two capacitors have a capacitance is 5 micro farad when connected in parallel and 1.2 micro farade when connected in series. calculate their capacitance
Answers
Two capacitors have a capacitance is 5mF when connected in parallel.
Cp = C1 + C2
5 = C1 + C2 ...............(1st equation)
Two capacitors have a capacitance is 1.2mF when connected in series.
1/Cs = 1/C1 + 1/C2
1/1.2 = (C2 + C1)/C1C2
1.2 = C1C2/(C1 + C2) ........(2nd equation)
Substitute value of (1st equation) in (2nd equation)
1.2 = C1C2/5
1.2*5 = C1C2
6.0 = C1C2
C1C2 = 6
C1 = 6/C2
Substitute value of C1 in (1st equation)
→ 5 = 6/C2 + C2
→ 5 = [6 + (C2)²]/C2
→ 5C2 = 6 + (C2)²
→ (C2)² - 5C2 + 6 = 0
→ (C2)² - 3C2 - 2C2 + 6 = 0
→ C2(C2 - 3) -2(C2 - 3) = 0
→ (C2 - 3) (C2 - 2) = 0
On comparing we get,
→ C2 = 3mF and 2mF
Now, substitute value of C2 in C1
→ C1 = 6/3 and C1 = 6/2
→ C1 = 2mF and 3mF
Therefore, the value of capacitors are 2mF and 3mF.
Explanation:
Two capacitors have a capacitance is 5mF when connected in parallel.
Cp = C1 + C2
5 C1 C2....... ...(1st equation)
Two capacitors have a capacitance is 1.2mF when connected in series.
1/Cs 1/C1 + 1/C2
1/1.2 = (C2 + C1)/C1C2
1.2 = C1C2/(C1+ C2) ........ (2nd equation)
Substitute value of (1st equation) in (2nd
equation)
1.2 = C1C2/5
1.2*5 C1C2
6.0 = C1C2
C1C2 = 6
C1 = 6/C2
Substitute value of C1 in (1st equation)
56/C2 + C2
- 5=[6+ (C2)2]/C2
→ 5C2 = 6 + (C2)²
→ (C2)² - 5C2+6=0
→ (C2)² - 3C2 - - 2C2+6=0 ->
→C2(C2 - 3) -2(C2-3)=0
→ (C2 - 3) (C2 - 2) = 0
On comparing we get,
→ C2 = 3mF and 2mF
Now, substitute value of C2 in C1
→ C1 6/3 and C1 = 6/2
→ C1 = 2mF and 3mF
Therefore, the value of capacitors are 2mF and 3mF.