A circuit is constructed with four resistors, one capacitor, one battery and a s

Question

A circuit is constructed with four resistors, one capacitor, one battery and a switch as shown. The values for the resistors are: R1 = R2 = 59 ?, R3 = 65 ? and R4 = 152 ?. The capacitance is C = 79 ?F 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 I4(0), the magnitude of the current through the resistor R4 just after the switch is closed?

2) What is Q(?), the charge on the capacitor after the switch has been closed for a very long time?

3)after the switch has been closed for a very long time, it is then opened. What is Q(topen), the charge on the capacitor at a time topen = 553 ?s after the switch was opened?

4) What is IC,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 arrow shown.

5) What is IC,max(open), 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.

Explanation / Answer

Hi,

1) When the switch is just closed, maximum current flows through the circuit and capacitor is almost as a short circuit, which makes R2 parallel to R3. Hence the resultant resistance of the circuit is given by

      Req = R1 + (R2||R3) + R4 = 211.907 ohms.

      Hence the total current flowing through the circuit = I = V / Req                                                                                    = 12 / 211.907

                                                                                    = 0.0566 A

      This is the same current that passes through R1, parallel branches R2 + Capacitor and R3, and R4. Hence same current flows through R4. => I4 = 0.0566A

2) The voltage drop across R3 and the parallel section (R2 series with capacitor) is given by V3 = I * R3 = 0.0566 * 65 = 3.679 V.

    This is the same voltage drop across R2 and the capacitor C. Hence after the switch is closed for a long time, the capacitor will be charged to its maximum value given by

   Q = C * Vc = C * V3 = 79 * 10^(-6) * 3.679 = 290.641 * 10^(-6) C

3) If the switch is opened now, the capacitor starts discharging through R2 and R3. Now the time constant will be t = (R2 + R3)C = 9796 * 10^(-6) = 0.009796 sec.

    Hence the charge on the capacitor after a time 553 * 10^(-6) sec is given from equation

         Q(t) = Q * e^(-t/t) = 290.641 * 10^(-6) * e^(-0.000553/0.009796)

                                    = 290.641 * 10^(-6) * e^(-0.0564)   C

             4) Current will be maximum when the switch is just closed during charging and it decays slowly. Hence the maximum current is given by

     Im = V3 / R2 = 3.679 / 59 = 0.062355 A

5) During discharge, the maximum current that flows is given by

       Im = V3 / (R2 + R3)   (as the effective resistance now is the sum of R2 and R3).

   => Im = 3.679 / 124 = 0.02966 A

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