A capacitor (without dielectric) is discharging through a resistor. At same instant a dielectric is inserted between the plates, then-
Text Solution
Verified by ExpertsThe correct answer is:
D
Let the capacitance before insertion of dielectric be C and the resistance be R.
∴ q = q 0
and i = 
Just after insertion of dielectric the capacitance increases.
The charge just after insertion of dielectric remains same, but the current decreases
⇒ and are false.
The energy stored in capacitor is
, hence energy decreases.
⇒ is false.
The time constant is RC and hence increases.
⇒ is true.
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