Curso-Taller para los representantes de barrios

∴ I =

=

∴ P _max = _{( )}

= _{( )}

∴ P _max =

R =

(∵ R = R₀)

The power supplied by D.C. voltage is P_i = VI = .

/ =

(∵ R = R₀) Percentage efficiency is

 = =

=4

5 = = 50%

Q.18- How much (in percent) has a filament diameter decreased due to

evaporation if the maintenance of the previous temperature required an increase of the voltage by  = 1.0%? The amount of the heat transferred from the filament into surrounding space is assumed to be proportional to the filament surface area.

Sol. At steady state, internal energy of filament becomes constant. So, the temperature becomes constant. From energy point of view, thermal power developed in filament is lost at heat in environment at steady state.

crackiitjee.in

According to given problem, thermal power Q loss is proportional to the surface area A of filament (cylindrical in shape) mathematically,

Q A or Q = kA or Q = k 2rl

Where r = radius of filament and l = length of filament The power supplied by the applied voltage V is

P = At steady state,

∴ P = Q

or = k 2rl ………..(i)

Here R = resistance of the filament = =

From equ. (i), we get

∴ = k 2rl

or = k 2rl

or = 2k

or = k0 (here k₀ = 2k = constant)

∴ V² =

Differentiating both sides with respect to r,

2V

= -

crackiitjee.in

Negative sign indicates that when radius of filament decreases, voltage increases.

∴ 2V .

/ = or 2V

= .∵

/ or 2V

= .∵ /

or 2 = or 2 =

=

∴ 100 = 2 . /

Hence, percentage decrease in diameter of filament is 100 = 2. / = 2η = 2 × 1% = 2%

Q.1. A conductor has a temperature –independent resistance R and a total capacity C. at the moment t = 0 it is connected to a dc voltage V. Find the time dependence of a conductor’s temperature T assuming the thermal power dissipated into surrounding spaces to vary as q = k(T - T₀), where k is a constant, T₀ is the environmental temperature (equal to the conductor’s temperature at the initial moment).

Sol. It is understood that before steady state, power supplied = Thermal power loss + change in internal energy.

crackiitjee.in

When a conductor receives electrical energy internal energy of conductor increases so, the temperature of conductor increases at t = 0, no temperature differences are found between conductor and environment. So at t = 0, heat loss is zero.

When t > 0, a part of electrical energy increases internal energy (temperature of conductor) and remaining part of electrical energy is lost as heat.

Due to this, temperature difference between conductor and environment increases. Hence, heat lost increases.

After some time, steady states come into play. At steady state, total electrical power supplied to conductor is lost as heat in environment.

The temperature of conductor becomes constant.

Before steady state:

Electrical power supplied = Thermal power lost + rate of change in internal energy

or = k (T - T₀) +

or = k (T - T₀)

(CT) or = k (T - T₀)+ C or C ∫

( )

= ∫

After integrating, we get T =

(1-e^-kt/c)

Q.2. A circuit shown in has resistances R₁ = 20 and R₂ = 30 . At what value of the resistance R_x will the thermal power generated in it be practically

crackiitjee.in

independent of small variations of the resistance ? The voltage between the points A and B is supposed to be constant in this case.

Sol. Power developed in resistance R_x to be maximum, the power generated in R_x is practically independent of small variation of resistance. For maximum power developed in R_x is,

= 0 In loop (1),

V - IR₁ - (I - I₁) R₂ = 0 ……(i) In loop (2),

- I₁R_X + (I - I₁) R₂ = 0 …….(ii) After solving Eq. (i) and (ii), we get

I₁ =

The thermal power generated in the resistance R_x is P = I₁² R_X= .

/² R_X For maximum value of p,

= 0 After solving, we get R_X =

On putting the values, we get R_X = 12

Q.3. in a circuit shown in . resistances R₁and R₂ are known, as well as emf’s ₁ and

2. The internal resistances of the sources are negligible. At what value of the

crackiitjee.in

resistance R will the thermal power generated in it be the highest ? What is it equal to?

