Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
The energy stored in the capacitor in the steady state is
Q. The energy stored in the capacitor in the steady state is
1736
185
Current Electricity
Report Error
A
338
μ
J
B
196
μ
J
C
98
μ
J
D
8
μ
J
Solution:
At steady state, capacitor acts as open circuit.
V
A
+
3
×
1
−
14
−
5
×
3
=
V
B
⇒
V
A
−
V
B
=
26
V
Energy stored in capacitor,
U
=
2
1
C
V
2
=
2
1
×
1
0
−
6
×
(
26
)
2
=
338
μ
J