Thermal effect of electric current - class-XII

This is because some of the electric energy is converted into heat energy, i.e., heating effect of electric current.

When a metal plate is getting heated, it may be due to the passage of
direct current, alternating current or even induced current through the
plate. As time varying magnetic field produces induced current  in the
plate, so both (a) and (b) are correct.

A conductor has its own resistance. Due to the flowing current, certain heat is produced in the conductor according to the Joule's heating effect  i.e   $H = I^2Rt$.

Answer is A.

A metallic conductor has a large number of free electrons in it. When a potential difference is applied across the ends of a metallic wire, the free electrons begin to drift from the low potential to the high potential region. These electrons collide with the positive ions (the atoms which have lost their electrons). In these collisions, energy of the electrons is transferred to the positive ions and they begin to vibrate more violently. As a result, heat is produced. Greater the number of electrons flowing per second, greater will be the rate of collisions and hence more heat is produced.
Hence, the temperature of a metal wire rises when an electric current is passed through it because collision of conduction electrons with the atoms of metal gives them energy which appears as heat.

As we know,

Amount of heat produced in wire = $I^2R$

$Power=i^2R$

Prefix Mega means $ 10^6 $.

Given, P$\displaystyle = 805 W, V= 230 V$
$\displaystyle \therefore =I \frac{P}{V}$
$\displaystyle = \frac{805}{230}= 3.5ampere.$

Electrical Power is defined as the electrical work done per unit time.

According to Joule's heating effect of current, heat energy is generated across a conductor due to the current flowing through it and hence its temperature increases.

Whenever an electric current is passed through a conductor, it gets heated up. And the amount of heat created is directly proportional to the square of the current.

Whenever an electric current is passed through a conductor,it gets heated up. This indicates that a part of electrical energy gets converted into heat energy.

An electric bulb works on the principle of hearing effect of current. It consists of a tungsten filament with a coating of thorium.

he electricity used over the lifetime of a single incandescent bulb costs 5 to 10 times the original purchase price of the bulb itself. Light Emitting Diode (LED) and Compact Fluorescent Lights (CFL) bulbs have revolutionized energy-efficient lighting. CFLs are simply miniature versions of full-sized fluorescents.

$1.$ Bulb work by heating up a small wire called a filament. Electric current heats the filament so much that it glows.

Electric bulb works by heating up a small wire called a filament. Electric current heats the filament so much that it glow. In fact, $90%$ of the electricity used by a electric bulb is converted to heat and this conversion will produce maximum wastage of electricity.

Total power P = V * I = 220 V * 5 A = 1100 W. Since each bulb is 20 W, the number of bulbs n = 1100 / 20 = 55.

Treating initially that there is no radiation and on applying voltage to the heater, the temperature of the heater rises. As the temperature rises, the radiation also increases.
As the input power($i^2r$) is constant, the output power should also be constant.
Initially temperature will be increasing and  radiation as well. When the heater reaches a certain temperature, the radiation power equals the input power and the temperature of the heater stops rising.
Thus temperature becomes constant after some time.  

As  heat produced depends upon resistance which has no relation with AC or DC but heat produced  also depends upon current as $H \propto\  I^2$ so for heating effect both A.C. and D.C. can be used, that's why appliances based on heating effect can work on AC as well as DC. hence correct option is C.

The right matches are given below.
Heat Generated$\rightarrow $Propotional to the square of current$\rightarrow $ $\frac { VIt }{ 4.18 } cal$  
Resistance in parallel$\rightarrow $Is used to reduce effective resistance in a circuit$\rightarrow $  $\frac { 1 }{ { R } _{ p } } =\frac { 1 }{ { R } _{ 1 } } +\frac { 1 }{ { R } _{ 2 } } $
Resistivity$\rightarrow $Depends on the material of the conductor$\rightarrow $  $\rho =\frac { RA }{ l } $
Ohm's law$\rightarrow $Gives relation between V and I$\rightarrow $V=IR

An electric iron does not use an electric motor .

$\displaystyle I = \frac {m}{Zt} = \frac {297 \times 10^{-6}}{3.3 \times 10^7 \times 15 \times 60}=1A$

$H = I^2Rt = 1 \times 10 \times 15 \times 60$
$=9000J = 9kJ$

The differential form of the Joule heating equation gives the power per unit volume.

