The circuit a and b in fig contain capacitors that can be discharged. Answer the
ID: 1966379 • Letter: T
Question
The circuit a and b in fig contain capacitors that can be discharged. Answer the following questions for both circuits.
a) When the capacitor is fully charges, the battery is removed from the circuit. What is the charge on the capacitor as a function of time as the capacitor is discharging? Assume the battery is removed at time t=0.
b) What is the current through the resistor as a function of time as the capacitor is discharging?
c) At what time when the capacitor is discharging will the current through the resistor be 1microA.?
Explanation / Answer
a) Charge on the capacitor as the capacitor is discharging q ( t ) = Q e - t / RC __________________________________________ b) Current though the resistor as the capacitor is discharging I ( t ) = ( - Q / RC ) e - t / RC = - ( V / R ) e - t / RC ___________________________________________ For circuit (a) c) Capacitance of the capacitor is C = 2 F = ( 2 F ) ( 10-6 F / 1 F ) = 2 *10-6 F Resistance of the resistor R = 5 Voltage across the ciruit V = 9 V Time constant = RC = ( 5 ) ( 2 *10-6 F ) = 10-5 s Initial current Io = V / R = ( 9 V / 5 ) = 1.8 A time when the capacitor is discharging will the current through the resistor be 1 A = 1 * 10 -6 A I = I o e - t / RC 1 *10-6 A = ( 1.8 A ) e - t / RC 5.55*10-7 = e - t / RC - 14.403 = - t / RC t = ( 14.403 ) ( 10-5 s ) = 1.44*10-4 s _______________________________________________________________________ For circuit (b) Capacitance of the capacitor is C = 2 F = ( 2 F ) ( 10-6 F / 1 F ) = 2 *10-6 F Resistance of the resistor R = 4 Voltage across the ciruit V = 1.5 V Time constant = RC = ( 4 ) ( 2 *10-6 F ) = 8*10-6 s Initial current Io = V / R = ( 1.5 V / 4 ) = 0.375 A time when the capacitor is discharging will the current through the resistor be 1 A = 1 * 10 -6 A I = I o e - t / RC 1 *10-6 A = (0.375 A ) e - t / RC 2.66*10-6 = e - t / RC - 14.403 = - t / RC t = (-12.83 ) ( 8*10-6 s ) = 1.02*10-4 s - 14.403 = - t / RC t = ( 14.403 ) ( 10-5 s ) = 1.44*10-4 s _______________________________________________________________________ For circuit (b) Capacitance of the capacitor is C = 2 F = ( 2 F ) ( 10-6 F / 1 F ) = 2 *10-6 F Resistance of the resistor R = 4 Voltage across the ciruit V = 1.5 V Time constant = RC = ( 4 ) ( 2 *10-6 F ) = 8*10-6 s Initial current Io = V / R = ( 1.5 V / 4 ) = 0.375 A time when the capacitor is discharging will the current through the resistor be 1 A = 1 * 10 -6 A I = I o e - t / RC 1 *10-6 A = (0.375 A ) e - t / RC 2.66*10-6 = e - t / RC - 14.403 = - t / RC t = (-12.83 ) ( 8*10-6 s ) = 1.02*10-4 s Capacitance of the capacitor is C = 2 F = ( 2 F ) ( 10-6 F / 1 F ) = 2 *10-6 F Resistance of the resistor R = 4 Voltage across the ciruit V = 1.5 V Time constant = RC = ( 4 ) ( 2 *10-6 F ) = 8*10-6 s Initial current Io = V / R = ( 1.5 V / 4 ) = 0.375 A time when the capacitor is discharging will the current through the resistor be 1 A = 1 * 10 -6 A I = I o e - t / RC 1 *10-6 A = (0.375 A ) e - t / RC 2.66*10-6 = e - t / RC - 14.403 = - t / RC t = (-12.83 ) ( 8*10-6 s ) = 1.02*10-4 s - 14.403 = - t / RC t = (-12.83 ) ( 8*10-6 s ) = 1.02*10-4 sRelated Questions
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