Irodov Solutions: Electric Current- 3 Notes | EduRev

Physics Class 12

JEE : Irodov Solutions: Electric Current- 3 Notes | EduRev

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Q. 184. Find a potential difference φA — φB  between the plates of a capacitor C in the circuit shown in Fig. 3.47 if the sources have emf's Irodov Solutions: Electric Current- 3 Notes | EduRev and Irodov Solutions: Electric Current- 3 Notes | EduRev and the resistances are equal to R1 = 10 Ω, R2 = 20 Ω, and R3 = 30 Ω. The internal resistances of the sources are negligible. 

Irodov Solutions: Electric Current- 3 Notes | EduRev

Solution. 184. Indicate the currents in all the branches using charge conservation as shown in the figure. Applying the loop rule (- Δ φ = 0) in the loops 12341 and 15781, we get

Irodov Solutions: Electric Current- 3 Notes | EduRev

Irodov Solutions: Electric Current- 3 Notes | EduRev

Solving Eqs. (1) and (2), we get

Irodov Solutions: Electric Current- 3 Notes | EduRev

Hence, the sought p.d.

Irodov Solutions: Electric Current- 3 Notes | EduRev

Irodov Solutions: Electric Current- 3 Notes | EduRev


Q. 185. Find the current flowing through the resistance R1 of the circuit shown in Fig. 3.48 if the resistances are equal to R1 =  10 Ω, R2 = 20 Ω, and R3 = 30 Ω, and the potentials of points 1, 2, and 3 are equal to φ1 =  10 V, φ2  = 6 V, and φ3 = 5 V 

Irodov Solutions: Electric Current- 3 Notes | EduRev

Solution. 185. Let us distribute the currents in the paths as shown in the figure.

Irodov Solutions: Electric Current- 3 Notes | EduRev

Irodov Solutions: Electric Current- 3 Notes | EduRev

Simplifying Eqs. (1) and (2) we get

Irodov Solutions: Electric Current- 3 Notes | EduRev


Q. 186. A constant voltage V = 25 V is maintained between points A and B of the circuit (Fig. 3.49). Find the magnitude and direction of the current flowing through the segment CD if the resistances are equal to R1 =  1.0 Ω, R2 = 2.0 Ω, R3 = 3.0 Ω, and R4 = 4.0 Ω. 

Irodov Solutions: Electric Current- 3 Notes | EduRev

Solution. 186. Current is as shown. From Kirchhoffs Second law 

Irodov Solutions: Electric Current- 3 Notes | EduRev

Irodov Solutions: Electric Current- 3 Notes | EduRev

Eliminating i2

Irodov Solutions: Electric Current- 3 Notes | EduRev

Hence 

Irodov Solutions: Electric Current- 3 Notes | EduRev
Irodov Solutions: Electric Current- 3 Notes | EduRev

or, Irodov Solutions: Electric Current- 3 Notes | EduRev

On substitution we get i3 = 1.0 A from C to D


Q. 187. Find the resistance between points A and B of the circuit shown in Fig. 3.50. 

Irodov Solutions: Electric Current- 3 Notes | EduRev

Solution. 187. From the symmetry of the problem, current flow is indicated, as shown in the figure.

Irodov Solutions: Electric Current- 3 Notes | EduRev

Irodov Solutions: Electric Current- 3 Notes | EduRev

Equivalent resistance between the terminals A and B using (1) and (2),

Irodov Solutions: Electric Current- 3 Notes | EduRev


Q. 188. Find how the voltage across the capacitor C varies with time t (Fig. 3.51) after the shorting of the switch Sw at the moment t = 0. 

Irodov Solutions: Electric Current- 3 Notes | EduRev

Solution. 188. Let, at any moment of time, charge on the plates be +qr and - q respectively, then voltage across the capacitor, φ = q/C      (1)

Irodov Solutions: Electric Current- 3 Notes | EduRev

Now, from charge conservation,

Irodov Solutions: Electric Current- 3 Notes | EduRev

In the loop 65146, using - Δ φ • 0.

