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(d) We know that impedencez=\frac{1}{\left(\sqrt{\left(w-\frac{1}{\omega l}\right)^{2}+\frac{1}{R^{2}}}\right)}z is maximum when square root is mmimum This can be achieved when,\begin{array}{l}\left(\omega c-\frac{1}{\omega L}\right)^{2}=0 \\\therefore \omega c=\frac{1}{\omega L} \\=\omega^{2}=\frac{1}{L C} \\=\omega=\frac{1}{\sqrt{L C}}\end{array} (b) Resonance angular frequency.\begin{array}{l}\omega_{R}=\frac{1}{\sqrt{L C}}=\frac{1}{(\sqrt{(1 \times 10-7) \times 0.3})} \\\therefore \omega_{R}=5773.5026 \mathrm{rad} / \mathrm{s}\end{array}(C) I_{\max }=\frac{11}{z}=\frac{240}{100} \Rightarrow I_{\max }=2-4 \mathrm{~A}(d) Amplitude of currents in resistir.I=\frac{V}{R}=\frac{240}{100} \Rightarrow P=2.4 \mathrm{~A}(2) Amplitude of current m mductance.\begin{array}{l}I=\frac{V}{X_{1}}=\frac{(240)}{(5773)(0.3)} \\I=0.1385 \mathrm{~A}\end{array} (f) As circuit is in resonance\begin{array}{c}\therefore X_{L}=X_{C} \\\therefore I=\frac{V}{X_{C}}=\frac{240}{(5773 \times 0.3)} \\I=0.1385 \mathrm{~A}\end{array} C3 Do Rate ...