Sol. This problem can be solved by maximum power transfer theorem.

The problem can be solved in following ways:

First: Show the original circuit.

Now : remove the resistor of resistances R from the circuit

In this step, we remove that resistance from the circuit in which maximum thermal power is generated.

Now: drew the circuit after short circuited the all cells(but internal resistances of cells remains in the circuit) of circuit.

The circuit of this step

The equivalent resistance of circuit in between point B and D is R_BD =

For maximum power generated in resistance R,the value of R.

= R_BD =

Now draw original circuit and find current through R.

In loop (1), ₁ - IR₁ - I₁R = 0 ….(i) In loop (2)

- (I - I₁) ₂ + I₁R = 0 ….(ii) After solving Eq.(i) and (ii), we get

crackiitjee.in

I₁ = ^{( )}

∴ Power generated in resistance R is P = R For P_max, R = R_BD =

∴ P_max = ⁽ _{( )} ⁾ .

/ = ^{( )}

( )

Q.4. A series-parallel combination battery consisting of a large number N = 300 of identical cells, each with an internal resistance r = 10 Ω. Find the number n of parallel groups consisting of an equal number of cells is connected in series, at which the external resistance generates the highest thermal power.

Sol. This problem can be solved in the following way

First: find equivalent emf and internal resistance of the combination of cells. Let each row contains number of shell

Let each row contains number of cells in series.

∴ The equivalent emf of each row is ₀ = n₀

Also, the equivalent internal resistance of each row is r₀ = n₀r The equivalent circuit is

Now connect the equivalent cell with external resistance R.

crackiitjee.in

From KVL

₀ - – IR = 0

∴ I =

or I =

(∵ = and ₀ = ) ∴ I =

The total number of cells = N = number of row × number of cells in each row N = nn₀ ∴ n₀ =

∴ I = ^{. /}

=

The power generated in external resistance R is P = R = . / R For maximum power, = 0

After solving, we get n =√

On putting the values, we get n = 3

Q.5. A capacitor of capacitance C = 5.00 F is connected to a source of constant emf  = 200 V. Then the switch S_w was thrown over from contact 1 to contact 2.

Find the amount of heat generated in a resistance R₁ = 500  if R₂ = 330 .

Sol.

crackiitjee.in

This problem can be solved in following ways : Now draw the circuit when switch is at position at 1.

The circuit is shown in . In this case, the circuit is in steady state. So, current in the circuit will be zero.

In loop (1)

 = = 0 ∴ q₀ = C  ………..(i) Now discuss the circuit at the position (2) of the circuit :

At t = 0, when the switch S_w was thrown over from contact 1 to contact 2, current in the circuit is zero. Now capacitor starts to discharge through resistances R₁ and R₂.

After long time, again circuit comes in steady state. At this steady state, charge on capacitor and current in circuit both are zero.

From energy point of view energy stored on capacitor appears as thermal energy in resistors R₁ and R₂.

But R₁ and R₂ are in series, so current in both are same.

∴ Δ H = I² Rt ∴ Δ H₁  R₁ and Δ H₂  R₂

∴ =

∴ Δ H₂ = Δ H₁ ……….. (ii) Total heat generated is Δ H = Δ H₁ + Δ H₂

According to conservation principle of energy, Δ H + Δ U = 0

crackiitjee.in

or Δ H + (U_f - U_i) = 0

∴ Δ H = U_i – U_f

∴ Δ H =

- 0 or Δ H₁ + Δ H₂ =

Δ H₁ + Δ H₁ = (from eqn (ii)) or Δ H₁ . / =

or Δ H₁ . / =

∴ Δ H₁ = _{( )(}

) = _{( )(} ^{( )}

) =

( ) = 60 mJ

Q.6. A glass plate totally fills up the gap between the electrodes of a parallel plate

In document RECONOCIMIENTO INTERSUBJETIVO DE NECESIDADES HUMANAS (página 152-158)