For semiconductors, the charge carrier density $n$ increases exponentially with temperature $T$ according to:

$$n(T) = n_0 e^{-E_g / (2 k_B T)}$$

Since electrical resistivity $\rho = \frac{m}{n e^2 \tau}$, as temperature increases, the increase in carrier concentration $n$ heavily dominates over the decrease in relaxation time $\tau$.
Consequently, the resistivity of a semiconductor decreases exponentially with an increase in temperature:

$$\rho(T) = \rho_0 e^{E_g / (2 k_B T)}$$

Graph 3 shows this characteristic non-linear, exponential decrease of resistivity with increasing temperature.

In a potentiometer, the emf of a cell is directly proportional to its balancing length:

$$E \propto l \implies \frac{E_1}{E_2} = \frac{l_1}{l_2}$$

Given:

• $E_1 = 1.5\,\text{V}$, $l_1 = 45.0\,\text{cm}$
• $l_2 = 75.0\,\text{cm}$

Substituting the values:

$$E_2 = E_1 \times \frac{l_2}{l_1} = 1.5 \times \frac{75.0}{45.0} = 1.5 \times \frac{5}{3} = 2.5\,\text{V}$$

First, calculate the individual resistance of the heater and the bulb:

$$R_H = \frac{V^2}{P_{H,\text{rated}}} = \frac{220^2}{750} = \frac{48400}{750} \approx 64.53\,\Omega$$ $$R_B = \frac{V^2}{P_{B,\text{rated}}} = \frac{220^2}{200} = \frac{48400}{200} = 242.0\,\Omega$$

Total resistance in series:

$$R_{\text{total}} = R_H + R_B = 64.53 + 242.0 = 306.53\,\Omega$$

Circuit current under $220\,\text{V}$ supply:

$$I = \frac{V}{R_{\text{total}}} = \frac{220}{306.53} \approx 0.7177\,\text{A}$$

Power consumed by the bulb:

$$P_B = I^2 R_B = (0.7177)^2 \times 242 \approx 0.5151 \times 242 \approx 124.8\,\text{W}$$

Power consumed by the heater:

$$P_H = I^2 R_H = (0.7177)^2 \times 64.53 \approx 0.5151 \times 64.53 \approx 33.25\,\text{W}$$