A 19.0-F capacitor and a 44.0-F capacitor are charged by being connected across

Question

A 19.0-F capacitor and a 44.0-F capacitor are charged by being connected across separate 45.0-V batteries. (a) Determine the resulting charge on each capacitor. (Give your answer to at least three significant figures.) 19.0-F capacitor .855 mC 44.0-F capacitor 1.980 mC (b) The capacitors are then disconnected from their batteries and connected to each other, with each negative plate connected to the other positive plate. What is the final charge of each capacitor? 19.0-F capacitor C 44.0-F capacitor C (c) What is the final potential difference across the 44.0-F capacitor? Comment

Explanation / Answer

(a) Charge on 19.0 uF capacitor, Q= C x V = 19x10^-6x45 = 8.55x10^-4 C = 0.855 mC.

Charge on 44.0 uF capacitor, Q= C x V = 44x10^-6x45 = 1.98x10^-3 C = 1.980 mC

(b) Final charge of each capacitor = (0.855+1.980)/2 = 1.4175 mC.

(c) Final potential difference across 44.0 uF capacitor, V = Q/C = (1.4175x10^-3)/(44.0x10^-6) = 32.216 V.

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