1. Problem 25.1: Calculate the charge transferred during a lightning strike. Given: Current $I = 25000$ A, time $t = 40 \mu s = 40 \times 10^{-6}$ s. Formula: Charge $Q = I \times t$. Calculation: $$Q = 25000 \times 40 \times 10^{-6} = 1$$ C. Answer: The charge transferred is 1 coulomb. 2. Problem 25.11 (a): Find resistivity $\rho$ at 20.0°C. Given: Length $L = 1.50$ m, diameter $d = 0.500$ cm $= 0.005$ m, voltage $V = 15.0$ V, current at 20°C $I_1 = 18.5$ A. Calculate cross-sectional area: $$A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.005}{2}\right)^2 = \pi \times (0.0025)^2 = 1.9635 \times 10^{-5} \text{ m}^2$$ Calculate resistance at 20°C: $$R_1 = \frac{V}{I_1} = \frac{15.0}{18.5} = 0.8108 \ \Omega$$ Use resistivity formula: $$R = \rho \frac{L}{A} \Rightarrow \rho = R \frac{A}{L} = 0.8108 \times \frac{1.9635 \times 10^{-5}}{1.50} = 1.061 \times 10^{-5} \ \Omega \cdot m$$ Answer: Resistivity at 20°C is approximately $1.06 \times 10^{-5} \ \Omega \cdot m$. 3. Problem 25.11 (b): Find temperature coefficient of resistivity $\alpha$ at 20°C. Given: Current at 92°C $I_2 = 17.2$ A, temperature change $\Delta T = 92 - 20 = 72$°C. Calculate resistance at 92°C: $$R_2 = \frac{V}{I_2} = \frac{15.0}{17.2} = 0.8721 \ \Omega$$ Use formula for resistance change with temperature: $$R_2 = R_1 (1 + \alpha \Delta T) \Rightarrow 1 + \alpha \Delta T = \frac{R_2}{R_1}$$ Calculate $\alpha$: $$\alpha = \frac{\frac{R_2}{R_1} - 1}{\Delta T} = \frac{\frac{0.8721}{0.8108} - 1}{72} = \frac{1.0759 - 1}{72} = \frac{0.0759}{72} = 0.001054 \ \text{°C}^{-1}$$ Answer: Temperature coefficient of resistivity is approximately $1.05 \times 10^{-3} \ \text{°C}^{-1}$. 4. Problem: Find internal resistance $r$ of battery. Given: emf $\varepsilon = 8.5$ V, terminal voltage $V = 7$ V, current $I = 6$ A. Formula: $$V = \varepsilon - Ir \Rightarrow r = \frac{\varepsilon - V}{I}$$ Calculation: $$r = \frac{8.5 - 7}{6} = \frac{1.5}{6} = 0.25 \ \Omega$$ Answer: Internal resistance is 0.25 ohms. 5. Problem 10 (a): Voltmeter reading when switch is OFF (open circuit). When switch is OFF, no current flows, so terminal voltage equals emf: $$V = \varepsilon = 12 \text{ V}$$ Answer: Voltmeter reads 12 V. 6. Problem 10 (b): Voltmeter reading when switch is ON (closed circuit). Given: $\varepsilon = 12$ V, internal resistance $r = 2.0 \ \Omega$, external resistance $R = 22 \ \Omega$. Calculate current: $$I = \frac{\varepsilon}{R + r} = \frac{12}{22 + 2} = \frac{12}{24} = 0.5 \text{ A}$$ Calculate terminal voltage: $$V = \varepsilon - Ir = 12 - 0.5 \times 2 = 12 - 1 = 11 \text{ V}$$ Answer: Voltmeter reads 11 V when switch is ON.