The circuit parameters in the diagram above have the followingvalues: E = 7 V, C

Question

The circuit parameters in the diagram above have the followingvalues:

E = 7 V, C = 7µF, L = 3 mH, R1 = 5 W, R2 = 7 W, and R3 = 7 W.

After it has been open for a long time, the switch is closed att = 0.

(a) Find the currents at t = 0+. (Current arrows in thecircuit diagram define conventions for positive directions).

I1(0+) = A    *
.583   OK

I2(0+) = A    *
0   OK

I3(0+) = A     *
.583   OK

HELP:  The presence of the inductorenforces I2(0+) = I2(0-) =0. The charge on the capacitor right after the switch is closedfollows from an analogous (but physically distinct) argument:Q(0+) = Q(0-) = 0. Hence the voltage drop of thecapacitor is zero at t = 0+.

(b) At what rate is I2 increasing rightafter the switch is closed?

dI2/dt(0+) = A/s    
0   NO

HELP:  From the structure of thecircuit and the knowledge I2(0+) = 0,VC(0+) = 0, you can see right away that thevoltage drop across the inductor equals the voltage drop acrossR3 at t = 0+.
HELP:  Use the above observationtogether with your results in part (a) and the basic equation forthe voltage drop across an inductor to solve for the currentderivative.

(c) After the switch has been closed for a long time, how muchenergy is stored in the inductor?

UL = J     *
.51E-3   OK

(d) After the switch has been closed for a long time, how muchenergy is stored in the capacitor?

UC = J     *
58.2E-6   O

(e) After the switch has been closed for a long time, it isopened at time T. Calculate the currentsI2 and I3 immediatelyafterward.

I2(T+) = A   
0   NO

I3(T+) = A    
0   NO

HELP:  The inductor enforcesI2(T+) =I2(T-), the latter being a quantityyou know from part (c).
HELP:  With the switch open, the circuitreduces to a simple one-loop geometry. The rest isstraightforward

Explanation / Answer

A. At t=0, I2 = 0 and V(C) = 0, so I3 = E/(R1+R3) = 8/13 A, thus I1= 8/13 A B. At t=0, I2 = 0 and V(C) = 0, so V(L) = ER3/(R1+R3) = 64/13 V,and LdI2/dt = V(L) ==> dI2/dt = V(L)/L = 1641.0 A/s C. At t=(large number), I3 = 0 and V(L) = 0, so I2 = E/(R1+R2) =8/11 A, and U = I2^2*L/2 = 7.9339E-4 J D. At t=(large number), I3 = 0 and V(L) = 0, so V(C) = ER2/(R1+R2)= 48/11 V, and U = V(C)^2*C/2 = 8.5686E-5 J E. Immediately after the switch is opened, current I2 continues toflow, but now it flows in the loop comprising L, C, R3 and R2. Thiscurrent is in the same (downward) direction through R2 as before,and in the upward direction through R3. I2 = 8/11 A, I3 = -8/11A

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