# NCERT Solution : Current Electricity

Q.10. (a) In a meter bridge the balance point is found to be at 39.5 cm from the left end A, if an unknown resistor X is in the left gap and known resistor Y of resistance 12.5 Ω is in the right gap. Determine the resistance of X. Why are the connection between resistance in a Wheatstone or metre bridge  made of thick copper strips ?

(b) Determine the balance point of the above bridge if X and Y are interchanged ?

(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge ? would the galvanometer show any current ?

Sol.

Here,  l=39.5cm ; R = X = ? ; S = Y = 12.5 Ω

$\displaystyle \frac{R}{S} = \frac{l}{100-l}$

$\displaystyle R = \frac{l}{100-l}S$

$\displaystyle R = \frac{39.5}{100-39.5}(12.5)$

R  = 8.16 Ω

⇒ X = 8.16 Ω

Thick copper strips are used to minimise resistance of the connections which are not accounted in the formula

(b) As X and Y are interchanged therefore, l1 and l2 (i.e, lengths are also interchanged. Hence,

l = 100 – 39.5 = 60.5 cm.

(c) The position of balance point remains unchanged. The galvanometer will show no current.

Q.11.A storage battery of emf 8.0 V and internal resistance of 0.5 Ω is being charged by a 120 V d.c. supply using a series resistor of 15.5 Ω What is the terminal voltage of the battery during charging ? What is the purpose of having a series resistor in the charging circuit?

Sol.

Here, e.m.f of the battery = 8.0 V ;

Voltage of d.c. supply = 120 V

Internal resistance of battery, r = 0.5 Ω ;

External resistance, R = 15.5 Ω

Since a storage battery of e.m.f 8 V is changed with a d.c. supply of 120 V, the effective e.m.f in the circuit is given by

e = 120 – 8 =112 V

Total resistance of the circuit = R + r =15.5 + 0.5 =16 Ω

Current in the circuit during charging is given by. $\displaystyle I =\frac{e}{R+r}$

I = 112/16 = 7 A

Voltage across R = IR =7.0 ×15.5 = 108.5 V

During charging, the voltage of the battery of the d.c. supply in a circuit must be equal to the sum of the voltage drop across R and terminal voltage of the battery

120 −108.5 = 11.5 V

The series resistor limits the current drawn from the external source of d.c. supply In its absence the current will be dangerously high.

Q.12. In a potentiometer arrangement, a cell of emf 1.25 gives a balance point at 35.0 cm length of the wire, If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell ?

Sol. Here,        E1 = 1.25 V  , l1 = 35.0 cm , E2 = ?  , l2 = 63.0 cm

$\displaystyle \frac{E_2}{E_1} = \frac{l_2}{l_1}$

$\displaystyle E_2 = \frac{l_2}{l_1}E_1$

$\displaystyle E_2 = \frac{63}{35}\times 1.25$

= 2.25 V

Q.13.The number density of free electrons in a copper conductor is estimated at 8.5 × 1028 m-3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end ? The area of cross- section of the wire is 2.0 × 10-6 m2 and it is carrying a current of 3.0 A.

Sol.

Here,   n = 8.5 × 1028 m-3 ; l = 3.0 m ; A = 2.0 ×10-6 m2 ; I = 3.0 A , t = ?

As.   I = n Ae vd

$\displaystyle v_d = \frac{I}{n e A}$

Now, $\displaystyle t = \frac{l}{v_d} = \frac{l neA}{I}$

$\displaystyle t = \frac{3\times 8.5\times 10^28 \times 1.6 \times 10^{-19}\times 2 \times 10^{-6}}{3}$

= 2.72 ×104 sec

= 7 hours 33 minutes

Q.14. The earth’s surface has a negative surface charge density of 10-9 Cm-2 . The potential difference of 400 k V between the top of the atmosphere and the surface result (due to low conductivity of the lower atmosphere) in a current of only 1800A over the entire globe. If there were no neutralise the earth’s surface ? (This never happens in practice because there is a mechanism to replenish charges namely the continual thunder storms and lightning in different parts of the globe). Radius of the earth = 6.37 ×106 m

Sol.

