Find-the-impedence-of-an-RC-circuit-with-R-10-and-C-10-F-at-an-angular-frequency-of-21800rads-1-

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

Find the impedence of an RC circuit

with R = 10 Ω and C = 10 μ F at an

angular frequency of {21800 rads}^{- 1}

Answered by ajfour last updated on 18/Apr/21

Z = \sqrt{R^{2} + \frac{1}{{(ω C)}^{2}}}

= \sqrt{100 + \frac{1}{{(21800 \times 10 \times 10^{- 6})}^{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 Ω

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

Thanks a lot prof ajour

Answered by physicstutes last updated on 18/Apr/21

impedance Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}

but X_{L} = 0 since we have an R - C circuit .

\Rightarrow Z = \sqrt{R^{2} + X_{C}^{2}}

where X_{C} = \frac{1}{2 π f C} = \frac{1}{ω C} = \frac{1}{(21800) (10 \times 10^{- 6})}

\Rightarrow Z = \sqrt{10^{2} + {[\frac{1}{(21800) (10 \times 10^{- 6})}]}^{2}} \approx 11 Ω

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

thank you sir physicstutes