ESSENTIAL REQUIREMENTS OF GOOD HEATING ELEMENT High-specific resistance so that small length of wire may be required to provide given amount of heat. High-melting point so that it can withstand for high temperature, a small increase in temperature will not destroy the element.

Heat generated $H = i^2 Rt$

CFL is Compact Fluorescent Lamp.

The electric bulb has a filament called tungsten when electricity passes through this filament, it heats up and glows. 

Total power = 10 tubes * 40 W = 400 W = 0.4 kW. Energy = 0.4 kW * 10 hours = 4 kWh. Cost = 4 kWh * 2.50 Rs/kWh = 10 Rs.

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

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.

Given:

Given :    $t = 6\ hrs$             

Given :   $P = 4.4$ kW             $t = 4$ h

Electrical energy consumed by an electric appliance is equal to product of its power rating and time for which it is used $E = P \times t$.

Electricity bill is measured in units where 1 unit is equal to kWh which is the unit of energy. Hence, when we pay for our electricity bill, we pay for the amount of electrical energy consumed by us.

Energy consumed, $E =2.5 \times 2 = 5\ kWh $
Cost, $C = 5 \times 0.60 = Rs\ 3 $

For bulbs :

$Energy\,\,=\,power\times time$

Total hours in 30 days $5\times 30=150hours$

So, $ energy=0.1\times 150=15J $

 $  $Power of 4 bulbs

 $ =4\times 15=60kW $

 $  $For LED:

Total hours in 30 days $=150hours$

$E=0.008\times 150=1.2J$

So, power of 4 LEDs $=1.2\times 4=4.8hours$

Saved energy 

  $ =\,\,60-4.8 $

 $ =55.2units $

Energy consumed $=10\times 50\times 10\times 30\times 3600 J$
$[1 Kwh=3600\times 1000J]$
$=\frac {10\times 50\times 10\times 30\times 3600}{3600\times 1000}kWh=150$

Power P is proportional to V^2. If V decreases by 2.5%, V_new = 0.975 V_old. P_new is proportional to (0.975)^2 = 0.9506. The decrease is approximately 5%.

An ohmmeter measures resistance at low voltage/current. A tungsten bulb's resistance increases significantly with temperature. When in use, the filament is hot, so its operating resistance is much higher than its cold resistance measured by an ohmmeter.

Given:
$H _1 = 500\ W$
$V _1 = 115\ V$
$V _2 = 110\ V$

From Joule's Law of heating,
$H _1 = \cfrac{(V _1)^2}{R}$ and $H _2 = \cfrac{(V _2)^2}{R}$
$\Rightarrow R = \cfrac{(V _1)^2}{H _1} = \cfrac{(V _2)^2}{H _2}$
$\Rightarrow H _2 = \cfrac{V _2^2}{V _1^2} H _1$
$\therefore H _2 = \cfrac{(110)^2}{(115)^2} (500)$
$\therefore H _2 = 457.46 W$

Heat lost per second per unit surface area of fuse wire is
$H = \frac{I^2 \rho }{2 \pi ^2 r^3}$
$\Rightarrow  I^2 \propto r^3$
$\Rightarrow  I \propto r^{\frac{3}{2}}$

Heat produced is given by
$Q = I^2 R t$

$\text{Energy = Power} \times \text{time}$
$\text{Energy} = 1000 \times 30 = 30000\ J = 3 \times {10}^{4}\ J$

We know that 

Power is given as: $P=VI$

The electric power in watts associated with a complete electric circuit or a circuit component represents the rate at which energy is converted from the electrical energy of the moving charges to some other form, e.g., heat, mechanical energy, or energy stored in electric fields or magnetic fields. 

For lamp, electrical energy consumed is $E _L = P _Lt =100\times 10 = 1\ kWh$

Given :       $P = 0.050\ kW$ 

Given :    $P = 1000$ W $=1$ kW              $t = 10$ h

Energy consumed per day is: $E = Pt$

$E = VIt$
  $= 12 \times 4 \times \dfrac{1}{6}$
  $= 8 \times {10}^{-3}   kwh$

Given :       $P=0.1$ kW           $t = 1$ hour per day

$I = 5A,   p.d = 200   V$
Power $= V \times I = 200 \times 5 = 1000   \omega$
For 2 fans $= 2 \times 1000 = 2000   \omega$.