Here,   r = 6.37 ×106 m   , σ = 10-9 Cm-2 , I = 1800 A

Area of the globe, A= 4 π r2

= 4 × 3.14 × ( 6.37 ×106 )2

= 509.64 × 1012 m2

As charge,  Q = σ × A = 10-9 × 509.64 × 1012

= 509.64 × 103 C

t = Q/I

= 509.64 × 103/1800

= 283.1 sec

Q:15 . (a) Six lead-acid type of secondary cells each of emf 2.0V and internal resistance 0.015 Ω  are joined in series to provide a supply to a resistance of 8.5 Ω . What are the current drawn from the supply and its terminal voltage ?

(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380Ω.What maximum current can be drawn from the cell ? Could the cell drive the starting motor of a car ?

Sol:

(a) Here, E= 2.0 V ; n = 6 ; r = 0.0015 Ω ; R = 8.5 Ω

$\displaystyle I = \frac{n E}{R + nr}$

$\displaystyle I = \frac{6\times 2}{8.5 + 6\times 0.0015}$

I = 1.4 A

(b) E = 1.9 V ; r = 380 Ω

$\displaystyle I_{max} = \frac{E}{r}$

$\displaystyle I_{max} = \frac{1.9}{380}$

Imax = 0.005 A

Q:16.Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter ? Hence explain why aluminium wires are preferred for overhead power cables, Given, For Al,  ρ1 = 2.63 × 10-8 Ω m,For Cu, ρ2 = 1.72 × 10-8 Ω m. Realtive density of Al = 2.7 , of Cu = 8.9

Sol: Given For aluminium wire ; R1 = R ; l1 = l ;Relative density d1 = 2.7

For copper wire, R2 = R ; l2 = l ; d2 = 8.9

Let A1 , A2 be the area of cross section for aluminium wire and copper wire.

We know, $\displaystyle R_1 = \rho_1 \frac{l_1}{A_1} = \frac{2.63\times 10^{-8} l}{A_1}$

and mass of the aluminium wire, m1 = A1 x l1 × d1= A1 x 1×2.7

Also, $\displaystyle R_2 = \rho_2 \frac{l_2}{A_2} = \frac{1.72\times 10^{-8} l}{A_2}$

Mass of copper wire, m2 = A2 x l2 × d2= A2 x 1× 8.9

Since two wires are of equal resistance R1 = R2

Hence , $l \displaystyle \frac{A_2}{A_1} = \frac{1.72}{2.63}$

We have , $\displaystyle \frac{m_2}{m_1} = 2.16$

Q:17 .Sol.
Since the ratio of voltage and current for different readings is same so Ohm’s law is valid to high accuracy. The resistivity of the alloy manganin is nearly independent of temperature.

Q:18. Answer the following question :
(a)A steady current flows in a metallic conductor of non-uniform cross-section. Explain which of these quantities is constant along the conductor : current, current density, electric field and drift speed?

(b)Is Ohm’s law universally applicable for all conducting elements ? If not, give examples of elements which do not obey Ohm’s law.

(c)A low voltage supply from which one needs high currents must have low internal resistance. Why ?

(d)a high tension (HT) supply of say 6 k V must have a very large internal resistance. Why ?

Sol.
(a)Only current through the conductor of non-uniform area of cross-section is constant as the remaining quantities vary inversely with the area of cross-section of the conductor.

(b)Ohm’s law is not applicable for non- ohmic elements. For example; vacuum tubes, semi-conducting diode, liquid electrolyte etc.

(c)As, Imax = e.m.f./ internal resistance, so for maximum current, internal resistance should be least.

(d)A high tension supply must have a large internal resistance otherwise, if accidently the circuit is shorted, the current drawn will exceed safety limit and will cause damage to circuit.

