Ncert Exemplar Problems with solutions Chapter 3 Current Electricity

Multiple Choice Questions (MCQs)

Q. 1 Consider a current carrying wire (current I) in the shape of a circle.

(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

Ans.  (b) Current per unit area (taken normal to the current), I/A, is called current density and is denoted by j. The SI units of the current density are A/m². The current density is also directed along E and is also a vector and the relationship is given by

J – sE

The j changes due to electric field produced by charges accumulated on the surface of wire.

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 this.

Q. 2 Two batteries of emf ε₁ and ε₂(ε₂ > ε₁) and internal resistances r₁ and r₂ respectively are connected in parallel as shown in figure.

(a) Two equivalent emf ε_eq of the two cells is between ε₁ and ε₂, i.e., ε₁ < ε_eq < ε₂

(b) The equivalent emf ε_eq is smaller than ε₁

(c) The ε_eq is given by ε_eq = ε₁ + ε₂ always

(d) ε_eq is independent of internal resistances r₁ and r₂

Ans.  (a) The equivalent emf of this combination is given by

ε_eq = 

This suggest that the equivalent emf ε_eq of the two cells is given by

ε₁ < ε_eq < ε₂

Q. 3 A resistance R is to be measured using a meter bridge, student chooses the standard resistance S to be 100Ω. He finds the null point at I₁ = 2.9 cm. He is told to attempt to improve the accuracy.

Which of the following is a useful way?

(a) He should measure I₁ more accurately

(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 given up hope of a more accurate measurement with a meter bridge

Ans.  (c) The percentage error in R can be minimised by adjusting the balance point near the middle of the bridge, i.e., when I₁ is close to 50 cm. This requires a suitable choice of S.

Since,   

Since here, ft :S :: 2.9: 97.1 imply that the S is nearly 33 times to that of ft. In orded to make this ratio 1:1. it is necessary to reduce the value of S nearly  times i.e., nearly 3Ω.

Q. 4 Two cells of emfs approximately 5 V and 10 V are to be accurately compared using a potentiometer of length 400 cm.

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

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

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

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

Ans.  (b) 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 5V and 10V, therefore, the potential drop along the potentiometer wire must be more than 10V.

Q. 5 A metal rod of length 10 cm and a rectangular cross–section of 1cm x  cm is connected to a battery across opposite faces. The resistance will be

(a) maximum when the battery is connected across 1 cm x  cm faces

(b) maximum when the battery is connected across 10 cm x  cm faces

(c) maximum when the battery is connected across 10 cm x  cm faces

(d) same irrespective of the three faces

Ans.  (a) The resistance of wire is given by

R = 

For greater value of R, l must be higher and A should be lower and it is possible only when the battery is connected across 1 cm x  cm (area of cross–section A).

Q. 6 Which of the following characteristics of electrons determines the current in a conductor?

(a) Drift velocity alone

(b) Thermal velocity alone

(c) Both drift velocity and thermal velocity

(d) Neither drift nor thermal velocity

Ans.  (a) The relationship between current and drift speed is given by

I = ne Av_d

Here, I is the current and v_d is the drift velocity.

So,  Iv_d

Thus, only drift velocity determines the current in a conductor.

Multiply Choice Questions (More Than One Options)

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)

Kirchhoff’s junction rule is also known as Kirchhoff’s current law which states that the algebraic sum of the currents flowing towards any point in an electric network is zero, i.e., charges are conserved in an electric network.

So, Kirchhoff’s junction rule is the reflection of conservation of charge

Q. 8 Consider a simple circuit shown in figure stands for a variable resistance R’./R’ can vary from R₀ to infinity, r is internal resistance of the battery (r<<R<<R,).

(a) Potential drop across AB is nearly constant as R’ is varied

(b) Current through R’ is nearly a constant as R’ is varied

(c) Current /depends sensitively on R’

(d) always

Ans.  (a, d)

Here, the potential drop is taking place across AB and r. Since the equivalent resistance of parallel combination of R and R is always less than R, therefore I ≥ always.

Note In parallel combination of resistances, the equivalent resistance is smaller than smallest resistance present in combination.

Q. 9 Temperature dependence of resistivity p(T) of semiconductors, insulators and metals is significantly based on the following factors

(a) number of charge carriers can change with temperature T

(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

Ans.  (a, b)

The resistivity of a metallic conductor is given by,

e = 

where n is number of charge carriers per unit volume which can change with temperature T and  is time interval between two successive collisions which decreases with the increase of temperature.

