Svar på KS2

Uppgift 1:

$R_T=5+10=15k\Omega$ 

$\tau=RC=15k\cdot15\mu=15\cdot10^3\cdot15\cdot10^{-6}=225\cdot10^{-3}=0,225s$ 

$v_C\left(t\right)=V_{Final}+\left(V_{Initial}-V_{Final}\right)e^{-\frac{t}{\tau}}=V_{Final}+\left(0-V_{Final}\right)e^{-\frac{t}{\tau}}=V_{Final}-V_{Final}e^{-\frac{t}{\tau}}=V_{Final}\left(1-e^{-\frac{t}{\tau}}\right)=20\left(1-e^{-\frac{t}{0,225}}\right)V$   

$i\left(t\right)=\frac{E-v_C\left(t\right)}{R_T}=\frac{20-20\left(1-e^{-\frac{t}{0,225}}\right)}{15k}=\frac{20-20+20e^{-\frac{t}{0,225}}}{15k}=\frac{20}{15}e^{-\frac{t}{0,225}}mA$ 

$v_R\left(t\right)=i\left(t\right)\cdot R_1=\frac{20}{15k}e^{-\frac{t}{0,225}}\cdot5k=\frac{20}{3}e^{-\frac{t}{0,225}}V$  

$v_C\left(0\right)=0v$   (based on theory)

$v_C\left(0,1\right)=20\left(1-e^{-\frac{0,1}{0,225}}\right)=20\left(1-e^{-0,44}\right)=20\left(1-0,641\right)=7,17V$  

$v_C\left(10\right)=20v$ (based on theory where 10s>5τ)

$v_C\left(\tau\right)=20\left(1-e^{-\frac{\tau}{\tau}}\right)=20\left(1-e^{-1}\right)=20\left(1-0,367\right)=20\cdot0,632=12,642V$  

Uppgift 2:

First we need to find reactance of all elements and transform the source form into phasor:

Since $e=21,21\sin\left(\omega t+60^\circ\right)$, then $E=\frac{21,21}{\sqrt{2}}\angle60^\circ=15\angle60^\circ$ . To transform it to rectangular form: $E=15\left(\cos60^\circ+\sin60^\circ\cdot i\right)=7.5+12.99i$  

$X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{1}{2\pi\cdot50\cdot637\mu}=5\Omega$  

$X_L=\omega L=2\pi fL=2\pi\cdot50\cdot32m=10\Omega$ 

Now we will nodal analysis with node V₁ on node where the capacitor, the inductor and the resistor are connected.

$-\frac{V_1-E}{Z_C}-\frac{V_1}{Z_L}-\frac{V_1}{R}=0$ 

$-\frac{V_1}{Z_C}+\frac{E}{Z_C}-\frac{V_1}{Z_L}-\frac{V_1}{R}=-\frac{V_1}{-5i}+\frac{7.5+12.99i}{-5i}-\frac{V_1}{10i}-\frac{V_1}{6}=\frac{V_1}{5i}-\frac{7.5+12.99i}{5i}-\frac{V_1}{10i}-\frac{V_1}{6}=0$     

$\frac{V_1}{5i}\cdot \:30i-\frac{7.5+12.99i}{5i}\cdot \:30i-\frac{V_1}{10i}\cdot \:30i-\frac{V_1}{6}\cdot \:30i=0$  (Multiply by LCM=30i)

$-5iV_1+3V_1-6\left(7.5+12.99i\right)=0$  (simplify)

$-5iV_1+3V_1-45-77.94i=0$  

$-5iV_1+3V_1=77.94i+45$   

$\frac{\left(-5i+3\right)V_1}{-5i+3}=\frac{77.94i}{-5i+3}+\frac{45}{-5i+3}$  

$V_1=-7.491 +13.495i$  

Now voltage on the resistor R is the same as V₁, so:

$V_R=-7.491 +13.495i=15.43\angle119.03^\circ$   V

$v(t)=15.43\sqrt{2}\sin\left(\omega t+119^\circ\right)=21.82\sin\left(\omega t+119^\circ\right)$   

We see here an interesting picture v_R(t): the amplitude of the voltage across an inductor is greater than the amplitude of the generator. This can be true in RLC circuits. As we know maximum current=(voltage from the generator)/(total impedance). So maximum voltage on the inductor is $V_L=\frac{I}{X_L}=\frac{\frac{e}{Z}}{X_L}=e\frac{X_L}{Z}$ 

Therefore if X_L is greater than Z, then the voltage across the inductor will be greater than e.

Uppgift 3:

For element X1:

Efficiency is measured as useful power divided by total power. The useful power is 15kW and we need to know the total power. So:

$P_1=\frac{15k}{0,9}=16,667kW$ 

$Q_1=0VAR$ 

$S_1=P_1=16,667kVA$ 

For element X2:

$P_2=10kW$ (given)

Since power factor is real power divided by apparent power, $S_2=\frac{P_2}{\cos\theta}=\frac{10k}{0,7}=14,286kVA$ 

$\theta=\cos^{-1}\left(0,7\right)=45,573^\circ$ 

$Q_2=S_2\sin\theta=14,286k\cdot0,714=10.2kVAR$ (inductive)

For element X3:

$S_3=20kVA$ (given)

$\theta=\cos^{-1}\left(0,8\right)=36,870^\circ$ 

$Q_3=S_3\sin\left(36,870^\circ\right)=20k\cdot0,6=12kVAR$ (inductive)

Since power factor is real power divided by apparent power, $P_3=S_3\cdot\cos\theta=20\cdot0,8=16kW$

 

Total real power: $P_T=P_1+P_2+P_3=16,667+10+16=42,667kW$  

Total reactive power: $Q_T=Q_1+Q_2+Q_3=0+10,2+12=22,2kVAR$  (inductive)

Total apparent factor: $S_T=\sqrt{P^2_T+Q^2_T}=\sqrt{42,667^2+22,2^2}=48,097kVA$  

(total power factor: $\cos\theta=\frac{P_T}{S_T}=\frac{42,667}{48,097}=0,887$   )

 

To raise power factor we need to add capacitor because the system is inductive

To raise power factor to 0,95, we need to find the total reactive power that will be in such system.

$\theta=\cos^{-1}\left(0,95\right)=18,195^\circ$

$S'_{improved}=\frac{P}{\cos\theta}=\frac{42,667}{0,95}=44,913kVA$  

$Q'_{imporved}=S'_{imporved}\sin\left(18,195^\circ\right)=44,913\cdot0,312=14,013kVAR$  

We need to remove $\Delta Q'_{imporved}=22,2-14,013=8,187kVAR$     

Alternative formula for reactive power is $Q=\frac{E^2}{X_C}$ , so $X_C=\frac{E^2}{Q}=\frac{E^2}{\Delta Q}=\frac{220^2}{8,187k}=5,912\Omega$  

Since $X_C=\frac{1}{2\pi fC}$ , then $C'=\frac{1}{2\pi fX_C}=\frac{1}{2\pi\cdot50\cdot5,912}=5,384\cdot10^{-4}=538\mu F$    

 

To raise power factor to 1, it means we need to remove entire Q_total

_{$X_C=\frac{220^2}{22,2k}=2,180\Omega$}  

_{$C''=\frac{1}{2\pi fX_C}=\frac{1}{2\pi\cdot50\cdot2,180}=1,460\cdot10^{-3}=1,5mF$}