Q:19.Choose the correct alternatives:

(a)Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.

(b)Alloys usually have much (lower/ higher) temperature coefficients of resistance than pure metals.

(c)The resistivity of the alloy manganin is (nearly independent of /increases rapidly ) with increase of temperature.

(d)The resistivity of a typical insulator (e.g. amber) is greater than that of a metal by a factor of the order of (1022 or 103).

Sol.
(a) greater

(b) lower

(c) nearly independent of

(d) 1022

Q:20. (a) Given n resistors each of resistance R. How will you combine them to get the (i) maximum (ii) minimum effective resistance ? What is the ratio of the maximum to minimum resistance ?

(b) Given the resistance of 1 Ω , 2 Ω , 3 Ω , how will you combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω (iii) 6Ω (iv) (6/11) Ω ?

Determine the equivalent resistance of networks shown in Fig. (a) and (b)

Sol.
(a) For maximum effective resistance, the n resistance must be connected in series.

Maximum effective resistance, in Rs = n R

For minimum effective resistance, the n resistors must be connected in parallel.

Maximum effective resistance, Rp = R⁄n

Rs/Rp = nR/(R⁄n) = n2

It is to be noted that(a) the effective resistance of parallel combination of resistor is less then the individual resistance and (b) the effective resistance of series combination of resistors is more then individual resistance.

Case (i) Parallel combination of 1 Ω and 2 Ω is connected in series with 3 Ω.

Equivalent resistance of 2 Ω and 3 Ω in parallel will be given by = Rp = (1×2)/(1+2) =2/3 Ω

Equivalent resistance of 2/3 Ω and 3 Ω in series =2/3+3 = 11/3 Ω

Case (ii) Parallel combination of 2 Ω and 3 Ω is connected in series with 1 Ω

Equivalent resistance of 2 Ω and 3 Ω in parallel = (2×3)/(2+3)=6/5 Ω

Equivalent resistance of 6/5 Ω and 1 Ω in series = 6/5+1=11/5 Ω

Case (iii) All the resistance are to be connected in series. Now

Equivalent resistance = 1 + 2 + 3 = 6 Ω

Case (iv) All the resistance are to be connected in parallel

Equivalent resistance (R) is given by 1/R=1/1+1/2+1/3=(6+3+2)/6= 11/6 or R = 6/11 Ω

The given net work is a series combination of 4 equal units. Each unit has 4 resistance in which, two resistance (1 Ω each in series ) are in parallel with two other resistance (2 Ω each in series ).

Effective resistance of two resistance ( each of 1 Ω ) in series = 1 + 1 = 2 Ω

Effective resistance of two resistance (each of 2 Ω ) in series = 2 + 2 = 4 Ω

If R_p is the resistance of one unit of resistance, then 1/R_p = 1/2+1/4= 3/4 or R_p= 4/(3 ) Ω

Total resistance of net work (4 such units ) = 4/3 ×4=16/3 Ω =5.33 Ω

For Fig. (b), the five resistance each of value R, are connected in series.

Their effective resistance = 5 R

Q:21. Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the following infinite net work ; each resistor has 1 Ω resistance.

Sol.
Let x be the equivalent resistance of infinite net work. Since the net work is infinite, therefore, the addition of one more unit of there resistance each of value of 1 Ω across the terminals will not alter the total resistance of net work, i . e ., it should remain x.

Therefore. the network would appear as shown in Fig. and its total resistance should remain x.