Q. 10 The measurement of an unknown resistance R is to be carried out using Wheatstones bridge as given in the figure below. Two students perform an experiment in two ways. The first students takes R₂ = 10Ω and R₁ = 5Ω. The other student takes R₂ = 1000Ω and R₁ = 500Ω. In the standard arm, both take R₃ = 5Ω.

Both find R =  ,R3 = 10Ω within errors.

(a) The errors of measurement of the two students are the same

(b) Errors of measurement do depend on the accuracy with which R₂ and R₁ can be measured

(c) If the student uses large values of R₂ and R₁ the currents through the arms will be feeble. This will make determination of null point accurately more difficult

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

Ans.  (b, c)

Given, for first student, R₂ = 10 Ω, R₁ = 5 Ω, R₃ = 5Ω

For second student, R₁ = 500 Ω2, R₃ = 5 Ω

Now, according to Wheatstone bridge rule,

Now putting all the values in Eq. (i), we get R = 10Ω for both students. Thus, we can analyse that the Wheatstone bridge is most sensitive and accurate if resistances are of same value.

Thus, the errors of measurement of the two students depend on the accuracy and sensitivity of the bridge, which inturn depends on the accuracy with which R₂ and R₁ can be measured.

When R₂ and R₁ are larger, the currents through the arms of bridge is very weak. This can make the determination of null point accurately more difficult.

Q. 11 In a meter bridge, the point D is a neutral point (figure).

(a) The meter bridge can have no other neutral. A point for this set of resistances

(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

Ans.  (a, c)

At neutral point, potential at 8 and neutral point are same. When jockey is placed at to the right of D, the potential drop across AD is more than potential drop across AB, which brings the potential of point D less than that of B, hence current flows from B to D.

Very Short Answer Type Questions

Q. 12 Is the motion of a charge across junction momentum conserving? Why or why not?

Ans.  When an electron approaches a junction, in addition to the uniform electric field E facing it normally. It keep the drift velocity fixed as drift velocity depend on £ by the relation drift velocity

vd = 

This result into accumulation of charges on the surface of wires at the junction. These produce additional electric field. These fields change the direction of momentum.

Thus, the motion of a charge across junction is not momentum conserving.

Q. 13 The relaxation time x 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 p with temperature. Elaborate why?

Ans.  Relaxation time is inversely proportional to the velocities of electrons and ions. The applied electric field produces the insignificant change in velocities of electrons at the order of 1 mm/s, whereas the change in temperature (T), affects velocities at the order of 10²/m/s. This decreases the relaxation time considerably in metals and consequently resistivity of metal or conductor increases as .

Q. 14 What are the advantages of the null–point method in a Wheatstone bridge? What additional measurements would be required to calculate R_unknown by any other method?

Ans.  The advantage of null point method in a Wheatstone bridge is that the resistance of galvanometer does not affect the balance point, there is no need to determine current in resistances and the internal resistance of a galvanometer.

It is easy and convenient method for observer.

The R_unknown can be 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.

Note The necessary and sufficient condition for balanced Wheatstone bridge is

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

Q. 15 What is the advantage of using thick metallic strips to join wires in a potentiometer?

Ans.  In potentiometer, the thick metallic strips are used as they have negligible resistance and need not to be counted in the length l₁ of the null point of potentiometer. It is for the convenience of experimenter as he measures only their lengths along the straight wires each of lengths 1 m.

This measurements is done with the help of centimetre scale or metre scale with accuracy.

Q. 16 For wiring in the home, one uses Cu wires or Al wires. What considerations are involved in this?

Ans.  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.

Q. 17 Why are alloys used for making standard resistance coils?

Ans.  Alloys have small value of temperature coefficient of resistance with less temperature sensitivity.

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

R p

Q. 18 Power P is to be delivered to a device via transmission cables having resistance R_c. If V is the voltage across R and I the current through it, find the power wasted and how can it be reduced.

Ans.  The power consumption in transmission lines is given by P = i² R_c , where R_c is the resistance of transmission lines. The power is given by

P = VI

The given power can be transmitted in two ways namely (i) at low voltage and high current or (ii) high voltage and low current. In power transmission at low voltage and high current more power is wasted as P ∞ i² whereas power transmission at high voltage and low current facilitates the power transmission with minimal power wastage.

The power wastage can be reduced by transmitting power at high voltage.

Q. 19 AB is a potentiometer wire (figure). If the value of R is increased, in which direction will the balance point J shift?

Ans.  With the increase of R, the current in main circuit decreases which in turn , decreases the potential difference across AB and hence potential gradient(k) across AB decreases.

Since, at neutral point, for given emf of cell, I increases as potential gradient (k) across AS has decreased because

E = k I

Thus, with the increase of I, the balance point neutral point will shift towards B.

