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**MULTIPLE CHOICE QUESTIONS - I**

**Q.1. Consider a current carrying wire (current I) in the shape of a circle. Note that as the current progresses along the wire, the direction of j (current density) changes in an exact manner, while the current I remain unaffected. The agent that is essentially responsible for is(a) Source of emf.(b) Electric field produced by charges accumulated on the surface of wire.(c) The charges just behind a given segment of wire which push them just the right way by repulsion.(d) The charges ahead.**

Current density (J) depends on conductivity Electric field (J = σE), current and length and area of cross-section. In our options only E i.e., electric field can be varied by the charges accumulated on the surface of wire.

(a) The equivalent emf ε

(b) The equivalent emf ε

(c) The ε

(d) ε

We know that equivalent emf ε

Clearly, part ‘c’ and ‘d’ are discarded by formula. This formula suggests that ε

(a) He should measure

(b) He should change S to 1000Ω and repeat the experiment.

(c) He should change S to 3Ω and repeat the experiment.

(d) He should give up hope of a more accurate measurement with a meter bridge.

Condition of balanced wheatstone bridge: The bridge is said to be balanced if the ratio of the resistances in same branch is equal R/S =l

Wheatstone bridge is an arrangement of four resistances which can be used to measure one unknown resistance of them in terms of rest.

The percentage error in R can be minimised by adjusting the balance point near the middle of the bridge, i.e., when /, is close to 50 cm. This requires a suitable choice of S.

Since , R/S =l

Since here, R : S = 2.9 : 97.1

then the value of S is nearly 33 times to that of R. In order to make this ratio 1:1, it is necessary to reduce the value of S nearly 1/33 times, i.e., nearly 3 Ω.

(a) The battery that runs the potentiometer should have voltage of 8V.

(b) The battery of potentiometer can have a voltage of 15V and R adjusted so that the potential drop across the wire slightly exceeds 10V.

(c) The first portion of 50 cm of wire itself should have a potential drop of 10V.

(d) Potentiometer is usually used for comparing resistances and not voltages.

In a potentiometer experiment, the emf of a cell can be measured if the potential drop along the potentiometer wire is more than the emf of the cell to be determined. Here,

values of emfs of two cells are given as 5 V and 10 V, therefore, the potential drop along

the potentiometer wire must be more than 10 V.

(a) Maximum when the battery is connected across 1 cm x 1/2 cm faces.

(b) Maximum when the battery is connected across 10 cm x 1 cm faces.

(c) Maximum when the battery is connected across 10 cm x 1/2 cm faces.

(d) Same irrespective of the three faces.

As we know The maximum resistance will be when the value of 1/A is maximum, i.e., ‘A’ must be minimum, it is minimum when area of cross section is 1 cm x 1/2 cm. Verifies option (a).

As I Anev

Although I also depends on n, the number of free electrons which increases on increasing temperature which makes more collision between electrons increases resistance or decrease current. So ans. ‘a’ verified.

**MULTIPLE CHOICE QUESTIONS - II**

**Q.7. Kirchhoff’s junction rule is a reflection of****(a) Conservation of current density vector.****(b) Conservation of charge.****(c) The fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction.****(d) The fact that there is no accumulation of charges at a junction.****Ans. **(b, d)**Solution.****Key concept: **Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction.

Or

Algebraic sum of the currents flowing towards any point in an electric network is zero, i.e., charges are conserved in an electric network.

The proof of this rule follows from the fact that when currents are steady, there is no accumulation of charges at any junction or at any point in a line. Thus, the total current flowing in, (which is the rate at which charge flows into the junction), must equal the total current flowing out.

Kirchhoffs junction rule is also known as Kirchhoff’s current law.

So, Kirchhoffs junction rule is the reflection of conservation of charge**Important point:** Sign convention of current from a junction: We are taking outgoing current from a junction as negative. And we are taking incoming current towards a junction as positive.**Q.8. Consider a simple circuit shown in Fig 3.2.stands for a variable resistance R ′. R ′ can vary from R _{0} to infinity. r is internal resistance of the battery (r<<R<<R_{0}).**

As r << R << R

(b) Time interval between two successive collisions can depend on T.

(c) Length of material can be a function of T.

(d) Mass of carriers is a function of T.

We know that resistivity (ρ) depends on mass of charge-carrier (m), relaxation time (τ). Length and mass cannot be function of T as the mass of a body is constant everywhere. So discards answer (d) and length of body changes negligibly with temperature discards answer (c).

As τ decreases on increasing T due to rise in speed of change-carriers and n increases on increasing temperature. So will affect the ρ or ρ is function of T verifies answers (a) and (b)

(c) If the student uses large values of R

(d) Wheatstone bridge is a very accurate instrument and has no errors of measurement.

