wiz-icon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

We wish to obtain a capacitance of 5 μF, by using some capacitors, each of 2 μF. Then, the minimum number of capacitors required is

A
3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Not possible
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 4
As we can only use 2 μF capacitors to achieve the total capacitance of 5 μF. So, the best combination that can be made is
5 μF = 2×2 μF+1 μF
For this, we need to connect two capacitors of 2 μF each in parallel and connect another two capacitors of 2 μF each in series and further connect both set in parallel to achieve the capacitance of 5 μF.
Its initial digramatic representation can be shown as

Upon further connecting all these capacitors in the systematic manner, the final circuit can be shown as below:
To acheive 1 μF we need to connect 2 capacitors of 2 μF each in series.
11μF=12μF+12μF
So, from the diagram above, we can say the minimum number of capacitors required to acheive the capacitance of 5 μF is 4.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Series and Parallel Combination of Capacitors
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon