Answer the following questions: A resistor is connected to an ac source generating an emf 'e' given…

Comparing the given alternating emf $e = 495\sin(100\pi t)$ with $e = e_{0}\sin(\omega t)$: Peak voltage, $e_{0} = 495\text{ V}$ rms voltage: $e_{\text{rms}} = \frac{e_{0}}{\sqrt{2}} = \frac{495}{\sqrt{2}} \approx 350\text{ V}$ Given resistance $R = 70\,\Omega$: $I_{\text{rms}} = \frac{e_{\text{rms}}}{R} = \frac{350\text{ V}}{70\,\Omega} = 5\text{ A}$ Final answer: 5 A

The self-inductance of an iron-cored solenoid is given by: $L = \frac{\mu_{r} \mu_{0} N^{2} A}{l}$ Given: $N = 25$ $l = 20\text{ cm} = 0\cdot2\text{ m}$ $A = 1 \times 10^{-4}\text{ m}^{2}$ $\mu_{0} = 4\pi \times 10^{-7}\text{ H m}^{-1}$ $\mu_{r} = 3000$ (from constants table) Substituting the values: $L = \frac{3000 \times (4 \times 3\cdot14 \times 10^{-7}\text{ H m}^{-1}) \times (25)^{2} \times (1 \times 10^{-4}\text{ m}^{2})}{0\cdot2\text{ m}}$ $L = \frac{3000 \times 1\cdot256 \times 10^{-6} \times 625 \times 10^{-4}}{0\cdot2} \approx 1\cdot18 \times 10^{-3}\text{ H} = 1\cdot18\text{ mH}$ Final answer: 1.18 \times 10^{-3} H

Soft iron has high magnetic permeability and a narrow hysteresis loop with low coercivity and retentivity, which minimizes hysteresis energy loss during repeated cycles of magnetization and demagnetization.