As the ratio of R

When R

(b) When the jockey contacts a point on meter wire left of D, current flows to B from the wire.

(c) When the jockey contacts a point on the meter wire to the right of D, current flows from B to the wire through galvanometer.

(d) When R is increased, the neutral point shifts to left.

**VERY SHORT ANSWER TYPE QUESTIONS**

**Q.12. Is the momentum conserved when charge crosses a junction in an electric circuit? Why or why not?****Ans.** In the circuit when an electron approaches a junction, in addition to the uniform E that faces it normally (which keep the drift velocity fixed), as drift velocity (v_{d}) is directly proportional to Electric field (E). That’s why there are accumulation of charges on the surface of wires at the junction.

These produce additional electric fields. These fields alter the direction of momentum. Thus, the motion of a charge across junction is not momentum conserving.**Q.13. The relaxation time τ is nearly independent of applied E field whereas it changes significantly with temperature T. First fact is (in part) responsible for Ohm’s law whereas the second fact leads to variation of ρ with temperature. Elaborate why?****Ans. **As the drift velocity increases, the relaxation time (τ) (average time between successive collision) decreases which increases the ρ by formula: p =

The drift velocity (v_{d}) changes of the order of one mm on increasing electric field, whereas

the drift velocity increases of the order of 10^{2}m/s when the number of free electrons (n)

increases on increasing temperature (T). So, due to increase in v_{d} the relaxation time (τ) considerably decreases in metal or conductor.**Q.14. What are the advantages of the null-point method in a Wheat stone bridge? What additional measurements would be required to calculate R _{unknown} by any other method?**

The R unknown can calculated applying Kirchhoff’s rules to the circuit. We would need additional accurate measurement of all the currents in resistances and galvanometer and internal resistance of the galvanometer.

where P and Q are ratio arms and R is known resistance and S is unknown resistance.

This measurements is done with the help of a centimetre scale or metre scale and leads to the accurate measurements.

The Cu wires or Al wires are used for wiring in the home. The main considerations are involved in this process are cost of metal and good conductivity of metal.

This keeps the resistance of the wire almost constant even in small temperature change. The alloys also have high resistivity and hence high resistance, because for given length and cross-section area of conductor (L and A are constant).

R α p

(i) If a constant power p is transmitted at low voltage (V) and high current (I). In this method high current will produce higher heat by H = I

(ii) If a constant or same power be transmitted at high voltage (V) and low current. It gives lower loss of power as heat. But need thicker insulation during transmission.

So to transmit high power at long distance, we use low current and high (132 kV) voltage to minimize heat losses through towers and thicket (long) insulator.

To transmit power supply at short distance, we can transmit power at low 440V, 220V, 11KV with higher current.

Since, at neutral point, for given emf of cell, l increases as potential gradient (k) across AB has decreased because E’ = kl

Thus, with the increase of l, which will result in increase in balance length. So, jockey J will shift towards B.

(i) Which terminal +or –ve of the cell E

(ii) Which terminal of the cell E

And clearly this is possible only when positive terminal of the cell E

(ii) If the current in auxiliary circuit increases, and potential difference across A and jockey J increases. Then also deflection in galvanometer is one sided.

And this is possible only when negative terminal of the cell E

So ...(I)

...(II)

Here E, r are constants. So

(from II)

and V ∝ R (from I)

With increase in R, P.D. across R is increased upto maximum value E.

**SHORT ANSWER TYPE QUESTIONS**

**Q.22. First a set of n equal resistors of R each are connected in series to a battery of emf E and internal resistance R. A current I is observed to flow. Then the n resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is ‘n’?****Ans. **When n resistance of each R Ω are connected in series and parallel then

R_{S} = R+R+R+.... n times ⇒ R_{S} = nR

When n resistors are connected in series connected with battery of emf E then current (I) flows. So

...(I)

And now n resistance are connected in parallel combination then current in circuit increased to 10 times of I

[Multiply by –n to both sides]

n^{2 }- 10 - 9n = 0

n^{2 }- 9n - 10 = 0

n^{2 }- 10n + 1n - 10 = 0

n(n-10) + 1(n-10) = 0

So, (n+1) (n-10) = 0

Or n = -1 is not possible or n = 10.

So, there are 10 resistors in combination.**Q.23. Let there be n resistors R _{1} ............R_{n} with R_{max} = max (R_{1}......... R_{n}) and R_{min} = min {R_{1} ..... R_{n}}. Show that when they are connected in parallel, the resultant resistance R_{P}< R_{min} and when they are connected in series, the resultant resistance R_{S} > R_{max}. Interpret the result physically.**

When resistors are connected in parallel then equivalent resistance

R

Multiplying both sides by R

Among R

Or

Or

So in parallel combination, the equivalent resistance R

When n resistance are connected in series then equivalent resistance-Rs=R

Here, in R.H.S. there must be a term R

R

Or R

So in series combination equivalent resistance is always greater than the maximum resistance (R

Current (I) in circuit,

For positive potential A is near to positive terminal of E

So potential between A and B = E

As current is flowing from B to A. So potential at B is larger than A.

