CBSE Class 12 Physics 2020 Outside Delhi Set 3 Paper

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Question : 10
Total: 13
You are given three capacitors of 2µF,3µF and 4µF, respectively.
(a) Form a combination of all these capacitors of equivalent capacitance ‌
13
3
µ
F
.
(b) What is the maximum and minimum value of the equivalent capacitance that can be obtained by connecting these capacitors ?
Solution:  
(a) 2µF and 4µF capacitors are to be connected in series. So, the effective capacitor will be
C‌eff ‌=‌
2×4
2+4
=‌
4
3
µ
F

With C‌eff ‌, the 3µF capacitor is to be connected in parallel. So, the equivalent capacitor will be
C‌eqv ‌=‌
4
3
+3
=‌
13
3
µ
F

(b) Maximum value of capacitance may be obtained by connecting the capacitors in parallel. In that case, the equivalent capacitance will be, (2+3+4)= 10µF.
Minimum value of capacitance may be obtained by connecting the capacitors in series. In that case, the equivalent capacitance will be CS.
‌
1
CS
‌
=‌
1
2
+‌
1
3
+‌
1
4
=‌
13
12

∴CS‌=‌
12
13
µ
F
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