A capacitance C 1 = 10.4 F is connected in series with a capacitance C 2 = 2.7 F

Question

A capacitance C1 = 10.4 F is connected in series with a capacitance C2 = 2.7F, and a potential difference of 225 V is applied across the pair.
Calculate the equivalent capacitance.

a. Calculate the equivalent capacitance.

b. What is the charge on C1?

c. What is the charge on C2?

d. What is the potential difference across C1?

e. What is the potential difference across C2?

f. Repeat for the same two capacitors but with them now connected in parallel. Calculate the equivalent capacitance.

g. What is the charge on C1?

h. What is the charge on C2?

i. What is the potential difference across C1?

j. What is the potential difference across C2?

Explanation / Answer

C1 = 10.4 uF    ,   C2 = 2.7 uF

a)

Series combination of C1 and C2 is given as

Ceq = C1 C2 / (C1 + C2) = 10.4 x 2.7 / (10.4 + 2.7) = 2.14 uF

b)

Q = Total charge coming from battery = Ceq V = 2.14 x 225 = 481.5 uC

so charge on C1 = Q1 = Q = 481.5 uC

c)

Q = Total charge coming from battery = Ceq V = 2.14 x 225 = 481.5 uC

so charge on C2 = Q2 = Q = 481.5 uC

d)

V1 = potential difference across C1 = Q1/C1 = 481.5 / 10.4 = 46.3 volts

e)

V2 = potential difference across C2 = Q2/C2 = 481.5 / 2.7 = 178.3 volts

f)

parallel combination of C1 and C2 is given as

Ceq = (C1 + C2) = (10.4 + 2.7) = 13.1 uF

g)

Charge on C1 = Q1 = C1V = 10.4 x 225 = 2340 uC

h)

Charge on C2 = Q2 = C2V = 2.7 x 225 = 607.5 uC

i )

v1 = 225

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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