Two capacitors, with capacitances of 4.00uF and 5.80 uF are connected in paralle

Question

Two capacitors, with capacitances of 4.00uF and 5.80 uF are connected in parallel across a 660-V supply line.
The charged capacitors are disconnected from the line and from each other, and then reconnected to each other with terminals of unlike sign together.
the charge q on 4.00=4560uC
the charge q on 5.80=3828uC

Find the final charge on the 4.00uF capacitor.
Find the final charge on the 4.00uF capacitor.
Find the final charge on the 5.80uF capacitor.
Find the final voltage across the 4.00uF capacitor.
Find the final voltage across the 5.80uF capacitor.

Explanation / Answer

initial charge of C1 = C1V1 = 4.0 uF *V1 VOLTS =4.3*10-6*V1 initial charge of C2 = C2V2 = 2.0 uF *V2 VOLTS = 5.8*10-6*V2 IF THEY ARE CONNECTED IN parallel THEN voltage IS SAME V1 =V2 V1 = V2 = 660 volts TOTAL INITIAL CHARGE = C2V2-C1V1= 5.8*10-6*(660) -4.3*10-6*(660) = (990)*10-6C IF They are then connected positive plate to negativeplate and negative to positivetive plate. CHARGE FLOWS FROM HIGHER POTENTIAL TO LOWER POTENTIAL UNTILCOMMON POTENTIAL IS ACHIEVED LET COMMON POTENTIAL BE V THEN C2V2-C1V1 = (C1+C2)V SOLVE FOR V = 98.01 Volts FINAL CHARGE ON C1 is q1 = C1*V PLUG THE VALUES OF C1 AND V WE GET THE ANSWER

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