A 2.60 M Ohm resistor and a 3.00 mu F capacitor are connected in series with an

Question

A 2.60 M Ohm resistor and a 3.00 mu F capacitor are connected in series with an ideal battery of E = 6.00 V. At 1.00 s after the connection is made, what is the rate at which the following take place? the charge of the capacitor increases C/s energy is stored in the capacitor W thermal energy appears in the resistor W energy is delivered by the battery W Did you use the charging equation for q? How is q related to the potential energy of the capacitor? How do you get a function for the current i from the function for q? How is the rate of dissipation in the resistor related to the current? How is the rate of energy supplied by the battery related to the current?

Explanation / Answer

(a) Charge on capacitor q = CV(1-e^(-t/RC)) By calculation we get q = 2.06 x 10^-6 C (b) Energy = q^2/2C = 2.06 x 10^-6 / 2*3*10^-6 = 0.34 J (c) i = V/Re^(-t/RC) By calculation, i = 2 x 10^-6 A So, P = i^2Rt P= 4 X 10^-12 *2.6 x 10^6 P = 1.04 x 10^-5 W

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