Current Electricity — NCERT Solutions

The maximum current is obtained when the external resistance is zero, so

$$I_{\max}=\frac{\varepsilon}{r}$$

Given $\varepsilon=12\,\text{V}$ and $r=0.4\,\Omega$,

$$I_{\max}=\frac{12}{0.4}=30\,\text{A}$$

Using $V=\varepsilon-Ir$, the terminal voltage is

$$V=10-(0.5)(3)=8.5\,\text{V}$$

Also, for the external resistor,

$$V=IR$$

so

$$R=\frac{V}{I}=\frac{8.5}{0.5}=17\,\Omega$$

But the battery drop across internal resistance is $Ir=1.5\,\text{V}$, so the total emf equation is

$$\varepsilon=I(R+r)$$

$$10=0.5(R+3)$$

$$R+3=20$$

$$R=17\,\Omega$$

So the resistance of the resistor is $17\,\Omega$, and the terminal voltage is $8.5\,\text{V}$.

Use

$$R_T=R_0[1+\alpha(T-T_0)]$$

Given $R_0=100\,\Omega$ at $T_0=27^\circ\text{C}$, $R_T=117\,\Omega$, and $\alpha=1.70\times10^{-4}\,^{\circ}\text{C}^{-1}$.

$$117=100[1+1.70\times10^{-4}(T-27)]$$

$$1.17=1+1.70\times10^{-4}(T-27)$$

$$0.17=1.70\times10^{-4}(T-27)$$

$$T-27=\frac{0.17}{1.70\times10^{-4}}=1000$$

$$T=1027^\circ\text{C}$$

So, the temperature is approximately 1028 °C.

Resistivity is

$$\rho=\frac{RA}{l}$$

Given $R=5.0\,\Omega$, $A=6.0\times10^{-7}\,\text{m}^2$, and $l=15\,\text{m}$:

$$\rho=\frac{5.0\times 6.0\times10^{-7}}{15}$$

$$\rho=\frac{30\times10^{-7}}{15}=2.0\times10^{-7}\,\Omega\text{ m}$$

So the resistivity is $2.0\times10^{-7}\,\Omega\text{ m}$.

For a small temperature range,

$$R_T=R_0[1+\alpha(T-T_0)]$$

Given $R_0=2.1\,\Omega$ at $T_0=27.5^\circ\text{C}$, and $R_T=2.7\,\Omega$ at $100^\circ\text{C}$.

$$2.7=2.1[1+\alpha(100-27.5)]$$

$$\frac{2.7}{2.1}=1+72.5\alpha$$

$$1.2857=1+72.5\alpha$$

$$72.5\alpha=0.2857$$

$$\alpha=\frac{0.2857}{72.5}\approx 0.00394\,^{\circ}\text{C}^{-1}$$

But the book’s standard method for temperature coefficient of resistivity from two resistance readings is to use

$$\alpha=\frac{R_2-R_1}{R_1(T_2-T_1)}$$

So,

$$\alpha=\frac{2.7-2.1}{2.1(100-27.5)} =\frac{0.6}{2.1\times72.5} \approx 3.94\times10^{-3}\,^{\circ}\text{C}^{-1}$$

Thus the temperature coefficient is $3.94\times10^{-3}\,^{\circ}\text{C}^{-1}$.