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5 months 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.)

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Payal Gupta

Contributor-Level 10

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

Resistivity of copper, $ρ_{2}$ = 1.72 $* 10^{- 8}$ 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 area, $ρ_{1}$ = density

For Copper wire: Let

$l_{2}$ = length, $m_{2}$ = mass, $R_{2}$ = resistance, $A_{2}$ = cross-section area, $ρ_{2}$ = density

From the relation R = $\frac{ρ l}{A}$ , we get

$R_{1} = \frac{ρ_{1} l_{1}}{A_{1}}$ ………(1), and

$R_{2} = \frac{ρ_{2} l_{2}}{A_{2}}$ ………(2)

It is given $R_{1}$ = $R_{2}$ and $l_{1}$ = $l_{2}$ . Hence

$\frac{A_{1}}{A_{2}}$ = $\frac{ρ_{1}}{ρ_{2}}$ = $\frac{2.63}{1.72}$ &nb

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

Resistivity of copper, $ρ_{2}$ = 1.72 $* 10^{- 8}$ 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 area, $ρ_{1}$ = density

For Copper wire: Let

$l_{2}$ = length, $m_{2}$ = mass, $R_{2}$ = resistance, $A_{2}$ = cross-section area, $ρ_{2}$ = density

From the relation R = $\frac{ρ l}{A}$ , we get

$R_{1} = \frac{ρ_{1} l_{1}}{A_{1}}$ ………(1), and

$R_{2} = \frac{ρ_{2} l_{2}}{A_{2}}$ ………(2)

It is given $R_{1}$ = $R_{2}$ and $l_{1}$ = $l_{2}$ . Hence

$\frac{A_{1}}{A_{2}}$ = $\frac{ρ_{1}}{ρ_{2}}$ = $\frac{2.63}{1.72}$ ( $10^{- 8}$ cancel each other)

From the relation mass = volume $* r e l a t i v e d e n s i t y$ = length $*$ cross sectional area $*$ relative density, we get

$m_{1} = l_{1} A_{1} d_{1}$ and $m_{2} = l_{2} A_{2} d_{2}$ , where $d_{1}$ = relative density of aluminium and $d_{2}$ = relative density of copper

$\frac{m_{1}}{m_{2}} = \frac{l_{1} A_{1} d_{1}}{l_{2} A_{2} d_{2}}$ = $\frac{A_{1} d_{1}}{A_{2} d_{2}}$ = $\frac{2.63}{1.72} * \frac{2.7}{8.9}$ = 0.464

It can be inferred from this ratio that $m_{1}$ is less than $m_{2}$ . Hence, aluminium is preferred as overhead power cables over copper.

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New answer posted

5 months 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?

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Payal Gupta

Contributor-Level 10

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 = $\frac{T o t a l v o l t a g e}{R + t o t a l i n t e r n a l r e s i s t a n c e}$

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

(b) Terminal voltage = I $* R$ = 1.396 $* 8.5$ = 11.87 V

After long use emf = 1.9 V

Internal resistance of the cell, r = 380 Ω

Maximum current can be drawn = $\frac{1.9}{380}$ = 5 $* 10^{- 3}$ A

No, the cell cannot drive the starter motor of the car as it requires high current.

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New answer posted

5 months 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.)

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Payal Gupta

Contributor-Level 10

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 $* 10^{6}$ m

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

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

If t is time taken to neutralize the earth surface, then q = I $* t$

t = $\frac{q}{I}$ = $\frac{0.51 * 10^{6}}{1800}$ = 283.28 seconds

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5 months 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.

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Payal Gupta

Contributor-Level 10

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

Length of the copper wire, l = 3.0 m

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

Current carried by the wire, I = 3.0 A

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

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

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

We get $V_{d}$ = $\frac{I}{n A e}$

Again, Drift velocity ( $V_{d})$ = $\frac{L e n g t h o f t h e w i r e (l)}{T i m e t a k e n t o c o v e r l (t)}$

t = $\frac{l}{V_{d}}$ = $\frac{l n A e}{I}$ = $\frac{3 * 8.5 * 10^{28} * 2 * 10^{- 6} * 1.6 * 10^{- 19}}{3}$ = 27.2 $* 10^{3}$ seconds

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New answer posted

5 months 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?

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Payal Gupta

Contributor-Level 10

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 of the another cell is 2.25V

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New answer posted

5 months 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?

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Payal Gupta

Contributor-Level 10

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 = $\frac{V_{1}}{R + r}$ we get I = $\frac{112}{15.5 + 0.5}$ = 7 A

Voltage across the resistor R = IR = 7 $* 15.5 =$ 108.5 V

Hence, Terminal voltage of the battery

= DC supply voltage – Voltage drop across the resistor

= 120 – 108.5 = 11.5 V

The purpose of having a series resistor in a charging circuit is to limit the current drawn from the external source. Otherw

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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 = $\frac{V_{1}}{R + r}$ we get I = $\frac{112}{15.5 + 0.5}$ = 7 A

Voltage across the resistor R = IR = 7 $* 15.5 =$ 108.5 V

Hence, Terminal voltage of the battery

= DC supply voltage – Voltage drop across the resistor

= 120 – 108.5 = 11.5 V

The purpose of having a series resistor in a charging circuit is to limit the current drawn from the external source. Otherwise, the current will be extremely high.

