A

Step 1: Calculate the resonance frequency (f₀).
The resonance frequency for an LCR circuit is given by:
$$ f_0 = \frac{1}{2 \pi \sqrt{LC}} $$
where L = 10 mH = 10 \times 10^{-3} H and C = 1 \mu F = 1 \times 10^{-6} F.
Plugging in the values:
$$ f_0 = \frac{1}{2 \pi \sqrt{10 \times 10^{-3} H \cdot 1 \times 10^{-6} F}} = \frac{1}{2 \pi \sqrt{10^{-8}}} = \frac{1}{2 \pi \cdot 10^{-4}} $$
$$ = \frac{10^4}{2 \pi} \approx 1591.55 Hz $$
Step 2: Calculate the frequency 10% lower than the resonance frequency.
Since we need a frequency that is 10% lower than the resonance frequency:
$$ f = f_0 \times 0.9 \approx 1591.55 \times 0.9 \approx 1432.40 Hz $$
Step 3: Calculate the impedance (Z) of the circuit at this frequency.
The impedance of the LCR series circuit is given by:
$$ Z = R + j \left( \omega L - \frac{1}{\omega C} \right) $$
where $$ \omega = 2 \pi f $$.
Calculate $$ \omega $$:
$$ \omega = 2 \pi \cdot 1432.40 \approx 8984.88 \, rad/s $$
Now calculate the inductive and capacitive reactances:
$$ X_L = \omega L = 8984.88 \cdot 10 \times 10^{-3} \approx 89.85 \Omega $$
$$ X_C = \frac{1}{\omega C} = \frac{1}{8984.88 \cdot 1 \times 10^{-6}} \approx 111.40 \Omega $$
Plugging these into the impedance formula:
$$ Z = 3 + j(89.85 - 111.40) = 3 - j21.55 $$
The magnitude of the impedance is:
$$ |Z| = \sqrt{3^2 + (-21.55)^2} \approx \sqrt{9 + 464.62} \approx \sqrt{473.62} \approx 21.77 \Omega $$
Step 4: Calculate the current amplitude (I₀).
Using Ohm's law: $$ I_0 = \frac{V_0}{|Z|} $$
where V₀ = 15 V (maximum voltage). So:
$$ I_0 = \frac{15}{21.77} \approx 0.688 A $$
Step 5: Calculate the average power dissipated per cycle.
The average power (P) in an AC circuit is given by:
$$ P = I_{rms}^2 R $$
where $$ I_{rms} = \frac{I_0}{\sqrt{2}} $$. Therefore:
$$ I_{rms} = \frac{0.688}{\sqrt{2}} \approx 0.486 A $$
Now calculating the power:
$$ P = (0.486)^2 \cdot 3 \approx 0.236 imes 3 \\approx 0.708 W $$
Final answer: The current amplitude is approximately 0.688 A and the average power dissipated per cycle is approximately 0.708 W.