Ncert Exercise with Solution Chapter 3 Current Electricity

3.1. The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

Ans. Here ε = 12 V, r = 0.4Ω

The current drawn from the battery will be maximum when the external resistance in the circuit is zero i.e., R = 0.

∴ I_max =

= 30 A.

3.2. A battery of emf 10 V and internal resistance 3Ω is connected to a resistor. If the current in the circuit is 05 A, what is the resistance of the resistor ? What is the terminal voltage of the battery when the circuit is closed ?

Ans. As

I =

or R + r =

∴ R = – r = – 3 = 17 Ω

Terminal voltage,

V = IR = 0.5 × 17 = 8.5 V.

3.3. (i) Three resistors of 1 Ω, 2 Ω and 3 Ω are combined in series. What is the total resistance of the combination ? (ii) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Ans. (i) R_S = R₁ + R₂+ R₃ = 6 Ω.

(ii) Current in the circuit, I = = 2 A

∴ Potential drops across different resistors are

V₁ = I R₁ = 2 × 1 = 2 V,

V₂ = I R₂ = 2 × 2 = 4 V,

V₃ = I R₃ = 2 × 3 = 6 V.

3.4. (i) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination ? (ii) If the combination is connected to a battery of emf 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.

Ans. (i)

∴ R_p = Ω.

(ii) Currents drawn through different resistors are

, ,

Total current drawn from the battery,

I = I₁ + I₂ + I₃ = 10 + 5 + 4 = 19 A.

3.5. At room temperature (27° C), the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that temperature coefficient of the resistor material is 1.70 × 10^-4°C^-1.

Ans. Here R₁ =100Ω, R₂ = 117Ω, t₁= 27°C, α = 1.70 × 10^-4 °C^-1

As α =

∴

∴ t₂ = 1000 + t₁ = 1000 + 27 = 1027° C.

3.6. A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10^-7 m² and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment ?

Ans. Here l = 15 m, A = 6.0 × 10^-7m², R = 5.0Ω

Resistivity,

= 2.0 × 10^-7Ω m.

3.7. A silver wire has a resistance of 2.1 Ω at 27.5° C, and a resistance of 2.7 Ω at 100° C. Determine the temperature coefficient of resistivity of silver.

Ans. Here R₁ = 2.1 Ω, t₁ = 27.5°C, R₂ = 2.7 Ω, t₂ = 100°C Temperature coefficient of resistivity of silver,

α =

= 0.00394° C^-1.

3.8. A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27° C ? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10^-4 °C^-1.

Ans. Here V = 230 V, I₁ = 3.2 A,

I₂ = 2.8 A, α = 1.70 × 10^-4 °C^-1

Resistance at room temperature,

R₁ =

Resistance at steady temperature,

R₂ =

Now α =

∴

∴ Steady temperature of element,

t₂ = 840.35 + 27 = 867.35° C.

3.9. Determine the current in each branch of the network shown in Fig.

$C:\Users\NISHANT KUMAR\AppData\Local\Packages\Microsoft.MicrosoftEdge_8wekyb3d8bbwe\TempState\Downloads\media\image113.jpeg$

Ans. Let I, I₁, I₂, I₃ be the currents as shown in Fig… We apply Kirchhoff’s second rule to different loops.

$C:\Users\NISHANT KUMAR\AppData\Local\Packages\Microsoft.MicrosoftEdge_8wekyb3d8bbwe\TempState\Downloads\media\image114.jpeg$

For loop ABDA,

10I₁ + 5I₃ – 5I₂ = 0

For loop BCDB,

5(I₁– I₃) – 10(I₂+ I₃) – 5I₃= 0

For loop ADCFGA,

or …(1)

…(2)

…(3)

Solving equations (1), (2) and (3), we get

Currents in different branches are

Total Current,

3.10. (i) In a metre bridge the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance ofX. Why are the connections between resistors in a Wheatstone or metre bridge made of thick copper strips ? (ii) Determine the balance point of the bridge

$C:\Users\NISHANT KUMAR\AppData\Local\Packages\Microsoft.MicrosoftEdge_8wekyb3d8bbwe\TempState\Downloads\media\image115.jpeg$

above if X and Y are interchanged, (iii) What happens if the galvanometer and cell are interchanged at the balance point of the bridge ? Would the galvanometer show any current ? [CBSE D 05]

Ans. Here l = 35.9 cm, R = X = 7, S = Y = 12.5 Ω

As S = × R ∴ 12.5 = × R

or R = = 8.16 Ω

Connections are made by thick copper strips to minimize the resistances of connections which are not accounted for in the above formula.

