Question

5. A resistance 250, inductance 200mH, and capacitance 1 uF capacitor are all connected in series, and across the combination is connected to a 500V, 60 Hz supply. Calculate (a) the current flowing, (b) the phase difference between the supply voltage and current, (c) the voltage across the coil and the capacitor. 6. An 300-uF capacitor is connected in series with 3.25-MQ resistance across a 300-V supply. The switch is closed. Calculate a) Time constant of the circuit b) Initial charging current and c) Voltage across the capacitor 10 milliseconds after the switch is closed

Answer 1

Solu The values provided in the question Rosestanco (R) = 25on Inductance (L) = 200 mat -200 *105 H capautance (c) = IME 1x10GE supply voltage (VB = 5oor, Go H I - noo cos [ 21 Go t] Vo = 500 case synt) v The circaut can as : be made The time transformed domain as : domain int circuit frequency has or to be phasor

500 cos ( 347 ) V → Vo = 500 10° 1 = 2:00x103 H 2 = jout * 377 X 200 XD I = jant 21 - 165.4190° c=1x18bf 1.c - The y x377* 1x 10-6 2659.52 c. 652.521-90° R = 250m ZR = 2500 as: can be transformed The circuit 76.4952 225 2. ? V DOLE

through the circuit The can current flowing be found as: IV 209 I ea = IR + 7 + Zc Therefore, Zega sanot 45.490 + 2652.52 1-90 = 250+ j 75 - j 2652.52 Zeg = 25o aj 2544.12 e Zea 25 89.22-9476 Therefore, I 500 10° E- 2589-22 1-84.46 b) The and I = 0.1931 |84-46 phase difference between the current supply voltage can be made as V = 500 10 I = 0.1931184.46° Phase difference = 0 ~ 84.46 phase difference = 84.46

the c) Voltage across capacitor can the be coil given and as: Voltage aur V = I IL in ou chance = (0.1931 134-46°) (45.41900) V = 14.56 117446 r. Voltage across} . Ve= II c capacet ance = (0.1931184.46°) (2652:52 1-90°) Ve = 512.21-5.har. 6) Solu The values provided in the question are Resistance CR ) = 3. 25 MR = 3.25 xroos Capacitan co (c) - 300 x 10-6F voltage source (4) - Boor.

The circuit can be made as: BooHF 3.25 MR BOOV a) Time can constant be found for the from the RC circuit formula as. Time constant (a) = RC = 3.2 n xlox 300 x 100 T arñ seconds. b) Initial charging current for the capacitor can be found as: At t=0, the voltage across capacitor is on and the circuit can be made as: voor 3.25 Mass f I initial

I initial = 300 3.25x100 I; = 92.3 1x10 6 A I i = 92.1 PA c) Voltage can be across capacitor at t=10ms obtained by the formula as. Ve Ct) = V(o) [ -e tolv At t=0, the voltage across capacitor js same as voltage source Therefore, v (s) : 300v. - 10x10-3 Then, V Coxig-3)= 300[1- e ann where l = 945 seconds. Va coms) = 0 = 300[1.096 6x10] V Cloms) = 3.07 69 mv

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