resistor & capacitor

[logistic_guy]

A \(\displaystyle 1.20\)-\(\displaystyle \text{k}\Omega\) resistor and a \(\displaystyle 6.8\)-\(\displaystyle \mu\text{F}\) capacitor are connected in series to an ac source. Calculate the impedance of the circuit if the source frequency is \(\displaystyle \bold{(a)} \ 60 \ \text{Hz};\) \(\displaystyle \bold{(b)} \ 60,000 \ \text{Hz}\).

 

[khansaheb]

A \(\displaystyle 1.20\)-\(\displaystyle \text{k}\Omega\) resistor and a \(\displaystyle 6.8\)-\(\displaystyle \mu\text{F}\) capacitor are connected in series to an ac source. Calculate the impedance of the circuit if the source frequency is \(\displaystyle \bold{(a)} \ 60 \ \text{Hz};\) \(\displaystyle \bold{(b)} \ 60,000 \ \text{Hz}\).

Draw a circuit diagram.

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[logistic_guy]

\(\displaystyle \bold{(a)}\)

The reactance of the capacitor is:

\(\displaystyle X_C = \frac{1}{2\pi f C}\)

The impedance is given by:

\(\displaystyle Z^2 = R^2 + X^2_C = R^2 + \frac{1}{4\pi^2 f^2 C^2}\)

Then,

\(\displaystyle Z = \sqrt{1200^2 + \frac{1}{4\pi^2 60^2 (6.8 \times 10^{-6})^2}} = \textcolor{blue}{1262 \ \Omega}\)

 

[logistic_guy]

\(\displaystyle \bold{(b)}\)

\(\displaystyle Z = \sqrt{1200^2 + \frac{1}{4\pi^2 60000^2 (6.8 \times 10^{-6})^2}} = \textcolor{blue}{1200 \ \Omega}\)