Question

In a series resonant circuit, having L, C and R as its elements, the resonant current is i. The power dissipated in circuit at resonance is ______.

Options

• `("i"^2"R")/((omega"L"-1//omega"C")`

• zero

• i²ωL

• i²R

Answer 1

In a series resonant circuit, having L, C and R as its elements, the resonant current is i. The power dissipated in circuit at resonance is i²R.

Explanation:

At resonance ωL = 1/ωC and i = E/R,

So power dissipated in circuit is P = i²R.