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JEE Mains 13-Apr-2023 Shift 2 Solved Paper
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© examsnet.com
Question : 50
Total: 90
In the network shown below, the charge accumulated in the capacitor in steady state will be :
[13-Apr-2023 shift 2]
4.8
µ
C
12
µ
C
7.2
µ
C
10.3
µ
C
Validate
Solution:
In steady state, no current will pass through capacitor hence capacitor will act as open circuit.
i
2
=
0
i
1
=
3
6
+
4
=
3
10
A
Potential difference on
6
Ω
resistor
=
6
×
3
10
=
1.8
volt capacitor will have same potential so charge
=
cv
=
4
×
1.8
=
7.2
µ
c
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