Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

JEE: Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

The document Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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Q. 166. The gap between the plates of a parallel-plate capacitor is filled up with two dielectric layers 1 and 2 with thicknesses d1 and d2, permittivities ε1 and ε2, and resistivities p1 and p2. A de voltage V is applied to the capacitor, with electric field directed from layer 1 to layer 2. Find σ, the surface density of extraneous charges at the boundary between the dielectric layers, and the condition under which σ = 0. 

Solution. 166. Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


Q. 167. An inhomogeneous poorly conducting medium fills up the space between plates 1 and 2 of a parallel-plate capacitor. Its permittivity and resistivity vary from values ε1, p1 at plate 1 to values ε2, p2  at plate 2. A de voltage is applied to the capacitor through which a steady current I flows from plate 1 to plate 2. Find the total extraneous charge in the given medium. 

Solution. 167.  By current conservation

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
This has the solution,

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Hence charge induced in the slice per unit area

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEEIrodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Thus,  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Hence total charge induced, is by integration,

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


Q. 168. The space between the plates of a parallel-plate capacitor is filled up with inhomogeneous poorly conducting medium whose resistivity varies linearly in the direction perpendicular to the plates. The ratio of the maximum value of resistivity to the minimum one is equal to η The gap width equals d. Find the volume density of the charge in the gap if a voltage V is applied to the capacitor. ε is assumed to be 1everywhere.

Solution. 168. As in the previous problem

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
where  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
By integration     Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Thus Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Thus volume density of charge present in the medium

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


Q. 169. A long round conductor of cross-sectional area S is made of material whose resistivity depends only on a distance r from the axis of the conductor as p = α/r2, where a is α constant. Find:
 (a) the resistance per unit length of such a conductor;
 (b) the electric field strength in the conductor due to which a current I flows through it.

Solution. 169. (a) Consider a cylinder of unit length and divide it into shells of radius r and thickness dr Different sections are in parallel. For a typical section,

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
Integrating, Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

or,  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

(b) Suppose the electric filed inside is  Ez = E0  (Z axis is along the axiz of the conductor). This electric field cannot depend on r in steady conditions when other components of E are absent, otherwise one violates the circulation theorem

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

The current through a section between radii (r + dr, r) is 

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
Thus  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Hence Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


Q. 170. A capacitor with capacitance C = 400 pF is connected via a resistance R = 650 Ω to a source of constant voltage V0. How soon will the voltage developed across the capacitor reach a value V = 0.90 V0

Solution. 170. The formula is,  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
or,  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
or,   Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


Q. 171. A capacitor filled with dielectric of permittivity e = 2.1 loses half the charge acquired during a time interval τ = 3.0 min. Assuming the charge to leak only through the dielectric filler, calculate its resistivity.

Solution. 171. The charge decays according to the foumula

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Here, RC = mean life = Half-life/ln 2
So, half life = T = R C In 2

But,  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Hence,   Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


Q. 172. A circuit consists of a source of a constant Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE and a resist ante R and a capacitor with capacitance C connected in series. The internal resistance of the source is negligible. At a moment t = 0 the capacitance of the capacitor is abruptly decreased η-fold. Find the current flowing through the circuit as a function of time t. 

Solution. 172. Suppose q is the charge at time t. Initially  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Then at time t,

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE(- sign because charge decreases) 

so   Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

or,  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

or,  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Hence,  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
Finally,  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


Q. 173. An ammeter and a voltmeter are connected in series to a battery with an emf Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE When a certain resistance is connected in parallel with the voltmeter, the readings of the latter decrease η = 2.0 times, whereas the readings of the ammeter increase the same number of times. Find the voltmeter readings after the connection of the resistance. 

Solution. 173. Let r = internal resistance of the battery. We shall take the resistance of the ammeter to be = 0 and that o f voltmeter to be G.

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

So,   Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE   (1)

After the voltmeter is shunted

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE     (2)

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE  (3)

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

From (2) and (3) we have

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE    (4)

From (1) and (4)

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Then (1) gives the required reading

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


Q. 174. Find a potential difference φ1 — φ2  between points 1 and 2 of the circuit shown in Fig. 3.39 if R1 = 10 Ω, R2 = 20 Ω, Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE 5.0 V, and Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE V. The internal resist- ances of the current sources are negligible.

