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 10^{22}.
- Ans. Given : Density of free electrons in a copper conductor, n = 8.5 × 10^{28} m^{– 3}
- Length of copper wire, l = 3.0 m
- Area of cross-section of the wire,
- A = 2.0 × 10^{– 6} m^{2}
- Current carried by the wire, I = 3.0 A,
- I = nAev_{d}
- 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′′
$$\therefore\text{R}^{‘}=2+\frac{R^{‘}}{R^{‘}+1}\space\bigg[\because R^{”}=\frac{R^{‘}}{R^{‘}+1}\bigg]\\(\text{R}^{‘})^{2}-2\text{R}^{‘}-2=0$$
$$\text{R}^{‘}=\frac{2\pm\sqrt{4+8}}{2}\\=\frac{2\pm\sqrt{12}}{2}=1\pm\sqrt{3}\\\text{R}^{”}=\frac{R^{‘}}{R^{‘}+1}\\\text{Negative value of R}^{′}\text{cannot be accepted. Hence, equivalent resistance,}\\\text{R}^{‘}=(1+\sqrt{3})=1+1.73=2.73\spaceΩ\\\text{Internal resistance of the circuit,}\\\text{r}=0.5\spaceΩ\\\text{Hence, total resistance of the given circuit}\\=2.73+0.5=3.23\spaceΩ\\\text{Supply voltage,}\text{V}=12\text{V}\\\text{According to Ohm’s law, current drawn from the source is given by the ratio,}\\ \text{I}=\frac{V}{R}=\frac{12}{3.23}=3.72\space A $$
$$\text{R}_2=\frac{230V}{2.68A}=85.8\spaceΩ\\\text{From}\space\text{R}_2=\text{R}_1[1+\alpha(\text{T}_2-\text{T}_1)]\\\text{(temperature dependence of resistance)}\\\alpha=1.70×10^{-4}\degree\text{C}^{-1},\\\text{T}_2-\text{T}_1=\frac{(85.8-75.3)}{(75.3)×1.70×10^{-4}}=820\degree\text{C}\\\Rarr\text{T}_2=(820+27.0)\degree\text{C}=847\degree C\\\text{Thus, the steady temperature of the heating element is 847 °C.}$$