Question
In a series RC circuit with an AC source, $\text{R}=300\Omega,\text{C}=25\mu\text{F},\in_0=50\text{V}$ and $\nu=\frac{50}{\pi}\text{Hz}.$ Find the peak current and the average power dissipated in the circuit.
✓
Answer
$\text{R}=300\Omega,\text{C}=25\mu\text{F}=25\times10^{-6}\text{F}$$\in_0=50\text{V},\nu=\frac{50}{\pi}\text{Hz}$
$\text{X}_\text{C}=\frac{1}{\omega\text{C}}$
$=\frac{1}{\frac{50}{\pi}\times2\pi\times25\times10^{-6}}=\frac{10^4}{25}$
$\text{Z}=\sqrt{\text{R}^2+{\text{X}_\text{C}}^{2}}$
$=\sqrt{(300)^2+\Big(\frac{10^4}{25}\Big)^2}$
$=\sqrt{(300)^2+(400)^2}=500$
$=\frac{50\times50\times300}{2\times500\times500}=\frac{3}{2}=1.5\omega$
$\text{X}_\text{C}=\frac{1}{\omega\text{C}}$
$=\frac{1}{\frac{50}{\pi}\times2\pi\times25\times10^{-6}}=\frac{10^4}{25}$
$\text{Z}=\sqrt{\text{R}^2+{\text{X}_\text{C}}^{2}}$
$=\sqrt{(300)^2+\Big(\frac{10^4}{25}\Big)^2}$
$=\sqrt{(300)^2+(400)^2}=500$
- Peak current $=\frac{\text{E}_0}{\text{Z}}=\frac{50}{500}=0.1\text{A}$
- Average Power dissipitated $=\text{E}_\text{rms}\text{I}_\text{rms}\cos\phi$
$=\frac{50\times50\times300}{2\times500\times500}=\frac{3}{2}=1.5\omega$
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