Physics 3 Marks Question

$\text{V}_1=10\times10^{-3}\text{V}$

$\text{R}=1\times10^3\Omega$

$\text{C}=10\times10^{-9}\text{F}$

1. $\text{X}_\text{C}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\text{fC}}$

$=\frac{1}{2\pi\times10\times10^3\times10\times10^{-9}}$

$=\frac{1}{2\pi\times10^{-4}}$

$=\frac{10^4}{2\pi}=\frac{5000}{\pi}$

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

$=\sqrt{\big(1\times10^3\big)^2+\Big(\frac{5000}{\pi}\Big)^2}$

$=\sqrt{10^6+\Big(\frac{5000}{\pi}\Big)^2}$

$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{\text{V}_1}{\text{Z}}$

$=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{5000}{\pi}\Big)^2}}$

2. $\text{X}_\text{C}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\text{fC}}$

$=\frac{1}{2\pi\times10^5\times10\times10^{-9}}$

$=\frac{1}{2\pi\times10^{-3}}=\frac{10^3}{2\pi}=\frac{500}{\pi}$

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

$=\sqrt{\big(10^3\big)^{2}+\Big(\frac{500}{\pi}\Big)^2}$

$=\sqrt{10^6+\Big(\frac{500}{\pi}\Big)^2}$

$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{\text{V}_1}{\text{Z}}$

$=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{500}{\pi}\Big)^2}}$

$\text{V}_0=\text{I}_0\text{X}_\text{C}=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{500}{\pi}\Big)^2}}\times\frac{500}{\pi}$

$=1.6124\text{V}\approx1.6\text{mV}$

3. $\text{f}=1\text{MHz}=10^6\text{Hz}$

$\text{X}_\text{C}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\text{fC}}$

$=\frac{1}{2\pi\times10^6\times10\times10^{-9}}$

$=\frac{1}{2\pi\times10^{-2}}=\frac{10^2}{2\pi}=\frac{50}{\pi}$

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

$=\sqrt{\big(10^3\big)^{2}+\Big(\frac{50}{\pi}\Big)^2}$

$=\sqrt{10^6+\Big(\frac{50}{\pi}\Big)^2}$

$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{\text{V}_1}{\text{Z}}$

$=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{50}{\pi}\Big)^2}}$

$\text{V}_0=\text{I}_0\text{X}_\text{C}=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{50}{\pi}\Big)^2}}\times\frac{50}{\pi}$

$\approx1.16\mu\text{V}$

4. $\text{f}=10\text{MHz}=10^7\text{Hz}$

$\text{X}_\text{C}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\text{fC}}$

$=\frac{1}{2\pi\times10^7\times10\times10^{-9}}$

$=\frac{1}{2\pi\times10^{-1}}=\frac{10}{2\pi}=\frac{5}{\pi}$

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

$=\sqrt{\big(10^3\big)^{2}+\Big(\frac{5}{\pi}\Big)^2}$

$=\sqrt{10^6+\Big(\frac{5}{\pi}\Big)^2}$

$\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{\text{V}_1}{\text{Z}}$

$=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{5}{\pi}\Big)^2}}$

$\text{V}_0=\text{I}_0\text{X}_\text{C}=\frac{10\times10^{-3}}{\sqrt{10^6+\Big(\frac{5}{\pi}\Big)^2}}\times\frac{5}{\pi}$

$\approx16\mu\text{V}$

$\text{r}=4\Omega,\text{I}_\text{rms}=6\text{A}$

$\text{R}=\frac{\text{E}}{\text{I}}=\frac{24}{6}=4\Omega$

Internal Resistance $=4\Omega$

Hence net resistance $=4+4=8\Omega$

$\therefore$ Current $=\frac{12}{8}=1.5\text{A}$

$\phi=\frac{\pi}{2}$ so $\text{P}=0.$

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

$\text{X}_\text{L}=\omega\text{L}$

$=100\times\frac{5}{1000}=0.5\Omega$

$=\text{i}=\frac{\in_0}{\text{X}_\text{L}}=\frac{10}{0.5}=20\text{A}$

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

$\text{X}_\text{L}=\omega\text{L}$

$=500\times\frac{5}{1000}=2.5\Omega$

$=\text{i}=\frac{\in_0}{\text{X}_\text{L}}=\frac{10}{2.5}=4\text{A}$

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

$\text{X}_\text{L}=\omega\text{L}$

$=1000\times\frac{5}{1000}=5\Omega$

$=\text{i}=\frac{\in_0}{\text{X}_\text{L}}=\frac{10}{5}=2\text{A}$

$=\text{I}=\text{I}_0\sin\omega\text{t}$

Peak value $\text{I}=\frac{\text{I}_0}{\sqrt{2}}$

$\frac{\text{I}_0}{\sqrt{2}}=\text{I}_0\sin\omega\text{t}$

$\Rightarrow\ \frac{1}{\sqrt{2}}=\sin\omega\text{t}=\sin\frac{\pi}{4}$

$\Rightarrow\ \frac{\pi}{4}=\omega\text{t}$

Or $\text{t}=\frac{\pi}{400}$

$=\frac{\pi}{4\times2\pi\text{f}}=\frac{1}{8\text{f}}$

$=\frac{1}{8\times50}=0.0025\text{s}=2.5\text{ms}$

$\text{R}=\frac{\text{V}^2}{\text{P}}=\frac{220\times220}{60}$

$=806.67$

$\in_0=\sqrt{2}\text{E}=1.414\times220$

$=311.08$

$\text{I}_0=\frac{\in_0}{\text{R}}=\frac{806.67}{311.08}$

$=0.385\approx0.39\text{A}$

$\text{i}^2\text{Rt}=\text{i}^2_\text{rms}\text{RT}$

$\Rightarrow\ \frac{\text{E}^2}{\text{R}^2}=\frac{\text{E}^2_\text{rms}}{\text{R}^2}$

$\Rightarrow\ \text{E}^2=\frac{\text{E}_0^2}{2}$

$\Rightarrow\ \text{E}_0{^2}=2\text{E}^2$

$\Rightarrow\ \text{E}_0{^2}=2\times12^2=2\times144$

$\Rightarrow \text{E}_0=\sqrt{2\times144}=16.97\approx17\text{V}$

$\text{C}=20\mu\text{F}=20\times10^{-6}\text{F}$

$\text{L}=1\text{H},\text{Z}=500$ $(\text{from}\ 14)$

$\in_0=50\text{V},\text{I}_0=\frac{\text{E}_0}{\text{Z}}=\frac{50}{500}=0.1\text{A}$

Electric Energy stored in Capacitor $=\Big(\frac{1}{2}\Big)\text{CV}^2$

$=\Big(\frac{1}{2}\Big)\times20\times10^{-6}\times50\times50$

$=25\times10^{-3}\text{J}=25\text{mJ}$

Magnetic field energy stored in the coil $=\Big(\frac{1}{2}\Big)\text{LI}_0{^2}$

$=\Big(\frac{1}{2}\Big)\times1\times(0.1)^{2}$

$=5\times10^{-3}\text{J}=5\text{mJ}$