The net potential difference across 1st cell V

Or

Or 2r

It is the required condition for the potential difference across 1st cell to be zero.

⇒

⇒

⇒

⇒

∴ R

Let the equivalent resistance, emf and internal resistance of above combination is R

Now the resistance and cells are again connected in a manner that their equivalent resistance, emf and internal resistance are nR

So the current remains same if the R, E and r of a circuits is increased by n times, i.e. nR, nE, nr.

**LONG ANSWER TYPE QUESTIONS**

**Q.28. Two cells of voltage 10V and 2V and internal resistances 10Ω and 5Ω respectively, are connected in parallel with the positive end of 10V battery connected to negative pole of 2V battery (Fig 3.8). Find the effective voltage and effective resistance of the combination.****Ans. **Applying junction rule at A*I*_{1 }= *I* + *I*_{2} .................(i)

Apply Kirchhoff’s loop rule on loop BCEF and loop ADEF

10 = *I*R + 10*I*_{1} ...............(2)

2 = 5*I*_{2} — *I*R............(3)

⇒ 2 = 5(*I*_{1} — 7) — *I*R [using (1)]............(4)

multiply (4) by 2, we get

4 = 10*I*_{1} - 10*I* - 2*I* R ..............(5)

Subtract (5) from (2), we get

⇒ 6 = 3*I*R + 10*I*

...........(6)

Let the effective potential difference due to both batteries is V_{eq}. It will be be across resistance R. So

V_{eq}=I(R+R_{eq}) ...........(7)

Where R_{eq} is the resistance of circuit except R

Comparing (6) and (7)

V_{eq} = 2 Volts and R_{eq }=**Q.29. A room has AC run for 5 hours a day at a voltage of 220V. The wiring of the room consists of Cu of 1 mm radius and a length of 10 m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions?[ρ _{cu} = 1.7 × 10_{Ωm}^{–8} , ρ_{Al} = 2.7 × 10^{–8} Ωm]**

∴ Energy consumed in 1 hr by AC and wiring =2 kWh

So total power of AC and wire = 2000 W

P=VI

Let P

P0= I

So loss of energy in wiring ≅ 4.4J/sec

The fractional loss due to heating of wires

as l

So power loss in Al wiring =7 Watt

The fractional loss due to Al wiring =

**Q.30. In an experiment with a potentiometer, V _{B} = 10V. R is adjusted to be 50Ω (Fig. 3.9). A student wanting to measure voltage E_{1} of a battery (approx. 8V) finds no null point possible. He then diminishes R to 10Ω and is able to locate the null point on the last (4^{th}) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.**

∴ Variable resistance, R=50 Ω

I is the current in primary circuit which is at E

=I (in primary circuit) …I

Potential difference across the wire of potentiometer

V’=IR’

From I

As with R=50 Ω resistance, null point cannot be obtained by 8 Volt.

So, V’< 8 Volt.

(No balance point)

As 50+R’ is positive so we can multiply above equation by positive number and we get 10R’<400+8R’

2R’ < 400

R’ < 200 ...III

Similarly, null point obtained by R=10 Ω . Then V”>8 at balance point So it is possible when

(as from I, R=10)

Similarly, multiply above equation by positive number 10+R’ to both sides

10R’ > 80 + 8R

2R’ > 80

R’ > 40 ...IV

As the null point is obtained on 4th segment or at 3/4 of total length So at 3/4 R’ (No balance point)

Or (At balance point)

So, 7.5R’ < 80+8R

-0.5R’< 80

-R’< 160

R’ > -160

R’ can never be negative so, -160 Ω is considered 160 Ω

So...V

Any R’ between 160 Ω and 200 Ω will achieve null point. Since the null point is on last 4th segment of potentiometer wire, sothe potential drop across 400cm wire > 8 Volt.

So, k.400cm> 8V (At balance point)

K > 2 Volt/m

As balance point is at 4

K.3<8 (No balance point)

So,

R = 6Ω

n = 10

V = 6V

e = 1.6 x 10

me = 9.1 x 10

L = 10cm = 10

⇒

per electron

Number of electrons (free) in wire = n (volume of wire)

= n x Al

∴ KE of all electrons

=1.78×10

So, to start flow of current I, the electrons will take energy from cell

= KE of all electrons= 1.78 x 10

P= 1 x 1 x 6=6 Joule per second

∵ Energy = P.t

Or

≌ 0.29 x 10

= 3 x 10

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