A 0.50- F and a 1.4- F capacitor ( C 1 and C 2, respectively) are connected in s

Question

A 0.50-F and a 1.4-F capacitor (C1 and C2, respectively) are connected in series to a 13-Vbattery.Express your answers using two significant figures separated by a comma.

Part A)Calculate the potential difference across each capacitor.

V1, V2= ___?V

Part B)Calculate the charge on each capasitor.

Q1, Q2= ____C

Part C)Calculate the potential difference across each capacitor assuming the two capacitors are in parallel

V1, V2= ___V

Part D) Calculate the charge on each capasitor assuming the two capacitors are in parallel

Q1, Q2=___?C

Explanation / Answer

here,

C1 = 0.50 uF

C2 = 1.4 uF

V = 13 V

(a)

equivalent capacitance , 1/C = 1/C1 + 1/C2

1/C = 1/1.4 + 1/0.5

C = 0.37 uF

total charge , Q = C*V

Q = 0.37 * 10^-6 * 13

Q = 4.81 * 10^-6 C

as charge remains constant in series

potential difference across C1 , V1 = Q /C1

V1 = 4.81 * 10^-6 / 0.5 * 10^-6

V1 = 9.62 V

potential difference across C2 , V2 = Q /C2

V2 = 4.81 * 10^-6 / 1.4 * 10^-6

V1 = 3.44 V

(b)

charge on each capacitor is equal , Q1 = Q2 = 4.81 * 10^-6 C

(c)

the potential difference across each capacitor assuming the two capacitors are in parallel is same

i.e V1 = V2 = Vbattery = 13 V

(d)

equivalent capacitance when they are in parallel

C = C1 + C2

C = 1.4 + 0.5

C = 1.9 uF

charge across C1 , Q1 = C1 * V

Q1 = 0.5 * 13 * 10^-6

Q1 = 6.5 * 10^-6

charge across C2 , Q2 = C2 * V

Q2 = 1.4 * 10^-6 * 13

Q2 = 18.2 * 10^-6 C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.