$E = {I}^{2}RT = {0.5}^{2} \times 10 \times 1800$
       $= 4500  J$

$E = {I}^{2}Rt$
$= {5}^{2} \times 2 \times \dfrac{1}{2}$
$= 2.5 \times {10}^{-2}  kwh$

$V = 200   V,   P = 750   \omega$.
$P = VI ,   750 = 200 \times I$
$I = 3.75   A$.

$E = \dfrac{{V}^{2}t}{R}$
  $= \dfrac{{\left(200\right)}^{2} \times 10}{480}  =  1008.3  J$

P.D $= 2.5   V$
Current $= 0.5   A$
$t = 2$ minute $= 2 \times 60 = 120  s$
$ E = V \times I \times t = 2.5 \times 0.5 \times 120 = 150   J$.

Total unit consumed in two hours $E=2.5\times 2=5,unit$

1 unit is $3.6\times 10^6J$

Resistance of the bulb $R = \dfrac{V^2}{P}$

$\displaystyle E= Power(in kW)\times time=\frac{4\times 100\times 6}{1000}$kWh$=2.4$kWh
Consumed in $30$ days$=30\times 2.4=72$
Total cost $=72\times 5=360$Rs.

Answer is C.

As the three bulbs are connected in parallel, the voltage applied across them would be the same, lets say its V.
Now, due to the fact that the power dissipated by the three bulbs is different the current flowing through them would be different and thus they will have different brightness.
Now, the brightness of a bulb is directly proportional to the power it uses or the amount of current it draws. 
Thus, in this case the brightness of bulb B (60 W) will be more than that of the bulb A (40 W) but less than that of the bulb C (100 W). That is, C has the maximum brightness and the bulb has has the least brightness. Bulb B in between.
Hence, the correct statement is option C.

$9\ units = 9kWh$
For the month of April (30 days), we get an average of $ \cfrac{9000}{30} =300 Wh$ energy per day.
For an appliance of $60\ W$, this clearly corresponds to $ t = \cfrac{E}{P} = \cfrac{300}{60} = 5\ h $

Energy consumed is given by:

Resistance of Ist bulb $={ R } _{ 1 }=\cfrac { { V }^{ 2 } }{ { P } _{ 1 } } $
Resistance of IInd bulb $={ R } _{ 2 }=\cfrac { { V }^{ 2 } }{ { P } _{ 2 } } $
when both bulbs are connected in series
$\quad { R } _{ eq }={ V }^{ 2 }\left[ \cfrac { 1 }{ { P } _{ 1 } } +\cfrac { 1 }{ { P } _{ 2 } }  \right] =\cfrac { { V }^{ 2 }\left( { P } _{ 1 }+{ P } _{ 2 } \right)  }{ { P } _{ 1 }{ P } _{ 2 } } $
Hence, power consumed $P=\cfrac { { V }^{ 2 } }{ R } =\cfrac { { V }^{ 2 } }{ { V }^{ 2 }\left( \cfrac { { P } _{ 1 }+{ P } _{ 2 } }{ { P } _{ 1 }{ P } _{ 2 } }  \right)  } =\cfrac { { P } _{ 1 }{ P } _{ 2 } }{ { P } _{ 1 }+{ P } _{ 2 } } \quad $

Power consumed by bulb is $P=\dfrac{V^2}{R}$

Power of heater $=P=\dfrac{V^2}{R}$

Answer is A.

The heat which is produced due to the flow of  current   within an electric wire, is expressed in Joules.
James Prescott Joule was an English physicist who studied the nature of heat and established its relationship to mechanical work. He therefore laid the foundation for the theory of conservation of energy, which later influenced the First Law of Thermodynamics. He also formulated the Joules laws which deal with the transfer of energy.

If an electric fan is switched on in a closed room, the air will be heated because due to motion of the fan, the speed of air molecules will increase. In fact, we feel cold due to evaporation of our sweat

Laws of heating were given by Joule.

By Joule's Law, heat produced

According to Joule's law,       $W = JQ$