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New answer posted•a year ago

Electricity Class 12th

Q.3.18 Answer the following questions:

(a) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, 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 very low internal resistance. Why?

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

10 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Q.3.18 (a) When a steady current flows in a metallic conductor of non-uniform cross section, only the current flowing is constant. Current density, Electric field and Drift speed are inversely proportional to the cross section area, hence not constant.

(b) Ohm's law is not applicable to all co

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New answer posted•a year ago

Electricity Class 12th

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

Current (A)

Voltage (V)

Current (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

138.2

2.0

39.4

8.0

158.0

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

3.17 From Ohm’s law, R = $\frac{V}{A}$

From the given table we get $\frac{3.94}{0.2}$ = $\frac{7.87}{0.4}$ = $\frac{11.8}{0.6}$ …………… $\frac{158}{8}$ $\approx 19.7 Ω$ = constant

Hence manganin is an ohmic conductor.

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New answer posted•a year ago

Electricity Class 12th

3.16 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. ( $ρ_{A l}$ = 2.63 × 10^–8 Ωm, $ρ_{C u}$ = 1.72 × 10^–8 Ω m, Relative density of Al = 2.7, of Cu = 8.9.)

12 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

3.16 Given, resistivity of aluminium, $ρ_{1}$ = 2.63 $\times 10^{- 8}$ Ωm

Resistivity of copper, $ρ_{2}$ = 1.72 $\times 10^{- 8}$ Ωm

Relative density of aluminium = 2.7

Relative density of copper = 8.9

For Aluminium wire: Let

$l_{1}$ = length, $m_{1}$ = mass, $R_{1}$ = resistance, $A_{1}$ = cross-section

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New answer posted•a year ago

Electricity Class 12th

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?

10 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

3.15 (a) Number of secondary cells, n = 6

emf of each secondary cell = 2 V

Internal resistance of each secondary cell, r = 0.015 Ω

Resistance of the resistor, R = 8.5 Ω

Current drawn from the supply, I is given as I = $\frac{T o t a l v o l t a g e}{R + t o t a l i n t e r n a l r e s i s t a n c e}$

= $\frac{6 \times 2}{8.5 + 6 \times 0.015}$ = 1.396 A

(b) Terminal voltage = I $\times R$ = 1.396 $\times 8.5$ = 11

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New answer posted•a year ago

Electricity Class 12th

3.14 The earth’s surface has a negative surface charge density of 10^–9 C m^–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? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 10⁶ m.)

8 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

3.14 Surface charge density of the earth, $σ$ = $10^{- 9}$ C $m^{- 2}$

Current over entire Globe, I = 1800 A

Radius of the earth, r = 6.37 $\times 10^{6}$ m

Hence, surface area of the earth, A = 4 $π r^{2}$ = 4 $\times π \times {(6.37 \times 10^{6})}^{2}$ = 5.1 $\times 10^{14}$ $m^{2}$

Charge on the earth surface, q = $σ$ $\times A$ = 0.51 $\times 10^{6}$ C

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New answer posted•a year ago

Electricity Class 12th

3.13 The number density of free electrons in a copper conductor estimated in Example 3.1 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.

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

3.13 Given, number of free electron in a copper conductor, n = 8.5 $\times 10^{28}$ $m^{- 3}$

Length of the copper wire, l = 3.0 m

The area of cross section, A = 2 $\times 10^{- 6}$ $m^{2}$

Current carried by the wire, I = 3.0 A

From the relation I = nAe $V_{d}$ where

e = Electric charge = 1.6 $\times 10^{- 19}$ C

$V_{d} = D r i f t v e l o c i t y$

We get $V_{d}$ =

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New answer posted•a year ago

Electricity Class 12th

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?

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

3.12 EMF of the cell, $E_{1}$ = 1.25 V

Let the EMF of the replaced cell be $E_{2}$

Existing balance point, $l_{1}$ = 35 cm

New balance point, $l_{2}$ = 63 cm

From the relation of balance condition, we get

$\frac{E_{1}}{E_{2}}$ = $\frac{l_{1}}{l_{2}}$ , we get $E_{2}$ = $\frac{E_{1} * l_{2}}{l_{1}}$ = $\frac{1.25 * 63}{35}$ = 2.25 V

Therefore the emf

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New answer posted•a year ago

Electricity Class 12th

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?

8 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

3.11 EMF of the battery, E = 8 V

Internal resistance of the battery, r = 0.5 Ω

DC supply voltage, V = 120 V

Resistance of the resistor, R = 15.5 Ω

Let the effective voltage in the circuit be $V_{1}$

Then $V_{1}$ = V – E = 120 – 8 = 112 V

If the current flown in the circuit be I then from the relation

I =&

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New answer posted•a year ago

Electricity Class 12th

3.10 (a) In a meter bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor S is of 12.5 Ω. Determine the resistance of R. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?

(b) Determine the balance point of the bridge above if R and S 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?

15 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

3.10 (a) Balance point from end A, $l_{1}$ = 39.5 cm

Resistance of the resistor, S = 12.5 Ω

Condition for the balance is given as,

$\frac{R}{S}$ = $\frac{l_{1}}{{100 - l}_{1}}$

$R = S \times \frac{l_{1}}{100 - l_{1}}$ = 12.5 $\times \frac{39.5}{100 - 39.5}$ = 8.16 Ω

$T h e c o n n e c t i o n b e t w e e n r e s i s t o r s i n a W h e a t s t o n e$ bridge is made of thick copper strips to minimize the resistance, which is not taken into consideration in the bridge