Airforce X Y / Indian Navy SSR Exam  >  Airforce X Y / Indian Navy SSR Notes  >  Irodov Solutions: Electric Current - 2

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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 - Airforce X Y / Indian Navy SSR

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR


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 - Airforce X Y / Indian Navy SSR
Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR
This has the solution,

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Hence charge induced in the slice per unit area

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSRIrodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Thus,  Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Hence total charge induced, is by integration,

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR


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 - Airforce X Y / Indian Navy SSR
where  Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR
By integration     Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Thus Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Thus volume density of charge present in the medium

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR
Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR


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 - Airforce X Y / Indian Navy SSR
Integrating, Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

or,  Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

(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 - Airforce X Y / Indian Navy SSR

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

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR
Thus  Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Hence Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR


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 - Airforce X Y / Indian Navy SSR
or,  Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR
or,   Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR


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 - Airforce X Y / Indian Navy SSR

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

But,  Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Hence,   Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR


Q. 172. A circuit consists of a source of a constant Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR 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 - Airforce X Y / Indian Navy SSR

Then at time t,

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR(- sign because charge decreases) 

so   Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR
Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

or,  Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

or,  Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Hence,  Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR
Finally,  Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR


Q. 173. An ammeter and a voltmeter are connected in series to a battery with an emf Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR 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 - Airforce X Y / Indian Navy SSR

So,   Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR   (1)

After the voltmeter is shunted

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR     (2)

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR  (3)

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

From (2) and (3) we have

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR    (4)

From (1) and (4)

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Then (1) gives the required reading

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR


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 - Airforce X Y / Indian Navy SSR 5.0 V, and Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR V. The internal resist- ances of the current sources are negligible.

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

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

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR
Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR


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 - Airforce X Y / Indian Navy SSR

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

or,  Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

So,  Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR 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 - Airforce X Y / Indian Navy SSR 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 - Airforce X Y / Indian Navy SSR

Solution. 176. (a) Current,  Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR


Q. 177. In the circuit shown in Fig. 3.41 the sources have emf's Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR V and Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR 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 - Airforce X Y / Indian Navy SSR 

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

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR


Q. 178. In the circuit shown in Fig. 3.42 the emf of the source is equal to Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR = 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 - Airforce X Y / Indian Navy SSR

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

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR
Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

So, current through the resistor R1,

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

and similary, current through the resistor R2,

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR


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 - Airforce X Y / Indian Navy SSR

Solution. 179. 

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR
Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR
Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR
Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSRIrodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

For  Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR


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 - Airforce X Y / Indian Navy SSR  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 - Airforce X Y / Indian Navy SSR

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

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR
Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

Thus one can replace the given arrangement of the cells by a single cell having the emf  Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR 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 - Airforce X Y / Indian Navy SSRand 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 - Airforce X Y / Indian Navy SSR

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

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

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

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR  (1)

and in the loop 23562,

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR  (2)

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

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 - Airforce X Y / Indian Navy SSR 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 - Airforce X Y / Indian Navy SSR

(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 - Airforce X Y / Indian Navy SSR
In the loops 1 BA 61 and B34AB from the loop rule, - Δ φ = 0, we get, respectively

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR


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 - Airforce X Y / Indian Navy SSR

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 - Airforce X Y / Indian Navy SSR

Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

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FAQs on Irodov Solutions: Electric Current - 2 - Airforce X Y / Indian Navy SSR

1. What is the formula to calculate electric current?
Ans. The formula to calculate electric current is I = Q/t, where I represents current, Q represents charge, and t represents time.
2. How is electric current measured?
Ans. Electric current is measured using an ammeter, which is connected in series to the circuit. The ammeter measures the flow of electric charge through a conductor.
3. What is the difference between AC and DC current?
Ans. AC (alternating current) and DC (direct current) are two types of electric currents. AC changes its direction periodically, while DC flows in only one direction. AC is commonly used in households and businesses, while DC is used in batteries and electronic devices.
4. How does resistance affect electric current?
Ans. Resistance restricts the flow of electric current in a circuit. According to Ohm's Law, the current flowing through a conductor is inversely proportional to the resistance. This means that as resistance increases, the current decreases.
5. What is the role of a conductor in an electric current?
Ans. Conductors, such as metals, allow electric current to flow through them easily. They have a low resistance, which means they offer little opposition to the flow of electrons. Conductors are essential in completing an electric circuit and enabling the flow of current.
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