A resistor with resistance 640 is in a series with a capacitor of capacitance 6.

Question

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

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