# NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity

Q. Two charges 2 µC and – 2 µC are placed at points A and B 6 cm apart.

• Q. Choose the correct alternative :
• (i) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
• (ii) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
• (iii) The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature.
• (iv) 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. (i) Alloys of metals usually have greater resistivity than that of their constituent metals.
• (ii) Alloys usually have lower temperature coefficients of resistance than pure metals.
• (iii) The resistivity of the alloy, manganin, is nearly independent of increase of temperature.
• (iv) The resistivity of a typical insulator is greater than that of a metal by a factor of the order of 1022.
Q. The number density of free electrons in a copper conductor estimated 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. Given : Density of free electrons in a copper conductor, n = 8.5 × 1028 m– 3
• Length of copper wire, l = 3.0 m
• Area of cross-section of the wire,
• A = 2.0 × 10– 6 m2
• Current carried by the wire, I = 3.0 A,
• I = nAevd
• Where, e = Electric charge = 1.6 × 10– 19 C

$$\text{v}_d=\text{Drift velocity}=\frac{\text{Lenght of the wire (l)}}{\text{Time taken to cover} (r)}\\\text{I}=\frac{nAel}{t}\\t=\frac{nAel}{I}\\=\frac{8.5×10^{28}×2×10^{-6}×1.6×10^{-19}×3}{3.0}\\=2.7×10^{4}s\\\text{Therefore, the time taken by an electron to drift from one end of the wire to the other is}\space2.7 × 10^{4} s.$$

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

• Ans. The resistance of each resistor connected in the given circuit, R = 1 Ω
• Equivalent resistance of the given circuit = R′
• The network is infinite. Hence, equivalent resistance is given by the relation,
• Hence equivalent resistance can get from redrawn circuit as below
• R′ = 2 + R′′

$$\therefore\text{R}^{‘}=2+\frac{R^{‘}}{R^{‘}+1}\space\bigg[\because R^{”}=\frac{R^{‘}}{R^{‘}+1}\bigg]\\(\text{R}^{‘})^{2}-2\text{R}^{‘}-2=0$$

$$\text{R}^{‘}=\frac{2\pm\sqrt{4+8}}{2}\\=\frac{2\pm\sqrt{12}}{2}=1\pm\sqrt{3}\\\text{R}^{”}=\frac{R^{‘}}{R^{‘}+1}\\\text{Negative value of R}^{′}\text{cannot be accepted. Hence, equivalent resistance,}\\\text{R}^{‘}=(1+\sqrt{3})=1+1.73=2.73\spaceΩ\\\text{Internal resistance of the circuit,}\\\text{r}=0.5\spaceΩ\\\text{Hence, total resistance of the given circuit}\\=2.73+0.5=3.23\spaceΩ\\\text{Supply voltage,}\text{V}=12\text{V}\\\text{According to Ohm’s law, current drawn from the source is given by the ratio,}\\ \text{I}=\frac{V}{R}=\frac{12}{3.23}=3.72\space A$$

Q. An electric toaster uses nichrome for its heating element. When a negligibly small current passes through it, its resistance at room temperature (27.0° C) is found to be 75.3 W. When the toaster is connected to a 230 V supply, the current settles, after a few seconds, to a steady value of 2.68 A. What is the steady temperature of the nichrome element ? The temperature coefficient of resistance of nichrome averaged over the temperature range involved, is 1.70 × 10– 4 °C– 1.
Ans. When the current through the element is very small, heating effects can be ignored and the temperature T1 of the element is the same as room temperature. When the toaster is connected to the supply, its initial current will be slightly higher than its steady value of 2.68 A. But due to heating effect of the current, the temperature will rise. This will cause an increase in resistance and a slight decrease in current. In a steady state resistance R2 at the steady temperature T2 is

$$\text{R}_2=\frac{230V}{2.68A}=85.8\spaceΩ\\\text{From}\space\text{R}_2=\text{R}_1[1+\alpha(\text{T}_2-\text{T}_1)]\\\text{(temperature dependence of resistance)}\\\alpha=1.70×10^{-4}\degree\text{C}^{-1},\\\text{T}_2-\text{T}_1=\frac{(85.8-75.3)}{(75.3)×1.70×10^{-4}}=820\degree\text{C}\\\Rarr\text{T}_2=(820+27.0)\degree\text{C}=847\degree C\\\text{Thus, the steady temperature of the heating element is 847 °C.}$$