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JEE Advanced 2021 Full Test 4 Paper 2
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© examsnet.com
Question : 17
Total: 54
Find charge supplied by the cell after the switch is closed (in
µ
C
)
Your Answer:
Validate
Solution:
The potential difference across 1 µF and 2 µF (uppermost) capacitor must sum up to
q
1
+
q
∕
2
2
=
6
⇒
q
=
24
5
µ
C
∴
Answer is
12
+
24
5
=
16.8
µ
C
© examsnet.com
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