Question 497: AC fundamentals

Question

For a series LCR circuit the power loss at resonance is

Accepted Answer

The impedance Z of a series LCR circuit is given by,
Z = $$\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} ight)}^2}} $$
At resonance, X
L
= X
C
, hence Z = R.
Let, supply voltage = V
R
= V
$$ 	herefore $$ R.M.S. current, I = $${V \over R}$$
$$ 	herefore $$ Power loss = VI = I
2
R

Answer

$${{{V^2}} \over {\left[ {\omega L - {1 \over {\omega C}}} \right]}}$$

Answer

$${I^2}L\omega $$

Answer

$${I^2}R$$

Answer

$${{{V^2}} \over {C\omega }}$$

AC fundamentals

For a series LCR circuit the power loss at resonance is

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