Irodov Solutions: Electric Current- 3 Notes | EduRev

[using (1) and (2)]

In the loop 25632, using - Δ φ = 0

Irodov Solutions: Electric Current- 3 Notes | EduRev

From (1) and (2),

Irodov Solutions: Electric Current- 3 Notes | EduRev  (5)

On integrating the expression (5) between suitable limits,

Irodov Solutions: Electric Current- 3 Notes | EduRev

Thus  Irodov Solutions: Electric Current- 3 Notes | EduRev


Q. 189. What amount of heat will be generated in a coil of resistance R due to a charge q passing through it if the current in the coil
 (a) decreases down to zero uniformly during a time interval Δt;
 (b) decreases down to zero halving its value every Δt seconds? 

Solution. 189. (a) As current i is linear function of time, and at t = 0 and Δt, it equals i0 and zero respectively, it may be represented as,

Irodov Solutions: Electric Current- 3 Notes | EduRev

Thus  Irodov Solutions: Electric Current- 3 Notes | EduRev

So,  Irodov Solutions: Electric Current- 3 Notes | EduRev

Hence,  Irodov Solutions: Electric Current- 3 Notes | EduRev

The heat generated.

Irodov Solutions: Electric Current- 3 Notes | EduRev

(b) Obviously the current through the coil is given by

Irodov Solutions: Electric Current- 3 Notes | EduRev

Then charge  Irodov Solutions: Electric Current- 3 Notes | EduRev

So,  Irodov Solutions: Electric Current- 3 Notes | EduRev

And hence, heat generated in the circuit in the time interval t [0, ∞],

Irodov Solutions: Electric Current- 3 Notes | EduRev


Q. 190. A de source with internal resistance R0 is loaded with three identical resistances R interconnected as shown in Fig. 3.52. At what value of R will the thermal power generated in this circuit be the highest? 

Irodov Solutions: Electric Current- 3 Notes | EduRev

Solution. 190. The equivalent circuit may be drawn as in the figure.

Irodov Solutions: Electric Current- 3 Notes | EduRev

Resistance o f the network  Irodov Solutions: Electric Current- 3 Notes | EduRev

Let, us assume that e.m.f. of the cell is then current

Irodov Solutions: Electric Current- 3 Notes | EduRev

Now, thermal power, generated in the circuit 

Irodov Solutions: Electric Current- 3 Notes | EduRev


Q. 191. Make sure that the current distribution over two resistances R1 and R2 connected in parallel corresponds to the minimum thermal power generated in this circuit. 

Solution. 191. We assume current conservation but not KirchhofTs second law. Then thermal power dissipated is

Irodov Solutions: Electric Current- 3 Notes | EduRev

Irodov Solutions: Electric Current- 3 Notes | EduRev

The resistances being positive we see that the power dissipated is minimum when

Irodov Solutions: Electric Current- 3 Notes | EduRev

This corresponds to usual distribution of currents over resistance joined is parallel.


Q. 192. A storage battery with emf Irodov Solutions: Electric Current- 3 Notes | EduRev loaded with an external resistance produces a current I = 1.0 A. In this case the potential difference between the terminals of the storage battery equals V = 2.0 V. Find the thermal power generated in the battery and the power developed in it by electric forces.

Solution. 192. Let, internal resistance of the cell be r, then

Irodov Solutions: Electric Current- 3 Notes | EduRev

Irodov Solutions: Electric Current- 3 Notes | EduRev    (1)

where i is the current in the circuit. We know that thermal power generated in the battery.

Irodov Solutions: Electric Current- 3 Notes | EduRev

Putting r from (1) in (2), we obtain,

Irodov Solutions: Electric Current- 3 Notes | EduRev

In a battery work is done by electric forces (whose origin lies in the chemical processes going on inside the cell). The work so done is stored and used in the electric circuit outside. Its magnitude just equals the power used in the electric circuit We can say that net power developed by the electric forces is

Irodov Solutions: Electric Current- 3 Notes | EduRev

Minus sign means that this is generated not consumed.


Q. 193. A voltage V is applied to a de electric motor. The armature winding resistance is equal to R. At what value of current flowing through the winding will the useful power of the motor be the highest? What is it equal to? What is the motor efficiency in this case? 