(ii) When X and Y are interchanged,

R = Y = 12.5 Ω, S = X = 8.16 Ω, l =?

As ∴ × 12.5

or 8.16 l = 1250 – 12.5l

or l , from the end

(iii) When the galvanometer and cell are interchanged at the balance point, the conditions of the balanced bridge are still satisfied and so again the galvanometer will not show any current.

3.11. A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc 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 ?

Ans. When the storage battery of 8.0 volt is charged with a dc supply of 120 V, the net emf in the circuit will be

ε‘ = 120 – 8.0 = 112 V

Current in the circuit during charging

I = = 7 A

The terminal voltage of the battery during charging,

V = ε + Ir = 8.0 + 7 × 0.5 = 11.5 V

The series resistor limits the current drawn from the external source. In its absence, the current will be dangerously high.

3.12. In a potentiometer arrangement, a cell of emf 1.25 V 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 ?

Ans. Here ε₁ = 1.25 V, l₁ = 35.0 cm, l₂ = 63.0 cm, ε₂= ?

∴

3.13. The number density of free electrons in a copper conductor is 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.

Ans. Here n = 8.5 × 10²⁸m^-3, l = 3 m,

A = 2.0 × 10^-6 m², e = 1.6 × 10^-19 C, I = 3.0 A

Drift speed,

v_d =

= ms^-1

= ms^-1 = 1.1 × 10 ⁴ ms^-1

Required time,

t = s = 2.73 × 10⁴ s ≃ 7.57 h.

3.14. The earth’s surface has a negative surface charge density of 10^-9 Cm^-2. The potential difference of 400 Kv between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralize the earth’s surface? (Radius of the earth = 6.37 × 10⁶ m).

Ans. Surface charge density,

σ = 10^-9Cm^-2

Radius of the earth,

R = 6.37 × 10⁶ m

Current, I = 1800 A

Total charge of the globe,

q = surface area × σ = 4π R²σ

= 4 × 3.14 × (6.37 × 10⁶)² × 10^-9

= 509.65 × 10³ C

Required time,

t = = = 283.13 s ≃ 283 s.

3.15. (a) Six lead-acid type of secondary cells each of emf 2.0 V 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 380 Ω. What maximum current can be drawn from the cell ? Could the cell drive the starting motor of a car?

Ans. (a) Here ε = 2 V, r = 0.015 Ω, R = 8.5 Ω, n = 6

When the cells are joined in series, the current is

I =

Terminal voltage,

V = IR = 1.4 × 8.5 = 11.9 V.

(b) Here ε = 1.9 V, r = 380Ω

I_max = = A = 0.005 A

This secondary cell cannot drive the starting motor of a car because that requires a large current of about 100 A for a few seconds.

3.16. Two wires of equal length, one of aluminum 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 : ρ_Al = 2.63 × 10^-8Ωm, ρ_Cu = 1.72 × 10^-8 Ωm, relative density of Al = 2.7 and that of Cu = 8.9.

Ans. Mass = volume × density = Al d

= .ld = [∵ R = ρ ]

As the two wires are of equal length and have the same resistance, their mass ratio will be

i.e., copper wire is 2.2 times heavier than aluminium wire. Since aluminum is lighter, it is preferred for long suspension of cables otherwise heavy cable may sag down due to its own weight.

3.17. What conclusion can you draw from the following observations on a resistor made of alloy manganin:

Current I(A)	Voltage V	Current I(A)	Voltage V
0.2	3.94	3.0	59.2
0.4	7.87	4.0	78.8
0.6	11.8	5.0	98.6
0.8	15.7	6.0	118.5
1.0	19.7	7.0	138.5
2.0	39.4	8.0	158.0

Ans. We plot a graph between current I (along y-axis) and voltage V (along x-axis) as shown in Fig.

