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An LCR circuit has L = 10 mH, R = 3 Omeg...

An LCR circuit has L = 10 mH, `R = 3 Omega` and `C = 1 mu F` connected in series to a source of `15cos omega t`.volt. Calculate the current ampliuted and the average power dissipated per cycle at a frequency 10% lower than the resonance frequency.

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To solve the problem step by step, we need to calculate the current amplitude and the average power dissipated per cycle in an LCR circuit. ### Given Data: - Inductance, \( L = 10 \, \text{mH} = 10 \times 10^{-3} \, \text{H} \) - Resistance, \( R = 3 \, \Omega \) - Capacitance, \( C = 1 \, \mu\text{F} = 1 \times 10^{-6} \, \text{F} \) - Voltage source, \( V(t) = 15 \cos(\omega t) \, \text{V} \) - Frequency is 10% lower than the resonance frequency. ...
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