(a) Calculate the charge on the positive plate of the capacitorjust after the sw

Question

(a) Calculate the charge on the positive plate of the capacitorjust after the switch is closed.

Q(0+) = C     *
0   OK

HELP:  Can the charge on a capacitorchange discontinuosly?

(b) Calculate the following currents immediately after theswitch is closed.

I1(0+) = A    *
.122   OK

I2(0+) = A     *
0   OK

(c) After the switch has been closed for a long time, calculatethe following quantities.

I1 = A    *
.042   OK

I2 = A    *
.042   OK

Q = C     *
3.29E-5   OK

(d) At some yet later time T, the switch is againopened. Calculate the following currents immediatelyafterward.

I1(T+) = A    *
0   OK

I3(T+) = A   
0   NO

HELP:  At time T (justbefore the switch is opened), what is the voltage across thecapacitor? Can the charge on the capacitor change in theinfinitesimal time it takes to open the switch?
HELP:  Write down Kirchoff's Laws forthe circuit when the switch is open at time T+, given thatyou know the voltage across the capacitor at this time.

(e) What is the time constant of the dischargingcircuit?

t = s    
.0003   NO

HELP:  Redraw the original circuit,discarding all circuit elements thataresuperfluous after the switchis opened

E = 5 V
R1 = 41 W
R2 = 79 W
C = 10 µF

Explanation / Answer

   d.   Current   I3 (T+ )  =   VC /R2       Here   voltageacrosscapacitor   VC   =   Q/ C                                                                =   3.29* 10-5 / 10 * 10-6                                                                =   3.29   V       I3 (T +)   =   3.29 / 79                      =   4.16* 10-2   A                                                                =   3.29   V       I3 (T +)   =   3.29 / 79                      =   4.16* 10-2   A    e.   Once the switch is open,the circuit contains only C and R2.    Time constant of discharging circuit             =   R2* C             =   79* 10 * 10-6             =   7.9* 10-4   s             =   7.9* 10-4   s

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