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Class 12th Electricity

3.21 Determine the current drawn from a 12V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.

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Payal Gupta

Contributor-Level 10

3.21 Let the equivalent resistance of the given circuit be R'. The equivalent resistance of an infinite network is given by

R' = 2 + $\frac{R'}{(R^{'} + 1)}$ or R' = $\frac{2 R^{'} + 2 + R'}{(R^{'} + 1)}$ or ${R'}^{2}$ + R' = 2R' + 2 + R'

${R'}^{2} - 2 R^{'} - 2 = 0$

R' = $\frac{2 \pm \sqrt{4 + 8}}{2}$ = 1 $\pm \sqrt 3$

Since R' cannot be negative, hence R' = 1+ $\sqrt 3$ = 2.73 Ω

Internal resistance, r = 0.5Ω

Total resistance = 2.73 + 0.5 = 3.23 Ω

Current drawn from the source = $\frac{12}{3.23}$ A= 3.72 A

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Class 12th Electricity

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

(c) Determine the equivalent resistance of networks shown in Fig. 3.31.

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Payal Gupta

Contributor-Level 10

3.20 Total number of resistors = n

Resistance of each resistor = R

When the resistors are connected in series, effective resistance $R_{e f f}$ is maximum. $R_{e f f}$ = nR

When n resistors are connected in parallel, the effective resistance, $R_{e f f}$ is minimum. $R_{e f f}$ = $\frac{R}{n}$ . The ratio of maximum to minimum resistance = $\frac{n R}{\frac{R}{n}}$ = $n^{2}$

Let us assume, $R_{1}$ = 1 Ω, $R_{2}$ = 2 Ω, $R_{3}$ = 3 Ω

Required equivalent resistance, R = $\frac{11}{3}$ Ω

From the circuit the equivalent resistance is given by

R = $\frac{2 * 1}{2 + 1}$ + 3 = $\frac{2}{3}$ + 3 = $\frac{11}{3}$ Ω

Required equivalent resistance, R = $\frac{11}{5}$ Ω

From the circuit, the equivalen

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3.20 Total number of resistors = n

Resistance of each resistor = R

When the resistors are connected in series, effective resistance $R_{e f f}$ is maximum. $R_{e f f}$ = nR

Let us assume, $R_{1}$ = 1 Ω, $R_{2}$ = 2 Ω, $R_{3}$ = 3 Ω

Required equivalent resistance, R = $\frac{11}{3}$ Ω

From the circuit the equivalent resistance is given by

R = $\frac{2 * 1}{2 + 1}$ + 3 = $\frac{2}{3}$ + 3 = $\frac{11}{3}$ Ω

Required equivalent resistance, R = $\frac{11}{5}$ Ω

From the circuit, the equivalent resistance is given by

R = $\frac{2 * 3}{2 + 3}$ + 1 = $\frac{6}{5}$ + 1 = $\frac{11}{5}$ Ω

Required equivalent resistance, R = 6 Ω

From the circuit, the equivalent resistance is given by

R = 1 + 2+ 3 = 6 Ω

Required equivalent resistance, R = $\frac{6}{11}$ Ω

From the circuit, the equivalent resistance is given by

R = $\frac{1 * 2 * 3}{1 * 2 + 2 * 3 + 3 * 1}$ = $\frac{6}{11}$ Ω

(a)

It can be observed from the given circuit that in the first small loop, two resistors of resistance 1 Ω each are connected in the series. Hence the equivalent resistance is = 1 + 1= 2 $Ω$ . At the bottom part 2 resistors of 2 Ω each are connected in a series and thus make equivalent resistance of 2 + 2 = 4 Ω. Thus the circuit can be redrawn as 2 Ω and 4 Ω resistances are connected in parallel. Hence the equivalent resistance of each loop is R = $\frac{2 * 4}{2 + 4}$ = $\frac{4}{3}$ Ω. All these 4 loop resistors are connected in series. Thus the total equivalent resistance is $\frac{4}{3}$ Ω $* 4 = \frac{16}{3}$ Ω

Here all 5 resistors are connected in series. So the equivalent resistance is

5 $* R = 5 R$

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Class 12th Electricity

Q.3.19 Choose the correct alternative:

(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 (10²²/10²³).

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Payal Gupta

Contributor-Level 10

3.19 (a) Alloys of metal usually have greater resistivity than that of their constituent metals.

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

(c) The resistivity of the alloy manganin is nearly independent of 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.

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Class 12th Electricity

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?

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Payal Gupta

Contributor-Level 10

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 conductors, vacuum diode semi-conductor is a non-ohmic conductor.

(c) According to ohm's law V = IR, voltage is directly proportional to current, hence to draw high current from a low voltage source, internal resistance ®, needs to be low.

(d) To prevent the drawing of extra current, which can cause short circuit, the internal resistance for

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Class 12th Electricity

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

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Payal Gupta

Contributor-Level 10

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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Class 12th Electricity

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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Class 12th Electricity

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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Class 12th Electricity

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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Class 12th Electricity

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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Class 12th Electricity

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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