A resistor with resistance 640 is in a series with a capacitor of capacitance 6.0 x 10^-6F. What capacitance must be placed in parallel with the original capacitance to change the capacitive time constant of the combination to three times its original value? What??? A resistor with resistance 640 is in a series with a capacitor of capacitance 6.0 x 10^-6F. What capacitance must be placed in parallel with the original capacitance to change the capacitive time constant of the combination to three times its original value? What???
Explanation / Answer
R = 640 C = 6*10^-6 F time constant = CR = 3840*10^-6 s Let C' be the the capacitance to be placed parallel to C such that new time constant ' = 3 R*(C+C') = 3*CR RC + RC' = 3CR RC' = 2RC C' = 2C = 12*10^-6 F