Solution. 193. As far as motor is concerned the power delivered is dissipated and can.be represented by a load, R0 . Thus

Irodov Solutions: Electric Current- 3 Notes | EduRev

Irodov Solutions: Electric Current- 3 Notes | EduRev

Irodov Solutions: Electric Current- 3 Notes | EduRev

This is maximum when R0 = R and the current / is then

Irodov Solutions: Electric Current- 3 Notes | EduRev

The maximum power delivered is

Irodov Solutions: Electric Current- 3 Notes | EduRev

The power input is Irodov Solutions: Electric Current- 3 Notes | EduRev  and its value when P is maximum is   Irodov Solutions: Electric Current- 3 Notes | EduRev

The efficiency then is  Irodov Solutions: Electric Current- 3 Notes | EduRev


Q. 194. How much (in per cent) has a filament diameter decreased due to evaporation if the maintenance of the previous temperature required an increase of the voltage by η = 1.0%? The amount of heat transferred from the filament into surrounding space is assumed to be proportional to the filament surface area. 

Solution. 194. If the wire diameter decreases by δ then by the information given

P = Power input  Irodov Solutions: Electric Current- 3 Notes | EduRev heat lost through the surface, H. 

Irodov Solutions: Electric Current- 3 Notes | EduRev like the surface area and 

Irodov Solutions: Electric Current- 3 Notes | EduRev

So,  Irodov Solutions: Electric Current- 3 Notes | EduRev

But  Irodov Solutions: Electric Current- 3 Notes | EduRev

Thus  Irodov Solutions: Electric Current- 3 Notes | EduRev


Q. 195. A conductor has a temperature-independent resistance R and a total heat capacity C. At the moment t = 0 it is connected to a de voltage V. Find the time dependence of a conductor's temperature T assuming the thermal power dissipated into surrounding space 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)

Solution. 195. The equation of heat balance is

Irodov Solutions: Electric Current- 3 Notes | EduRev

Put  Irodov Solutions: Electric Current- 3 Notes | EduRev

So,  Irodov Solutions: Electric Current- 3 Notes | EduRev

or,  Irodov Solutions: Electric Current- 3 Notes | EduRev

or,  Irodov Solutions: Electric Current- 3 Notes | EduRev

where A is a constant. Clearly

Irodov Solutions: Electric Current- 3 Notes | EduRev


Q. 196. A circuit shown in Fig. 3.53 has resistances R1 =  20Ω and R2 = 30 Q. At what value of the resistance Rx  will the thermal power generated in it be practically independent of small variations of that resistance? The voltage between the points A and B is supposed to be constant in this case. 

Irodov Solutions: Electric Current- 3 Notes | EduRev

Solution. 196.  Irodov Solutions: Electric Current- 3 Notes | EduRev

Now, thermal power generated in the resistance Rx,

Irodov Solutions: Electric Current- 3 Notes | EduRev

Irodov Solutions: Electric Current- 3 Notes | EduRev

For P to be independent of Rx,

Irodov Solutions: Electric Current- 3 Notes | EduRev


Q. 197. In a circuit shown in Fig. 3.54 resistances R1 and R2 are known, as well as emf's Irodov Solutions: Electric Current- 3 Notes | EduRev The internal resistances of the sources are negligible. At what value of the resistance R will the thermal power generated in it be the highest? What is it equal to? 

Irodov Solutions: Electric Current- 3 Notes | EduRev

Solution. 197. Indicate the currents in the circuit as shown in the figure. Appying loop rule in the closed loop 12561, - Δφ = 0 we get

Irodov Solutions: Electric Current- 3 Notes | EduRev

Irodov Solutions: Electric Current- 3 Notes | EduRev

and in the loop 23452,

Irodov Solutions: Electric Current- 3 Notes | EduRev

Solving (1) and (2), we get,

Irodov Solutions: Electric Current- 3 Notes | EduRev

So, thermal power, generated in the resistance R,

Irodov Solutions: Electric Current- 3 Notes | EduRev

For P to be maximum,  Irodov Solutions: Electric Current- 3 Notes | EduRev

Irodov Solutions: Electric Current- 3 Notes | EduRev

Hence,  Irodov Solutions: Electric Current- 3 Notes | EduRev


Q. 198. A series-parallel combination battery consisting of a large number N = 300 of identical cells, each with an internal resistance r = 0.3 Ω, is loaded with an external resistance R = 10 Ω. Find the number n of parallel groups consisting of an equal number of cells connected in series, at which the external resistance generates the highest thermal power.

