NCERT Solution Physics Class 12 Chapter 3 Current Electricity

NCERT Solution Physics Class 12 Chapter 3 Current Electricity

NCERT Solution Physics Class 12 Chapter 3 Current Electricity all questions and answers. Physics Class 12 3rd Chapter Current Electricity exercise solution and experts answer. As one of online learning platforms, we (netex.) are excited to offer the NCERT Solution Physics Class 12 Chapter 3. This solution is designed to help students who are looking to brush up on their physics concepts on Chapter 3 Current Electricity.

 

3.1) 

 

3.3) 

 

3.4) 

 

3.5) 

 

3.8) 

 

3.9) 

 

3.10) 

 

3.12) 

 

In case you are missed :- Previous Chapter Solution

 

 

3.15)

 

3.16)

 

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

From the given table ratio of voltage with current is a constant and equal to  = 3.94/0.2 = 59.2/3 = 19.7 as ratio of voltage and current is constant which is conclusion from ohms law hence mangan in can be considered as a ohmic conductor which obeys ohms law.

From ohms law,

Resistance = voltage / current = 19.7Ω.

 

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?

ANSWER-

(A) We know that current is independent of the area while all other quantities current density, electric field and drift velocity depends upon area of conductor. As given conductor is non-uniform cross section hence only current is constant.

(b)Ohm’s law is not universally applicable for all conducting elements. Semiconductor does not obey ohm’s law.

(c) According to ohm’s law i =v/r. For getting maximum current value voltage should be very large. If voltage is low then resistance should be very less as it is in denominator.

(d)A high tension (HT) supply 6 kV must have very large internal resistance to limit the current within safety level.

 

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).

ANSWER-

(a) Alloys of metals 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.

 

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.

To get the maximum effective resistance in circuit all n resistances of each resistance R are to be connected in series and maximum resistance is given by RMAX = nR.

To get the minimum effective resistance in circuit all n resistances of each resistance R are to be connected in parallel and minimum resistance is obtain given by RMIN= R/n

The ratio of maximum and minimum resistance is given by –

RMAX / RMIN = nr/ R/n = n2

Given resistances are R1 = 1Ω, R2 = 2Ω And R3 = 3Ω.

(i) To get an equivalent resistance of (11/3) .lets connect them as shown in figure (i). Here R1 = 1Ω, R2 = 2Ω are connected in parallel and this combination is connected in series with R3= 3Ω.

REFF = 2 x ½+1 + 3 = 2/3 + 3 = 11/3 Ω

(ii) To get an equivalent resistance of (11/5Ω) .Lets connect them as shown in figure (ii). Here R2 = 2Ω, R3= 3Ω are connected in parallel and this combination is connected in series with R1= 1Ω.

REFF = 2×3/2+3 +1 = 6/5 + 1 = 11/5 Ω

 

(iii) To get an equivalent resistance of (6Ω) .Lets connect them as shown in figure (iii). Here R1 =1Ω,

R2= 2Ω andR3 = 3Ω are connected in series.

REFF = 1+2+3 = 6Ω.

 

(iv) To get an equivalent resistance of (6/11Ω) .Lets connect them as shown in figure (iv). Here R1 =1Ω,

R2= 2Ω andR3 = 3Ω are connected in parallel.

REFF = 1x2x3/1×2+2×3+3×1 = 6/11 Ω

 

From figure 3.31(a),

Two resistances of each 1Ω connected in series hence they constitute the effective resistance of 2Ω. and Two resistances of each 2Ω connected in series hence they constitute the effective resistance of

2+2 = 4Ω. And this combination of 2Ω and 4Ω connected in parallel to each other  in all 4 loops.

Equivalent parallel resistance of each loop is

R = 2×4/2+4 = 8/6 = 4/3 Ω

 

3.21) 

 

3.22) 

 

3.23) 

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Updated: April 15, 2023 — 3:15 pm

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