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Question Number 147418 by jlewis last updated on 20/Jul/21
$$\mathrm{In}\:\mathrm{an}\:\mathrm{R}{LC}\:\mathrm{series}\:\mathrm{circuit}, \\ $$$$\mathrm{R}=\mathrm{1kilo}\:\mathrm{ohms},\mathrm{L}=\mathrm{0}.\mathrm{2H},\mathrm{C}=\mathrm{1} \mathrm{F}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{voltage}\:\mathrm{source}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}: \\ $$$$\left(\mathrm{V}=\mathrm{150}\:\mathrm{sin}\:\mathrm{377t}\:\right)\mathrm{V}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{peak}\:\mathrm{current}\:\mathrm{delivered}\:\mathrm{by}\:\mathrm{the}\: \\ $$$$\mathrm{source}? \\ $$
Answered by Olaf_Thorendsen last updated on 20/Jul/21
$$\mathrm{I}\:=\:\frac{\mathrm{V}}{\:\sqrt{\mathrm{R}^{\mathrm{2}} +\left(\omega\mathrm{L}−\frac{\mathrm{1}}{\omega\mathrm{C}}\right)^{\mathrm{2}} }} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{150}}{\:\sqrt{\mathrm{1000}^{\mathrm{2}} +\left(\mathrm{377}×\mathrm{0}.\mathrm{2}−\frac{\mathrm{1}}{\mathrm{377}×\mathrm{10}^{−\mathrm{6}} }\right)^{\mathrm{2}} }} \\ $$$$\mathrm{I}\:=\:\mathrm{54},\mathrm{3}\:{m}\mathrm{A} \\ $$
Commented by jlewis last updated on 20/Jul/21
$$\mathrm{thanks}\:\mathrm{Sir} \\ $$