Solution. 198. Let, there are x number of cells, connected in series in each of the n parallel groups

Irodov Solutions: Electric Current- 3 Notes | EduRev    (1)

Now, for any one of the loop, consisting of x cells and the resistor R, from loop rule

Irodov Solutions: Electric Current- 3 Notes | EduRev

Irodov Solutions: Electric Current- 3 Notes | EduRev

Irodov Solutions: Electric Current- 3 Notes | EduRev

Heat generated in the resistor R,

Irodov Solutions: Electric Current- 3 Notes | EduRev    (2)

and for (2 to be maximum,   Irodov Solutions: Electric Current- 3 Notes | EduRev  which yields

Irodov Solutions: Electric Current- 3 Notes | EduRev


Q. 199. A capacitor of capacitance C = 5.00 µF is connected to a source of constant emf Irodov Solutions: Electric Current- 3 Notes | EduRev(Fig. 3.55). Then the switch Sw was thrown over from contact 1 to contact 2. Find the amount of heat generated in a resistance R1 = 500 Ω if R2 = 330 Ω.

Irodov Solutions: Electric Current- 3 Notes | EduRev

Solution. 199. When switch 1 is closed, maximum chaige accumulated on the capacitor, 

Irodov Solutions: Electric Current- 3 Notes | EduRev   (1)

Irodov Solutions: Electric Current- 3 Notes | EduRev

and when switch 2 is closed, at any arbitrary instant of time,

Irodov Solutions: Electric Current- 3 Notes | EduRev

because capacitor is discharging. 

Irodov Solutions: Electric Current- 3 Notes | EduRev

Integrating, we get

Irodov Solutions: Electric Current- 3 Notes | EduRev   (2)

Differentiating with respect to time,

Irodov Solutions: Electric Current- 3 Notes | EduRev

or, Irodov Solutions: Electric Current- 3 Notes | EduRev

Negative sign is ignored, as we are not interested in the direction of the current.

thus.  Irodov Solutions: Electric Current- 3 Notes | EduRev   (3)

When the switch (Sw) is at the position 1, the charge (maximum) accumalated on the capacitor is,

Irodov Solutions: Electric Current- 3 Notes | EduRev

When the Sw is thrown to position 2, the capacitor starts discharging and as a result the electric energy stored in the capacitor totally turns into heat energy tho’ the resistors R1 and R2 (during a very long interval of time). Thus from the energy conservation, the total heat liberated tho the resistors.

Irodov Solutions: Electric Current- 3 Notes | EduRev

During Hie process of discharging of the capacitor, the current tho’ the resistors and R2 is the same at all the moments of time, thus

Irodov Solutions: Electric Current- 3 Notes | EduRev

So,  Irodov Solutions: Electric Current- 3 Notes | EduRev

Hence  Irodov Solutions: Electric Current- 3 Notes | EduRev


Q. 200. Between the plates of a parallel-plate capacitor there is a metallic plate whose thickness takes η = 0.60 of the capacitor gap. When that plate is absent the capacitor has a capacity C = 20 nF. The capacitor is connected to a de voltage source V = 100 V. The metallic plate is slowly extracted from the gap. Find:
 (a) the energy increment of the capacitor;
 (b) the mechanical work performed in the process of plate extraction.

Solution. 200. When the plate is absent the capacity of the condenser is

Irodov Solutions: Electric Current- 3 Notes | EduRev

When it is present, the capacity is

Irodov Solutions: Electric Current- 3 Notes | EduRev

(a) The energy increment is clearly.

Irodov Solutions: Electric Current- 3 Notes | EduRev

(b) The charge on the plate is

Irodov Solutions: Electric Current- 3 Notes | EduRev

A charge  Irodov Solutions: Electric Current- 3 Notes | EduRev has flown through the battery charging it and withdrawing Irodov Solutions: Electric Current- 3 Notes | EduRev units of energy from the system into the battery. The energy of the capacitor has decreased by just half of this. The remaining half i.e.  Irodov Solutions: Electric Current- 3 Notes | EduRev must be the work done by the external agent in withdrawing the plate. This ensures conservation of energy.

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