Five capacitor C1 = 2 uf , C2 = 4 uf , C3 = 6 uf , C4 = 5 uf and C5 = 10 uf are connected in Series and parallel as shown in Fig . Determine effective capacitance of above Fig .
Answers
Answer:
Given,
C1= 2 uf, C2 = 4 uf, C3 = 6 uf, C4= 5 uf C5 = 10 uf
Effective capacitance when connected in series
= 1/C1 +1/C2 +1/C3 +1/C4 +1/C5
= 1/2 +1/4 +1/6 +1/5 +1/10
= (30+15+10+12+6)/60
= 73/60 uf
Effective capacitance when connected in parallel = C1+C2+C3+C4+C5
= 2+4+6+5+10
= 27 uf
Answer:
The effective capacitance in series will be 0.822F and in parallel it will be 27F.
Explanation:
Case I:
Consider five capacitors of capacitance C1,C2, C3, C4 and C5connected in series. Let V be the potential difference across the series combination. Each capacitor will carry the same amount of charge q. Let V1,V2,V3,V4be the potential difference across the capacitors C1,C2,C3,C4,C5respectively. Thus V=V1+V2+V3+V4+V5.The potential across each capacitor will be,
V1=q/C1;V2=q/C2;V3=q/C3,V4=q/C4,V5=q/C5;
V=q/C1+q/C2+q/C3+q/C4=q(1/C1+1/C2+1/C3+1/C4+1/C5)
when CS will be the effective capacitance of the series combination, it should acquire a charge q when a voltage V is applied across it.
V=q(1/C1+1/C2+1/C3+1/C4+1/C5)
∴1/Cs=(1/C1+1/C2+1/C3+1/C4+1/C5)
now by substituting the value of C1,C2,C3, C4 and C5we will get,
Cs=0.822F
Case II:
Capacitors C1,C2,C3,C4 and C5connected in parallel. The same potential is applied across all the three capacitors. Let Q1,Q2,Q3,Q4 and Q5are charges on the three plates of capacitors such that
Q1=C1V
Q2=C2V
Q3=C3V
Q4=C4V
Q5=C5V
The total charge stored is Q=Q1+Q2+Q3+Q4+Q5and equivalent capacitance is Ceq=Q/V
So,Ceq=Q1+Q2+Q3+Q4+Q5/V
Ceq=C1+C2+C3+C4+C5
Ceq=27F