Q. 20 While doing an experiment with potentiometer (figure) it was found that the deflection is one sided and (i) the deflection decreased while moving from one and A of the wire, to the end R; (ii) the deflection increased, while the jockey was moved towards the end D.

(i) Which terminal positive or negative of the cell E₁ is connected at X in case (i) and how is E₁, related to E?

(ii) Which terminal of the cell E₁ is connected at X in case (1 in 1)?

Ans.  (i) The deflection in galvanometer is one sided and the deflection decreased, while moving from one end ‘A’ of the wire to the end ‘B’, thus imply that current in auxiliary circuit (lower circuit containing primary cell) decreases, while potential difference across A and jockey increases.

This is possible only when positive terminal of the cell E₁ is connected at X and E₁ >E.

(ii) The deflection in galvanometer is one sided and the deflection increased, while moving from one end A of the wire to the end 6, this imply that current in auxiliary circuit (lower circuit containing primary cell) increases, while potential difference across A and jockey increases.

This is possible only when negative terminal of the cell E₁, is connected at X.

Q. 21 A cell of emf E and internal resistance r is connected across an external resistance R. Plot a graph showing the variation of potential differential across R, versus R.

Ans.  The graphical relationship between voltage across R and the resistance R is given below

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.  In series combination of resistors, current I is given by I = 

whereas in parallel combination current 10 I is given by

Now, according to problem,

Q. 23  Let there be n resistors R₁…….R_n with R_max = max(R₁………R_n) and R_min = min{R₁ 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.

Ans.  When all resistances are connected in parallel, the resultant resistance R_p is given by

On multiplying both sides by R_min we have

Here, in RHS, there exist one term  = 1 and other terms are positive, so we have

This shows that the resultant resistance R_p < R_min.

Thus, in parallel combination, the equivalent resistance of resistors is less than the minimum resistance available in combination of resistors . Now, in series combination, the equivalent resistant is given by

R_s = R₁ + …………… + R_n

Here, in RHS, there exist one term having resistance R_max.

So, we have

or  R_s = R₁ + … + R_max+.. + ..+R_n

R_s = R₁ + … + R_max… + R_n = R_max + … (R₁ + …+)R_n

or  R_s ≥ R_max

R_s = R_max (R₁ + …….. + R_n)

Thus, in series combination, the equivalent resistance of resistors is greater than the maximum resistance available in combination of resistors. Physical interpretation

In Fig. (b), R_min provides an equivalent route as in Fig. (a) for current. But in addition there are (n–1) routes by the remaining (n –1) resistors. Current in Fig (b) is greater than current in Fig. (a). Effective resistance in Fig. (b) < R_mjn. Second circuit evidently affords a greater resistance.

(c)   (d)

In Fig. (d), R_max provides an equivalent route as in Fig. (c) for current. Current in Fig. (d) < current in Fig. (c). Effective resistance in Fig. (d) > R_max. Second circuit evidently affords a greater resistance.

Q. 24  The circuit in figure shows two cells connected in opposition to each other. Cell E₁ is of emf 6V and internal resistance 2Ω the cell E₂ is of emf 4V and internal resistance 8Ω. Find the potential difference between the points A and B.

Ans.  Applying Ohm’s law.

Effective resistance = 2Ω + 8Ω = 10Ω and effective emf of two cells = 6 – 4 = 2V, so the electric current is given by

along anti–clockwise direction, since E₁ > E₂.

The direction of flow of current is always from high potential to low potential. Therefore V_b> Va

  V_B–4V– (0.2) x 8 = V_A

Therefore,  V_B – V_A = 3.6V

Q. 25  Two cells of same emf E but internal resistance r₁ and r₂ are connected in series to an external resistor R (figure). What should be the value of R so that the potential difference across the terminals of the first cell becomes zero?

Ans.  Applying Ohm’s law,

Effective resistance = R + r₁ + r₂ and effective emf of two cells = E + E = 2E, so the electric current is given by

The potential difference across the terminals of the first cell and putting it equal to zero.

V₁ = E – Ir₁ = E –

or 

r₁ + r₂ + R = 2r₁  R = r₁– r₂

This is the required relation.

Q. 26  Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1mm. Conductor B is a hollow tube of outer diameter 2mm and inner diameter 1mm.

Find the ratio of resistance R_A to R_B.

Ans.  The resistance of first conductor

The resistance of second conductor,

Now, the ratio of two resistors is given by

 = 3 : 1

Q. 27  Suppose there is a circuit consisting of only resistances and batteries. Suppose one is to double (or increase it to n–times) all voltages and all resistances. Show that currents are unaltered. Do this for circuit of Examples 3,7 in the NCERT Text Book for Class XII.

