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Question-8607




Question Number 8607 by tawakalitu last updated on 17/Oct/16
Answered by ridwan balatif last updated on 18/Oct/16
23.  C=0.50μF=5×10^(−7) F            L=0.050H            f=((100)/π)            ω=2πf            ω=200 rad/s       (i) X_C =(1/(ω×C))=(1/(200×5×10^(−7) ))=10^4 Ω        (ii) X_L =ω×L=200×0.05=10Ω
$$\mathrm{23}.\:\:\mathrm{C}=\mathrm{0}.\mathrm{50}\mu\mathrm{F}=\mathrm{5}×\mathrm{10}^{−\mathrm{7}} \mathrm{F} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{L}=\mathrm{0}.\mathrm{050H} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{f}=\frac{\mathrm{100}}{\pi} \\ $$$$\:\:\:\:\:\:\:\:\:\:\omega=\mathrm{2}\pi\mathrm{f} \\ $$$$\:\:\:\:\:\:\:\:\:\:\omega=\mathrm{200}\:\mathrm{rad}/\mathrm{s} \\ $$$$\:\:\:\:\:\left(\mathrm{i}\right)\:\mathrm{X}_{\mathrm{C}} =\frac{\mathrm{1}}{\omega×\mathrm{C}}=\frac{\mathrm{1}}{\mathrm{200}×\mathrm{5}×\mathrm{10}^{−\mathrm{7}} }=\mathrm{10}^{\mathrm{4}} \Omega \\ $$$$\:\:\:\:\:\:\left(\mathrm{ii}\right)\:\mathrm{X}_{\mathrm{L}} =\omega×\mathrm{L}=\mathrm{200}×\mathrm{0}.\mathrm{05}=\mathrm{10}\Omega \\ $$
Commented by tawakalitu last updated on 18/Oct/16
Thank you for your help.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{your}\:\mathrm{help}. \\ $$

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