Here the parallel combination of x and 1 Ω is in series with the two resistors of 1 Ω each. The resistance of parallel combination is

$\displaystyle \frac{1}{R_p} = \frac{1}{x} + \frac{1}{1}$

$\displaystyle R_p = \frac{x}{x + 1}$

Total resistance of net work will be given by

$\displaystyle x = 1 + 1 + \frac{x}{x + 1}$

$\displaystyle x = 2 + \frac{x}{x + 1}$

$\displaystyle x^2 – 2x -2 = 0$

$\displaystyle x = \frac{2 \pm \sqrt{12}}{2}$

x = 1 ± √3

The value of resistance can not be negative, therefore, the resistance of net work

= 1 + √3 = 1 + 1.73

= 2.73 Ω

Total resistance of the circuit = 2.73 + 0.5 = 3.23 Ω

Current drawn, I = 12/3.23 =3.72 amp

Q: 22. Figure shows a Potentiometer with a cell of 2.0 V and internal resistance 0.4 Ω maintaining a potential drop across the resistor emf of 1.02 V ( for very moderate currents up to a few amperes) gives a balance point at 67.3 cm. Length of the standard cell, a very high resistance of 600 k Ω is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown e.m.f E and the balance point found similarly turns out to be at 82.3 cm length of the wire.

(a) What is the value of E ?

(b) What does the high resistance of 600 kΩ have ?

(c) Is the balance point affected by the high resistance ?

(d) Is the balance point affected by the internal resistance of the driver cell ?

(e) Would the method work in the absolute situation, if the driver cell of the Potentiometer had an emf of 1.0 V instead of 2.0 V ?

(f) Would the circuit work for determining extremely small emf, say of the order of a few mV (such as the typical emf of a thermocouple) ? If not, how will you modify the circuit ?

Sol:

(a) Here, E1 = 1.02 V ; l1 =67.3 cm ; E2 = ?; l2 = 82.3 cm

Since, $\displaystyle \frac{E_2}{E_1} = \frac{l_2}{l_1}$

$\displaystyle E_2 = \frac{l_2}{l_1} \times E_1$

$\displaystyle E_2 = \frac{82.3}{67.3} \times 1.02$

=1.247 V

The purpose of using high resistance of 600 k Ω is to allow very small current through the galvanometer when the movable contact is far from the balance point.

No, the balance point is not affected by the presence of this resistance.

No, the balance point is not affected by the internal resistance of the driver cell.

No, the method will not work as the balance point will not be obtained on the potentiometer wire if the e.m.f of the driver cell is less than the e.m.f of the other cell.

The circuit will not work for measuring extremely small e.m.f because in that case, the balance point will be just close to the end A. To modify the circuit, we have to use a suitable high resistance in series with the cell of 2.0 V. This would decrease the current in the potentiometer wire. Therefore, potential difference/cm of wire will decrease. Then only extremely small e.m.f can be measured.

Q:23. Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with given cell of e.m.f. E ?

Sol.

Here, l1 = 58.3cm ;l2 = 68 cm ;R = 10 Ω ; X= ?

Let I be the current in the potentiometer wire and E1 and E2 be the potential drops across R and X respectively. When connected in circuit by closing respective key. Then

$\displaystyle \frac{E_2}{E_1} = \frac{l_2}{l_1}$

$\displaystyle \frac{I X}{I R} = \frac{l_2}{l_1}$

$\displaystyle X = \frac{l_2}{l_1} \times R$

X = 11.75 Ω

If there is no balance point with given cell of e.m.f. E , it means potential drop across R or X is greater than the potential drop across the potentiometer wire AB. In order to obtain the balance point, the potential drops across R and X are to be reduced, which is possible by reducing the current in R and X. For that, either a suitable resistance should be put in series with R and X or a cell of smaller e.m.f. Ω should be used. Another possible way is to increase the potential droop across the potentiometer wire by increasing the voltage of driver cell.

Q:24. The Fig. shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm . length of the potentiometer. Determine the internal resistance of the cell.

Sol:

Here, l1 = 76 cm ; l2 = 64.8 cm ;

r = ? ; R = 9.5 Ω

Now, $\displaystyle r = (\frac{l_1}{l_2}- 1) \times R$

$\displaystyle r = (\frac{76}{64.8}- 1) \times 9.5$

r = 1.68 Ω

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