$C:\Users\NISHANT KUMAR\AppData\Local\Packages\Microsoft.MicrosoftEdge_8wekyb3d8bbwe\TempState\Downloads\media\image116.jpeg$

Since the V-I graph is almost a straight line, therefore, manganin resistor is an ohmic resistor for given ranges of voltage and current. As the current increases from 0 to 8 A, the temperature increases but the resistance of manganin does not change. This indicates that the temperature coefficient of resistivity of manganin alloy is negligibly small.

3.18. Answer the following questions:

(a) A steady current flows in a metallic conductor of non-uniform cross-section. Say which of these quantities is constant along the conductor: current, current density, electric field, drift speed? [CBSE D 15C, 17]

(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 current must have very low internal resistance. Wiry ?

(d) Why a high tension (H.T.) supply of say 6 kV must have a very large internal resistance ?

Ans. (a) Only current is constant because it is given to be steady. Other quantities : current density, electric field and drift speed vary inversely with area of cross-section.

(b) No, Ohm’s law is not universally applicable for all conducting elements. Examples of non-ohmic elements are vacuum diode, semiconductor diode, thyristor, gas discharge tube, electrolytic solution, etc.

I_max =

Clearly, I_max will be large if r is small.

(d) If the internal resistance is not very large, then the current will exceed the safety limits in case the circuit is short-circuited accidentally.

3.19. Choose the correct alternative :

(a) Alloys of metals usually have (greater/ lesser) 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 (10²²/10³).

Ans. (a) greater (b) lower (c) is nearly independent of (d) 10²².

3.20. (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 1Ω, 2Ω, 3Ω, how will you combine them to get an equivalent resistance of:

(i) Ω (ii) Ω (iii) 6 Ω (iv) Ω?

[CBSE F 15]

$C:\Users\NISHANT KUMAR\AppData\Local\Packages\Microsoft.MicrosoftEdge_8wekyb3d8bbwe\TempState\Downloads\media\image117.jpeg$

$C:\Users\NISHANT KUMAR\AppData\Local\Packages\Microsoft.MicrosoftEdge_8wekyb3d8bbwe\TempState\Downloads\media\image118.jpeg$

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

∴ Maximum effective resistance,

R_s = nR

For minimum effective resistance, all the n resistors must be connected in parallel. It is given by

= + + + ….. n terms =

∴ Minimum effective resistance,

R_p =

Ratio of the maximum to minimum resistance is

(b) Here R₁ = 1Ω, R₂ = 2 Ω, R₃ = 3 Ω

(i) When parallel combination of 1Ω and 2 Ω resistors is connected in series with 3Ω resistor [Fig. 3.318(a)], the equivalent resistance is

R =

= + 3 = + 3 = Ω.

(ii) When parallel combination of 2 Ω and 3 Ω resistors is connected in series with 1 Ω resistor [Fig. (b)], the equivalent resistance is

R = .

$C:\Users\NISHANT KUMAR\AppData\Local\Packages\Microsoft.MicrosoftEdge_8wekyb3d8bbwe\TempState\Downloads\media\image119.jpeg$

(iii) When the three resistances are connected in series [Fig.(c)], the equivalent resistance is

R = R₁ + R₂+ R₃= (1 + 2 + 3)Ω = 6Ω.

(iv) When all the resistances are connected in parallel [(d)],

= +

∴ Equivalent resistance, R = Ω.

(c) The network shown in Fig. (a) is a series combination of four identical units. One such unit is shown in Fig.(a) and it is equivalent to a parallel combination of two resistances of 2 Ω and 4 Ω as shown in Fig. (b).

$C:\Users\NISHANT KUMAR\AppData\Local\Packages\Microsoft.MicrosoftEdge_8wekyb3d8bbwe\TempState\Downloads\media\image120.jpeg$

Resistance R of one such unit is given by

= + = =

or R = Ω

∴ Resistance of the total network (4 such units)

= 4 × = Ω.

