C = 12 Farads fully charged from 0 to 24 Volts through 10 Ohm Resistance. How mu

Question

C = 12 Farads fully charged from 0 to 24 Volts through 10 Ohm Resistance. How much energy E in joules is contained in C? V1 is 240 volts RMS at zero phase, at 377 radians/second. Impedances are in Ohms. What is the phasor value of current i? specify magnitude and phase angle. In part (g) above, inductance value L is doubled. What is phasor value of I? V = 240 Volts; RI = R2 = R3 = 60 Ohms. C = 20 Milli Farads. Switches SI and S2 closed for long time; steady state reached. What are the values of currents II through RI, IC through C, and output voltage V0 in the steady state condition; S2 still closed, SI opened at time t =0. What are the values of l1, lC, and V0 soon after at att = 0+; What is the time-constant, tau under part (b)? under conditions of part (b) is Capacitor [charging, discharging, does not matter] After new steady state reached under part (b), S2 also opened. What are l1, lc, and V0, soon after S2 opened.

Explanation / Answer

(f) C = 12 F

V = 24 v

Now, stored energy = E = (1/2)*C*V^2 = 0.5 * 12 * 24 = 144 J

(g) Here R = 50 ohm

Xl = j*25 ohm

Xc = -j*50 ohm

So Z = R + Xl + Xc = 50 + j*25 - j*50 = 50 -j*25

V1 = 240 V

so, I = V1/Z = 240 / ( 50 - j*25) = 3.84 + j*1.92 = 4.2933 ( 13.2812 degree)

So, Magnitude = 4.2933 A

angle = 13.2812 degree

(h) now If L is double so the Xl is double

.: Xl = j*50 ohm

So, Z = 50 + j*50 -j*50 = 50 ohm

So, I = V1/Z = 240/50 = 4.8 A

So, Magnitude of I = 4.8 A

angle of I = 0 degree

(a) Here S1 and S2 is closed for long time so capacitor is worked as open circuit.

Now , total resistance = Req = R1 + R2 || R3 = 60 + 60 || 60 = 60 + 30 = 90 ohm

So, I1 = V/Req = 240/90 = 2.6667 A

Ic = 0 A

Now, I3 = current through R3 = I1/2 = 2.6667/2 = 1.3334 A

So, V0 = I3*R3 = 1.3334*60 = 80.004 V

(b) I1 = 0 A ( because S1 is open and no current is flow through open circuit)

Now Req = R2 || R3 = 60 || 60 = 30 ohm

Now, capacitor is charged to V0 = 80.004 V = Vc

So , Ic = Vc/Req = 80.004/30 = 2.6668 A

Now I3 = Ic/2 = 2.6668/2 = 1.3334 A

So, V0 = I3*R3 = 1.3334*60 = 80.004 V

(d) capacitor is discharging

(e) Now S1 and S2 both open so no current is flow through any resistor.

So, I1 = 0 A

Ic = 0 A

V0 = 0 V

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