Thank you for reporting, we will resolve it shortly
Q.
If in the circuit shown, voltmeter reads 100 V then L is
Alternating Current
Solution:
This is feasible only at resonance.
$ \therefore \quad 500 = \frac{1}{2\pi\sqrt{LC}}$ or $\frac{1}{2\pi\sqrt{L\times10\times10^{-6}}} = 500$
On solving, we get $L = 0.01 \,H$