A circuit is constructed with four resistors one capacitor o

A circuit is constructed with four resistors, one capacitor, one battery and a switch as shown. The values for the resistors are: R_1 = R_2 = 58 Ohm. R_3 = 103 Ohm and R_4 = 97 Ohm. The capacitance is C = 58 muF and the battery voltage is V = 12 V. The positive terminal of the battery is indicated with a + sign. 1) The switch has been open for a long time when at time t = 0. the switch is closed, what is I_4(0). the magnitude of the current through the resistor R_4 just after the switch is closed? A 2) What is Q(infinity), the charge on the capacitor after the switch has been closed for a very long time? muC 3) After the switch has been closed for a very long time, it is then opened. What is Q(t_open), the charge on the capacitor at a time t_open = 639 mus after the switch was opened? muC 4) what is I_C,max(closed), the current that flows through the capacitor whose magnitude is maximum during the time when the switch is closed? A positive value for the current is defined to be in the direction of the arrcw shown. A 5) What is I_c,max(closed), the current that flows through the capacitor whose magnitude is maximum during the time when the switch is open? A positive value for the current is defined to be in the direction of the arrow shown. A

Solution

1)

I4(0) = V/Req

         = V/(R1 + R4 + 1/(1/R2 + 1/R3))

   = 12/(58+97+1/(1/58+1/103))

        =  0.0624 A

2)

No current flow through R2: so, the current is,

i = V/Req = V/(R1 + R4 + R3) = 12/(58+97+103) = 0.0465 A

voltage: V3 = i R3 = 0.0465*103 = 4.79 V

charge is, Q = C V3 = 58 microC * 4.79 = 277.86 uC

3)

The charge is,

   Q = Qo e^-t/(R2+R3)C

       = [277.86X10^-6] e^{-(639X10-6)/(58+103)(58X10-6)}

          = 2.6X10^-6 uC

4)

the current is,

    I = V3/R2 = 4.79/103 = 0.0465 A