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Solution. 174. Assume the current flow, as shown. Then potentials are as shown. Thus,

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


Q. 175. Two sources of current of equal emf are connected in series and have different internal resistances R1 and R2. (R2 > R1). Find the external resistance R at which the potential difference across the terminals of one of the sources (which one in particular?) becomes equal to zero.

Solution. 175. Let, us consider the current i, flowing through the circuit, as shown in the figure. Applying loop rule for the circuit, - Δ φ = 0

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

or,  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

So,  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE which is the required resistance.


Q. 176. N sources of current with different emf's are connected as shown in Fig. 3.40. The emf's of the sources are proportional to their internal resistances, i.e. Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE where a is an assigned constant. The lead wire resistance is negligible. Find:
 (a) the current in the circuit;
 (b) the potential difference between points A and B dividing the circuit in n and N — n links. 

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Solution. 176. (a) Current,  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


Q. 177. In the circuit shown in Fig. 3.41 the sources have emf's Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE V and Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE and the resistances have the values R1 = 10 Ω and R2 = 20 Ω. The internal resistances of the sources are negligible. Find a potential difference φA  — φB  between the plates A and B of the capacitor C.

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE 

Solution. 177. As the capacitor is fully charged, no current flows through it So, current

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


Q. 178. In the circuit shown in Fig. 3.42 the emf of the source is equal to Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE = 5.0 V and the resistances are equal to R1 = 4.0 Ω and R2  = 6.0 Ω. The internal resistance of the source equals R = 0.10 Ω. Find the currents flowing through the resistances R1 and R2

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Solution. 178. Let us make the current distribution, as shown in the figure.

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

So, current through the resistor R1,

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

and similary, current through the resistor R2,

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


Q. 179. Fig. 3.43 illustrates a potentiometric circuit by means of which we can vary a voltage V applied to a certain device possessing a resistance R. The potentiometer has a length l and a resistance R0, and voltage V0 is applied to its terminals. Find the voltage V fed to the device as a function of distance x. Analyse separately the case R ≫ R0

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Solution. 179. 

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEEIrodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

For  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


Q. 180. Find the emf and the internal resistance of a source which is equivalent to two batteries connected in parallel whose emf's are equal to Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE  and internal resistances to R1 and R2

Solution. 180. Let us connect a load of resistance K between the points A and B (Fig.)

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

From the loop rule, Δ φ = 0, we obtain

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Thus one can replace the given arrangement of the cells by a single cell having the emf  Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE and internal resistance R0.


Q. 181. Find the magnitude and direction of the current flowing through the resistance R in the circuit shown in Fig. 3.44 if the emf's of the sources are equal to Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEEand the resistances are equal to R1 =10 Ω, R2 = 20 Ω, R = 5.0 Ω. The internal resistances of the sources are negligible. 

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Solution. 181. Make the current distribution, as shown in the diagram

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Now, in the loop 12341, applying - Δφ = 0

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE  (1)

and in the loop 23562,

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE  (2)

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

and it is directed from left to the right


Q. 182. In the circuit shown in Fig. 3.45 the sources have emf's Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE and the resistances are equal to R1 = 10 Ω, R2 = 20 Ω, R3 = 30 Ω. The internal resistances of the sources are negligible. Find:

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

(a) the current flowing through the resistance R1;
 (b) a potential difference φA  — φB between the points A and B.

Solution. 182. At first indicate the currents in the branches using charge conservation (which also includes the point rule).

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
In the loops 1 BA 61 and B34AB from the loop rule, - Δ φ = 0, we get, respectively

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


Q. 183. Find the current flowing through the resistance R in the circuit shown in Fig. 3.46. The internal resistances of the batteries are negligible.

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Solution. 183. Indicate the currents in all the branches using charge conservation as shown in the figure. Applying loop rule, - Δφ = 0 in the loops 1A781, 1B681 and B456B, respectively, we get

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

The document Irodov Solutions: Electric Current - 2 Notes | Study I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
All you need of JEE at this link: JEE

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