In an L–C circuit shown in figure,
C = 1F, L = 4H.
At time t = 0, charge in the capacitor is 4C and it is decreasing at a rate of 
.
(i) Maximum charge in the capacitor can be–
Text Solution
Verified by Experts(i) Sol; (ii) Sol; (iii) Sol
Ans.
(i)
Sol. At the given instant
i = 
= 
A
From conservation of energy
= 
+ 
Li 2 or q max = 
= 
= 6C
(ii)
Sol. We have to find time taken from q = 4 C to
q = 6C. This we can find by calculating time from 6C to 4C and then subtract it from 
Here, ω = 
= 
rad/s
∴ 
= cos 
or t = 2 cos
–1 
T = 
= (4π) sec
∴ Desired time is 
– t = 
sec
(iii)
Sol. i max = (q max ω) = (6) 
= 3A [as v max = ωA]
i = ω 
or 
= 
q = 
q max = 0.86 q max
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