12.) Two capacitors, C 1 = 4.00 µF and C 2 =15.0 µF, are connected in parallelan

Question

12.) Two capacitors, C1 = 4.00 µF and C2 =15.0 µF, are connected in paralleland the resulting combination is connected to a 9.00 V battery.
(a) What is the equivalent capacitance of thecombination?
_______________ µF

(b) What is the potential difference across each capacitor?
C1 = __________ V
C2 = __________ V

(c) What is the charge stored on each capacitor?
C1 = __________ µC
C2 = __________ µC (a) What is the equivalent capacitance of thecombination?
_______________ µF

(b) What is the potential difference across each capacitor?
C1 = __________ V
C2 = __________ V

(c) What is the charge stored on each capacitor?
C1 = __________ µC
C2 = __________ µC

Explanation / Answer

given that C1= 4.00F C2 = 15.0F they are connected parallel Ceq = C1 + C2           =19F V = 9.00v they are connected in parallel so potential differencein each capacitor is same which is C2 = 9.0v C1= 9.0v charge on 4.00F capacitor Q = C1V      = 4.00F * 9.0 v=36C Q = C2V      = 15.00F * 9.0 v=135C      = 15.00F * 9.0 v=135C

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