Ans.  Let the effective internal resistance of the battery is R_eff the effective external resistance R and the effective voltage of the battery is V_eff.

Applying Ohm’s law,

Then current through R is given by

If all the resistances and the effective voltage are increased n–times, then we have

and  R^new = nR

Then, the new current is given by

Thus, current remains the same.

Long Answer Type Questions

Q. 28  Two cells of voltage 10V and 2V and 10Ω internal resistances 10Ω and 5Ω respectively, are connected in parallel with the positive end of 10V battery connected to negative pole of 2V battery (figure). Find the effective voltage and effective resistance of the combination.

Ans.  Applying Kirchhoff’s junction rule, I₁ = I + I₂

Applying Kirchhoff’s II law /loop rule applied in outer loop containing 10V cell and resistance ft, we have

10 = IR+ 10I₁   …(i)

Applying Kirchhoff II law/ loop rule applied in outer loop containing 2V cell and resistance ft, we have

2 = 5I₂– RI = 5(I₁ –I)–RI

or  4 = 10I1 – 10I – 2RI  …(ii)

Solving Eqs. (i) and (ii), gives

 6 = 3RI +10I

Also, the external resistance is ft . The Ohm’s law states that

V = I(R + R_eff)

On comparing, we have V = 2V and effective internal resistance

Since, the effective internal resistance (R_eff) of two cells is, being the parallel combination of 5Ω and 10Ω. The equivalent circuit is given below

Q. 29  A room has AC run for 5 a day at a voltage of 220V. The wiring of the room consists of Cu of 1 mm radius and a length of 10m. 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?

[p_Cu = 11.7 x 10–⁸ Ωm, p _Al = 2.7 x 10–⁸ Ωm]

Ans.  Power consumption in a day i.e., in 5 = 10 units

Or power consumption per hour = 2units

Or power consumption = 2units = 2kW = 2000J/s

Also, we know that power consumption in resistor,

P = V x l

 2000W = 220V x l or l = 9A

Now, the resistance of wire is given by 

where, A is cross–sectional area of conductor.

Power consumption in first current carrying wire is given by

P = l²R

J/s = 4 J/s

The fractional loss due to the joule heating in first wire =  x 100 = 02%

Power loss in Al wire = _l = 1.6x 4 = 64 J/s

The fractional loss due to the joule heating in second wire =  x 100 = 0.32%

Q. 30  In an experiment with a potentiometer, V_B = 10V. R is adjusted to be 50Ω (figure). A student wanting to measure voltage E₁ 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 (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

Ans.  Let R’ be the resistance of the potentiometer wire.

Effective resistance of potentiometer and variable resistor (R = 50Ω) is given by = 50Ω + R’ Effective voltage applied across potentiometer = 10V.

The current through the main circuit,

Potential difference across wire of potentiometer,

Since with 50 Ω resistor, null point is not obtained it’s possible only when

 10R'<400+8R’

2R‘ < 400 or R‘< 200Ω.

Similarly with 10Ω resistor, null point is obtained its possible only when

  2R’>80

 R’> 40

  7.5R'<80+8R’

R’>160

 160<R'<200.

Any R’ between 160Ω and 200 Ω will achieve.

Since, the null point on the last (4th) segment of the potentiometer, therefore potential drop across 400 cm of wire > 8V.

This imply that potential gradient

k x 400 cm > 8V

or  k x 4m >8V

k >2V/m

Similarly, potential drop across 300 cm wire < 8V.

k x 300cm < 8V

or  k x 3m < 8V

k<22/3V/m

Thus,  V/m> k > 2V / m

Q. 31 (a) Consider circuit in figure. How much energy is absorbed by electrons from the initial state of no current (Ignore thermal motion) to the state of drift velocity ?

(b) Electrons give up energy at the rate of RI² per second to the thermal energy. What time scale would number associate with energy in problem (a)? n = number of electron/volume = 10²⁹ /m³. Length of circuit = 10 cm cross–section = A = (1 mm)².

Ans.  (a) By Ohm’s law, current I is given by

I = 6V/6Ω = 1A

But,  I = net A v_d

or  

On substituting the values

For, n = number of electron/volume = 10²⁹/m³

length of circuit = 10cm, cross–section = A = (1mm)²

=  × 10–4 m/s

Therefore, the energy absorbed in the form of KE is given by

KE =  m_e v_d² x nAI

=  x 9.1 x 10³¹ x  x 10²⁰ x10⁸ x 10⁶ x 10¹

= 2 x 10^–17J

(b) Power loss is given by P = I² R = 6 x 1² = 6W = 6J/s

Since,  P =

Therefore,  E = P x t

or t =