A capacitance C 1 = 9.9 ?F is connected in series with a capacitance C 2 = 4.6 ?
ID: 2290662 • Letter: A
Question
A capacitance C1 = 9.9 ?F is connected in series with a capacitance C2 = 4.6 ?F, and a potential difference of 275 V is applied across the pair.
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?
Repeat for the same two capacitors but with them now connected in parallel.
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?
Explanation / Answer
In series
a)
equivalent Capacitance
1/Ceq=1/9.9 +1/4.6
Ceq=3.14 uF or 3.14*10-6 F
b)
Total Charge
Q=CeqV=275*(3.14*10-6)
Q=8.64*10-4 F or 863.68 uF
Charge on C1
Q1=Q=8.64*10-4 F or 863.68 uF
c)
Charge on C2 is
Q2=Q=8.64*10-4 F or 863.68 uF
d)
Voltage across C1 is
V1=Q1/C1=863.68/9.9
V1=87.24 Volts
e)
Voltage acorss C2 is
V2=Q2/C2=863.68/4.6
V2=187.76 Volts
----------------------------------------------------------------------------------------------------------------------------------------------------------
In parallel
a)
equivalent capacitance
Ceq=9.9+4.6=14.5 uF or 1.45*10-5 F
b)
Charge on C1 is
Q1=C1V =9.9*10-6*275
Q1=2.72*10-3 C or 2722.5 uC
c)
Charge on C2
Q2=C2V =4.6*10-6*275
Q1=1.265*10-3 C or 1265 uC
d)
Voltage across C1 is
V1=275 Volts
e)
Voltage across C2 is
V2=275 Volts
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.