Physics 5 Marks Questions

$\text{E}=6.5\text{V},\text{f}=\frac{30}{\pi}\text{Hz}$

$\text{Z}=\sqrt{\text{R}^2+{\text{X}_\text{L}}^{2}}=\sqrt{\text{R}^2+(2\pi\text{fL})^2}$

$\text{Power}=\text{V}_\text{rms}\text{I}_\text{rms}\cos\phi$

$=6.5\times\frac{6.5}{\text{Z}}\times\frac{\text{R}}{\text{Z}}$

$=\frac{6.5\times6.5\times10}{\Big[\sqrt{\text{R}^2+(2\pi\text{f})^2}\Big]}$

$=\frac{6.5\times6.5\times10}{10^2+\Big(2\pi\times\frac{30}{\pi}\times0.4\Big)^2}$

$=\frac{6.5\times6.5\times10}{100+576}=0.625=\frac{5}{8}\omega$

$\text{H}=\int\limits_0^\text{H}=\int\frac{\text{E}_0\sin^2\omega\text{t}}{\text{R}}\text{dt}$

$=\frac{144}{100}\int\sin^2\omega\text{t}\text{ dt}=1.44\int\Big(\frac{1-\cos2\omega\text{t}}{2}\Big)\text{dt}$

$=\frac{1.44}{2}\Big[\int\limits_0^{10^{-3}}\text{dt}-\int\limits_0^{10^{-3}}\cos2\omega\text{t}\text{ dt}$

$=0.72\Bigg[10^{-3}-\Big(\frac{\sin2\omega\text{t}}{2\omega}\Big)_0^{10^{-3}}\Bigg]$

$=0.72\Big[\frac{1}{1000}-\frac{1}{500\pi}\Big]$

$=\frac{(\pi-2)}{1000\pi}\times0.72=0.0002614$

$=2.61\times10^{-4}{\text{J}}$

$\text{E}=(10\text{V})=\sin\omega\text{t}$

1. $\text{I}=\frac{\text{E}_0}{\text{X}_\text{C}}=\frac{\text{E}_0}{\Big(\frac{1}{\omega\text{C}}\Big)}$

$=\frac{10}{\Big(\frac{1}{10\times10^{-5}}\Big)}=1\times10^{-3}\text{A}$

2. $\omega=100\text{s}^{-1}$

$\text{I}=\frac{\text{E}_0}{\Big(\frac{1}{\omega\text{C}}\Big)}$

$=\frac{10}{\Big(\frac{1}{100\times10^{-5}}\Big)}=1\times10^{-2}\text{A}=0.01\text{A}$

3. $\omega=500\text{s}^{-1}$

$\text{I}=\frac{\text{E}_0}{\Big(\frac{1}{\omega\text{C}}\Big)}$

$=\frac{10}{\Big(\frac{1}{500\times10^{-5}}\Big)}=5\times10^{-2}\text{A}=0.05\text{A}$

4. $\omega=1000\text{s}^{-1}$

$\text{I}=\frac{\text{E}_0}{\Big(\frac{1}{\omega\text{C}}\Big)}$

$=\frac{10}{\Big(\frac{1}{1000\times10^{-5}}\Big)}=1\times10^{-1}\text{A}=0.1\text{A}$

$\in_0=50\text{V},\nu=\frac{50}{\pi}\text{Hz}$

$\text{X}_\text{C}=\frac{1}{\omega\text{C}}$

$=\frac{1}{\frac{50}{\pi}\times2\pi\times25\times10^{-6}}=\frac{10^4}{25}$

$\text{Z}=\sqrt{\text{R}^2+{\text{X}_\text{C}}^{2}}$

$=\sqrt{(300)^2+\Big(\frac{10^4}{25}\Big)^2}$

$=\sqrt{(300)^2+(400)^2}=500$

1. Peak current $=\frac{\text{E}_0}{\text{Z}}=\frac{50}{500}=0.1\text{A}$
2. Average Power dissipitated $=\text{E}_\text{rms}\text{I}_\text{rms}\cos\phi$

$=\frac{\text{E}_0}{\sqrt{2}}\times\frac{\text{E}_0}{\sqrt{2}\text{Z}}\times\frac{\text{R}}{\text{Z}}=\frac{\text{E}_0{^2}}{2\text{Z}^2}$

$=\frac{50\times50\times300}{2\times500\times500}=\frac{3}{2}=1.5\omega$

$\text{L}=1\text{ henry},\text{E}=50\text{V},\nu=\frac{50}{\pi}\text{Hz}$

1. $\text{I}_0=\frac{\text{E}_0}{\text{Z}}$

$\text{Z}=\sqrt{\text{R}^2+(\text{X}_\text{C}-{\text{X}_\text{L})}^{2}}$

$=\sqrt{(300)^2+\Big(\frac{1}{2\pi\text{fC}}-2\pi\text{fL}\Big)^2}$

$=\sqrt{(300)^2+\bigg(\frac{1}{2\pi\times\frac{50}{\pi}\times20\times10^{-6}}-2\pi\times\frac{50}{\pi}\times1\bigg)^2}$

$=\sqrt{(300)^2+\Big(\frac{10^4}{20}-100\Big)^2}=500$

$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{50}{500}=0.1\text{A}$

2. Potential across the capacitor = i₀ × X_C = 0.1 × 500 = 50V.

Potential difference across the resistor = i₀ × R = 0.1 × 300 = 30V.

Potential difference across the inductor = i₀ × X_L = 0.1 × 100 = 10V.

Rms. potential = 50V.

Net sum of all potential drops = 50V + 30V + 10V = 90V.

Sum or potential drops > R.M.S potential applied.