bulb of $30\ W$ consumes $0.72$ units of energy in a day. Find the number of hours it is working in a day?

physics electric current thermal effect of electric current heating effect of electric current electric current and its effects

  1. $6\ hrs$

  2. $12\ hrs$

  3. $18\ hrs$

  4. $24\ hrs$

Show answer
Correct Option: D
Explanation:
Given :   $P =0.030$ kW
Energy consumed per day       $E= 0.72\ \text{units}$$ = 0.72\ kWh$
Using   $E = P\times t$
$\therefore$   $0.72 = 0.03\times t$               
$\implies t = 24\ hrs$

A current I flows through a uniform wire of diameter d when the mean electron drift velocity is
V. The same current will flow through a wire of diameter d/2 made of the same material if the
mean drift velocity of the electron is:

physics electric current drift velocity and mobility drift speed drift velocity & mobility

  1. v/4

  2. v/2

  3. 2v

  4. 4v

Show answer
Correct Option: D
Explanation:

Let I be the current flow through wire. 

Let V be the drift velocity of wire which has diameter d. 
Let A be the area of wire. 
Consider a wire of same material but diameter is half
Therefore, Area =$\dfrac{A}{4}$
Since, $I=NeAV$
N is same for both wire because both of them made of same material. 
$I _{1}=I _{2}$
$NeA _{1}V _{1}=NeA _{2}V _{2}$
$AV=\dfrac{A}{4}V _2$
$V _{2}=4V$