(ii) The network shown in Fig.(b) is a series combination of 5 resistors, each of resistance R.

∴ Equivalent resistance = 5 R.

3.21. Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the following infinite network. Each resistor has 1 Ω resistance.

$C:\Users\NISHANT KUMAR\AppData\Local\Packages\Microsoft.MicrosoftEdge_8wekyb3d8bbwe\TempState\Downloads\media\image121.jpeg$

Ans. Let the equivalent resistance of the infinite network be X. This network consists of infinite units of three resistors of 1 Ω, 1 Ω, 1 Ω. The addition of one more such unit across AB will not affect the total resistance. The network obtained by adding one more unit would appear as shown in Fig. 3.321.

$C:\Users\NISHANT KUMAR\AppData\Local\Packages\Microsoft.MicrosoftEdge_8wekyb3d8bbwe\TempState\Downloads\media\image122.jpeg$

Resistance between A and B

= Resistance equivalent to parallel combination of X and 1Ω

= =

Resistance between P and Q

= 1 + + 1 = 2 +

This must be equal to the original resistance X.

∴ X = 2 +

or X² – 2X – 2 = 0

or X = 1 ±

As the value of resistance cannot be negative, so

X = 1 + = 2.732 Ω

Current, I = = =

= 3.713 A

3.22. Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emfofl.02 V (for very moderate currents upto a few A) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from 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 emf ε and the balance point found similarly turns out to be at 82.3 cm length of the wire.

(a) What is the value of ε ?

$C:\Users\NISHANT KUMAR\AppData\Local\Packages\Microsoft.MicrosoftEdge_8wekyb3d8bbwe\TempState\Downloads\media\image123.jpeg$

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

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

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

(e) Would the method work in the above 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 well 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 ?

Ans. (a) ε₁ = 1.02 V, l₁ = 67.3 cm, ε₂ = ε = ?, l₂ = 82.3 cm

Formula for the comparison of emfs by potentiometer is

∴

or ε = × 1.02 = 1.25 V.

(b) High resistance of 600 kΩ protects the galvanometer for positions far away from the balance point, by decreasing current through it.

(c) No, balance point is not affected by high resistance because no current flows through the standard cell at the balance point.

(d) Yes, the balance point is affected by the internal resistance of the driver cell. The internal resistance affects the current through the potentiometer wire, so changes the potential gradient and hence affects the balance point.

(e) No, the arrangement will not work. If ε is greater than the emf of the driver cell of the potentiometer, there will be no balance point on the wire AB.

(f) The circuit as it is would be unsuitable, because the balance point (for ε of the order of a few mV) will be very close to the end A and the percentage error in measurement will be very large. The circuit is modified by putting a suitable resistor R in series with the wire AB so that potential drop across AB is only slightly greater than the emf to be measured. Then the balance point will be at larger length of the wire and the percentage error will be much smaller.

3.23. 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 685 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell ε ?

$C:\Users\NISHANT KUMAR\AppData\Local\Packages\Microsoft.MicrosoftEdge_8wekyb3d8bbwe\TempState\Downloads\media\image124.jpeg$

Ans. Here R = 10.0Ω, l₁ = 58.3cm, X = ?, l₂ = 68.5cm

Let ε₁ and ε₂ be the potential drops across R and X respectively and I be the current in potentiometer wire.

Then

But ∴ =

or X = .R = × 10 = 11.75 Ω

If there is no balance point, it means potential drops across R or X are greater than the potential drop across the potentiometer wire AB. We should reduce current in the outside circuit (and hence potential drops across R and X) suitably by putting a series resistor.

3.24. Figure 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 wire. Determine the internal resistance of the cell.

$C:\Users\NISHANT KUMAR\AppData\Local\Packages\Microsoft.MicrosoftEdge_8wekyb3d8bbwe\TempState\Downloads\media\image125.jpeg$

Ans. Here l₁ = 76.3 cm, l₂ = 64.8 cm, R = 9.5 Ω

The formula for the internal resistance of a cell by potentiometer method is

r = R