A 5 microfarad capacitor is being discharged through an 8 x10^5 ohm resistor. (A

Question

A 5 microfarad capacitor is being discharged through an 8 x10^5 ohm resistor. (A) After how many seconds does the chargeon the capacitor drop to 1/4 of its original value? (B) If theoriginal voltage is 6V, what is the voltage across the capacitorwhen it's charge is 1/4 of its original charge? A 5 microfarad capacitor is being discharged through an 8 x10^5 ohm resistor. (A) After how many seconds does the chargeon the capacitor drop to 1/4 of its original value? (B) If theoriginal voltage is 6V, what is the voltage across the capacitorwhen it's charge is 1/4 of its original charge?

Explanation / Answer

In RC circuit charge at time t is q = Q [ e-t / RC ] Given q = Q / 4 So, 1/4 = [ e -t / RC ] -t / RC = -1.38629           t =1.38629 RC where R = reistance = 8 * 10 ^ 5 ohm            C= capacitance = 5 * 10 ^ -6 F substitue values weget t value (B). Voltage V = V o [ e -t / RC] where V o= 6 V and   time t = t    calculatedabove substitue values weget answer

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