Question please answer fast c) Determine the voltage across the load R for the supply voltage e(t) applied to the circuit shown in Figure Q1(c). Consider: e(t)=100+30sin(300t+π/6)+20sin900t+15sin(1500t−π/6)+10sin2100t (17 marks) Figure Q1(c)

P. 1e(t)\begin{array}{l}e(t)=\frac{100}{v_{1}}+30 \cdot \overbrace{\sin \left(300 t+\frac{\pi}{6}\right)}^{v_{2}}+\overbrace{20 \sin 900 t}^{v}+\overbrace{15 \sin \left(1500 t-\frac{\pi}{6}\right)}^{v 4} \\+\underbrace{10 \sin 2100 t}_{r_{5}} \\\end{array}e(t) is consist of voltage sources having d e(w=0) and A C component. A C sources also has diffrent frequancyHence voltage at across to will be calculated using superposition theorem\begin{array}{l}z_{1}=\gamma+j \omega L=10+j \omega\left(50 \times 10^{-3}\right)=10+j 0.05 \omega \\\bar{R}_{2}=\frac{\pi R \times \frac{1}{j \omega c}}{R+\frac{1}{j \omega c}}=\frac{R}{1+j \omega R C}=\frac{100}{1+j \omega \times 100 \times 100 \times 10^{-6}} \\z_{2}=\frac{100}{1+j 0.01 \omega} \\\therefore V_{0}=\frac{z_{2}}{z_{1}+z_{2}} e(t) \\\end{array}Honnorke.05\begin{array}{l}V_{0}=\frac{100 / 1+j 0.01 \omega}{(10+j 0.05 \omega)+\frac{100}{1+j 0.01 \omega}} e(t) \\V_{0}=\frac{100}{\left(110-0.0005 \omega^{2}\right)+j 0.15 \omega} e(t)\end{array}(p-2)case: 1 When e_{1}=100 \mathrm{~V} is active \omega=0\begin{array}{l}V_{01}=\frac{100}{110-j 0} \times 100 \\V_{01}=90.91 \mathrm{~V}\end{array}Case -2 When enly e_{2}=30 \sin \left(300 t+\frac{\pi}{6}\right) is active w=300\begin{aligned}V_{02} & =\frac{100}{\left[110-0(0.0005)(300)^{2}\right]+j 0.15 \times(300)} \times 30 \sin \left(300+\frac{\pi}{6}\right) \\& =\frac{100}{65+j 45} \times 30 \sin \left(300 t+90^{\circ}\right) \\& =\frac{100}{79.057 / 34.7^{\circ}} \times 30 \sin (300 t+30) \cdot\left[x+3 y=\sqrt{x^{2}+y^{2}+6+\frac{1}{x}}\right) \\V_{02} & =9795\left(300 t-4.7^{\circ}\right) \quad\left[\frac{A L \theta_{1}}{\theta \angle \theta_{2}}=\frac{A}{\theta} \angle \theta_{1}-\theta_{2}\right]\end{aligned}Case -3 when only e_{3}=20 \sin 900 t is active (\omega=900)\begin{aligned}V_{03} & =\frac{100}{\left(110-0.0005 \times(900)^{2}+j 015 \times(900)\right.} \times 20 \sin 900 t \\& =\frac{100}{-295+J 135} \times 20 \sin 900 t \\& =\frac{100}{324 \angle 155.4^{\circ}} \times 20 \sin 900 t \\V_{03} & \left.=6.1734 \sin 900 t+1554^{\circ}\right)\end{aligned}P3Case -4, when only e_{4}=15 \sin \left(1500 t-\frac{\pi}{6}\right) is adtive \left(\begin{array}{c}\omega=1500 \\ \text { radlse }\end{array}\right.\begin{aligned}V_{04} & =\frac{100}{\left[110-0.0005(1500)^{2}+j 0.15(1500)\right.} \times 15 \sin \left(1500 t-3^{\circ}\right) \\& =\frac{100}{-1015+j 225} \times 15 \sin \left(1500 t-30^{\circ}\right) \\& =\frac{100}{1039.64 L-1675^{\circ}} \times 15 \sin \left(1500 t-30^{\circ}\right) \\V_{84} & =1.44 \sin \left(1500 t+137.5^{\circ}\right) \mathrm{V}\end{aligned}Case-5: When only e_{5}=10 \sin 200 \mathrm{t} is active (Here \omega=2100 rad/s )\begin{aligned}V_{05} & =\frac{100}{\left[110-0.0005(2100)^{2}\right]+j 0.15(2100)} \times 10 \sin 200 t \\& =\frac{100}{-2095+j 315} \times 10 \sin 2100 t \\& =\frac{100}{2118.55 L-179.85^{\circ}} \times 10 \sin 2100 \mathrm{t} \\V_{05} & =0.472 \sin \left(2100 t+179.85^{\circ}\right)\end{aligned}from superposition theoren, voltage across R\begin{array}{l}v_{0}=v_{01}+v_{02}+v_{03}+v_{04}+v_{05} \\v_{0}=90.91+37.95\left(300 t-47^{\circ}\right)+6.173 \sin \left(900 t+1554^{\circ}\right) \\\quad+1.44 \sin \left(1500 t+137.5^{\circ}\right)+0472 \sin \left(2100 t+179.5^{\circ}\right)\end{array}.               ***** PLEASE ud83dude4f PLEASE ud83dude4f UPVOTE ***** ...