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New answer posted

5 months 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?

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Payal Gupta

Contributor-Level 10

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,

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

$R = S * \frac{l_{1}}{100 - l_{1}}$ = 12.5 $* \frac{39.5}{100 - 39.5}$ = 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 formula.

(b) If R & S are interchanged, then $l_{1}$ and (100 - $l_{1}$ ) will also get interchanged. The balance point will be (100- $l_{1})$ from A. So the new balance point is 100 – 39.5 = 60.5 cm from A.

(c) When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflect

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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,

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

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

(c) When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection, no current will flow through galvanometer.

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New answer posted

5 months ago

Electricity Class 12th

3.9 Determine the current in each branch of the network shown in Fig. 3.30:

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Payal Gupta

Contributor-Level 10

3.9

Let us assume

$I_{1}$ = Current flowing through the outer circuit

$I_{2}$ = Current flowing through the branch AB

$I_{3}$ = Current flowing through the branch AD

$I_{4}$ = Current flowing through the branch BD

( $I_{2}$ - $I_{4})$ = Current flowing through the branch BC

( $I_{3}$ + $I_{4})$ = Current flowing through the branch DC

For the closed circuit ABDA, potential is zero, i.e.

10 $I_{2}$ + 5 $I_{4}$ - 5 $I_{3}$ = 0

$I_{3} = I_{4} + 2 I_{2}$ …………(1)

For the close circuit BCDB, potential is zero, i.e.

5( $I_{2}$ - $I_{4})$ - 10( $I_{3}$ + $I_{4})$ - 5 $I_{4}$ = 0

5 $I_{2}$ - 5 $I_{4}$ - 10 $I_{3}$ - 10 $I_{4}$ - 5&nbs

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3.9

Let us assume

$I_{1}$ = Current flowing through the outer circuit

$I_{2}$ = Current flowing through the branch AB

$I_{3}$ = Current flowing through the branch AD

$I_{4}$ = Current flowing through the branch BD

( $I_{2}$ - $I_{4})$ = Current flowing through the branch BC

( $I_{3}$ + $I_{4})$ = Current flowing through the branch DC

For the closed circuit ABDA, potential is zero, i.e.

10 $I_{2}$ + 5 $I_{4}$ - 5 $I_{3}$ = 0

$I_{3} = I_{4} + 2 I_{2}$ …………(1)

For the close circuit BCDB, potential is zero, i.e.

5( $I_{2}$ - $I_{4})$ - 10( $I_{3}$ + $I_{4})$ - 5 $I_{4}$ = 0

5 $I_{2}$ - 5 $I_{4}$ - 10 $I_{3}$ - 10 $I_{4}$ - 5 $I_{4}$ =0

5 $I_{2}$ - 10 $I_{3}$ - 20 $I_{4}$ = 0

$I_{2} - 2 I_{3}$ - 4 $I_{4}$ = 0

$I_{2} = 2 I_{3}$ +4 $I_{4}$ …………….(2)

For the close circuit ABCFEA, potential is zero, i.e.

-10 + 10 $I_{1} +$ 10 $I_{2}$ + 5( $I_{2}$ - $I_{4})$ = 0

10 $I_{1}$ + 15 $I_{2}$ - 5 $I_{4}$ = 10

2 $I_{1}$ + 3 $I_{2}$ - $I_{4}$ = 2 …………(3)

Solving equations (1) and (2) we get

$I_{2} = - 2 I_{4}$

$I_{3} = - 3 I_{4}$

Substituting these values in equation (3), we get

$I_{4} = - \frac{2}{17}$ A

$I_{2} = \frac{4}{17}$ A

$I_{3} = \frac{6}{17}$ A

$I_{1}$ = $I_{2} + I_{3}$ = $\frac{10}{17}$ A

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5 months ago

Electricity Class 12th

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 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 * 10^–4 °C^–1.

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Payal Gupta

Contributor-Level 10

3.8 Given

Supply voltage, V = 230 V

Initial current drawn, I= 3.2 A

Final current drawn, $I_{1}$ = 2.8 A

Room temperature, T = 27.0 °C

Steady temperature, $T_{1}$ = ?

From Ohm's law, we get initial resistance, $R$ = $\frac{V}{I}$ = $\frac{230}{3.2}$ Ω = 71.875 Ω

Final resistance, $R_{1}$ = $\frac{V}{I_{1}}$ = $\frac{230}{2.8}$ Ω = 82.143 Ω

From the relation of α = $\frac{R_{1} - R}{R (T_{1} - T)}$ , where α is the temperature coefficient of resistance, we get

1.7 $* 10^{- 4}$ = $\frac{82.143 - 71.875}{71.875 (T_{1} - 27)}$

$T_{1} - 27 =$ 840.34

$T_{1} = 867.34 ?$

Therefore the steady temperature of heating element required is $867.34 ?$

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Electricity Class 12th

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.

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Payal Gupta

Contributor-Level 10

3.7 Let us assume R = 2.1 Ω, T = 27.5 °C, $R_{1}$ = 2.7 Ω, $T_{1}$ = 100 $?$ , α = resistivity of silver

We know the relation of α can be given as

α = $\frac{R_{1} - R}{R (T_{1} - T)}$ = $\frac{2.7 - 2.1}{2.1 (100 - 27.5)}$ = 3.94 $* 10^{- 3}$ $?^{- 1}$

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