Two capacitance of 6\(\mu\)F and 8\(\mu\)F are connected in series. What additional capacitance must be connected in series with this combination to give a total of 3\(\mu\)F

De sage

1 year ago

let unknown be =R

1/6+1/8+1/R=1/3

14/48+1/R=1/3

7/24+1/R=1/3
factorize out

7R+24/24R=1/3
cross multiply
24R×1=3(7R+24)

24R= 21R+72
CLT
24R-21R=72

3R=72

divide both sides by 3
R=72/3

R=24

Velapearl

2 months ago

The question says:
You have 6μF and 8μF in series
Then you add another capacitor in series
Final total becomes 3μF
Find the new capacitor.
Step 3: First, combine 6μF and 8μF
Since they are in series, we use:
1/C = 1/6 + 1/8
Don’t rush. Do it calmly.
LCM is 24:
1/6 = 4/24
1/8 = 3/24
So:
1/C = 7/24
Flip it:
C = 24/7
So those two behave like 24/7 μF
Step 4: Now add the unknown capacitor
Let’s call it X
Still in series, so:
1/Total = 1/(24/7) + 1/X
Total is 3μF, so:
1/3 = 7/24 + 1/X
Step 5: Solve slowly
Convert 1/3:
1/3 = 8/24
So:
8/24 = 7/24 + 1/X
Subtract:
1/X = 1/24
Flip it:
X = 24 μF
Final Answer: 24 μF

DrUche

11 years ago

Incomplete question!

Rose_west

3 months ago

as in very interesting

aliyujibrinmagaji

4 months ago

Confused

Velapearl

2 months ago

or this one


1/3 = 1/6 + 1/8 + 1/x
Combine the easy ones first:
1/6 + 1/8
LCM = 24
= 4/24 + 3/24
= 7/24
So now:
1/3 = 7/24 + 1/x
Convert 1/3:
1/3 = 8/24
So:
8/24 = 7/24 + 1/x
Subtract:
1/x = 1/24
Flip:
x = 24 μF

Rose_west

4 months ago

thank you 💕

Myschool Kelly

9 years ago

Thanks for your contributions. The question has been modified.