JEE MAIN - Physics (2021 - 20th July Evening Shift - No. 24)
A series LCR circuit of R = 5$$\Omega$$, L = 20 mH and C = 0.5 $$\mu$$F is connected across an AC supply of 250 V, having variable frequency. The power dissipated at resonance condition is ______________ $$\times$$ 102 W.
Answer
125
Explanation
XL = XC (due to resonance)
Z = R so $${i_{rms}} = {V \over Z} = {V \over R}$$
$${{{V^2}} \over R} = {{250 \times 250} \over 5} = 125 \times {10^2}W$$
Z = R so $${i_{rms}} = {V \over Z} = {V \over R}$$
$${{{V^2}} \over R} = {{250 \times 250} \over 5} = 125 \times {10^2}W$$
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