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Question Number 138764 by otchereabdullai@gmail.com last updated on 17/Apr/21

$$\mathrm{Find}\mathrm{the}\mathrm{impedence}\mathrm{of}\mathrm{an}\mathrm{RC}\mathrm{circuit}$$$$\mathrm{with}\mathrm{R}=10\mathrm{\Omega }\mathrm{and}\mathrm{C}=10\mu \mathrm{F}\mathrm{at}\mathrm{an}$$$$\mathrm{angular}\mathrm{frequency}\mathrm{of}{21800\mathrm{rads}}^{-1}$$

Answered by ajfour last updated on 18/Apr/21

$$Z=\sqrt{R^2+\frac{1}{{\left(\omega C\right)}^2}}$$$$=\sqrt{100+\frac{1}{{\left(21800\times 10\times {10}^{-6}\right)}^2}}$$$$=\sqrt{100+\frac{1}{{(0.218)}^2}}$$$$\frac{1}{0.218}=\frac{1000}{218}=\frac{500}{109}=\frac{5}{1.09}$$$$=5(1-0.09)=4.55$$$$Z=\sqrt{100+{(4.55)}^2}$$$$Z=\sqrt{100+16+4.4+0.3025}$$$$Z=\sqrt{120.7025}$$$$Z\approx 11\mathrm{\Omega }$$

Commented by otchereabdullai@gmail.com last updated on 18/Apr/21

$$\mathrm{Thanks}\mathrm{a}\mathrm{lot}\mathrm{prof}\mathrm{ajour}$$

Answered by physicstutes last updated on 18/Apr/21

$$\mathrm{impedance}\mathrm{Z}=\sqrt{R^2+{(X_L-X_C)}^2}$$$$\mathrm{but}{X}_L=0\mathrm{since}\mathrm{we}\mathrm{have}\mathrm{an}\mathrm{R}-\mathrm{C}\mathrm{circuit}.$$$$\Rightarrow Z=\sqrt{R^2+X_C^2}$$$$\mathrm{where}{X}_C=\frac{1}{2\pi fC}=\frac{1}{\omega C}=\frac{1}{\left(21800\right)\left(10\times {10}^{-6}\right)}$$$$\Rightarrow Z=\sqrt{{10}^2+{\left[\frac{1}{\left(21800\right)\left(10\times {10}^{-6}\right)}\right]}^2}\approx 11\mathrm{\Omega }$$

Commented by otchereabdullai@gmail.com last updated on 18/Apr/21

$$\mathrm{thank}\mathrm{you}\mathrm{sir}\mathrm{physicstutes}$$

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