(a) Derivation of the expression for impedance 2 Plot of impedance with frequency (b) Phase difference between voltage across inductor and capacitor (c) Reason and calculation of self induction (a) $\;\overset{\rightarrow}{|V|} = V_m$ $|V_R| = V_{Rm}$ $|V_L| = V_{Lm}$ From the figure, the pythagoras theorem gives $V_m^2 = V_{Rm}^2 + (V_{Lm} - V_{Cm})^2$ $V_{Rm} = i_m R, V_{Lm} = i_m X_L, V_{Cm} = i_m X_C$ $V_m = i_m Z$ $(V_m)^2 = (i_m Z)^2 = (i_m R)^2 + (i_m X_L - i_m X_C)$ or, $Z^2 = R^2 + (X_L - X_C)^2$ $Z = \sqrt{R^2 + (X_L - X_C)^2}$ [Note: award these two marks, If a student does it correctly for the other case i.e., $(V_C > V_D) ]$ (b) Phase difference between voltage across inductor and the capacitor at resonance is $180^{\circ}$. (c) Inductor will offer an additional impedance to $a c$ due to its self inductance. $R = \frac{V_{rms}}{I_{rms}} = \frac{200}{1} = 200 \Omega$ Impedance of the inductor $Z = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{200}{0.5} = 400 \Omega$ Since, $Z = \sqrt{R^2 + X_L^2}$ $\; \therefore (400)^2 - (200)^2 = X_L^2$ $\; X_L = \sqrt{600 \times 200} = 346.4 \Omega$ Inductance $(L) = \frac{X_L}{\omega} = \frac{364.4}{2 \times 3.14 \times 50} = 1.1 \text{ H}$ OR (a) Working Principle: When the alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux in secondary and induces an emf in it. It works on the principle of mutual induction. Four sources of energy loss: (i) Flux leakage between primary and secondary windings. (ii) Resistance of the windings. (iii) Production of eddy currents in the iron core. (iv) Magnetization of the core. (b) Total resistance of the line $\; = \text{length} \times \text{resistance per unit length}$ $\; = 40 \text{ km} \times 0.5 \Omega / \text{km}$ $\; = 20 \Omega$ Current flowing in the line $I = \frac{P}{V}$$I = \frac{1200 \times 10^3}{4000}$$\; = 300 \text{ A}$Line loss in the form of heat$P = I^2 R$$\; = (300)^2 \times 20$$\; = 1800 \text{ kW}$