Series circuit: same current $I$ through both inductors
$E_P = \frac{1}{2}L_P I^2 = \frac{1}{2}\times4\times I^2 = 2I^2$
$E_Q = \frac{1}{2}L_Q I^2 = \frac{1}{2}\times2\times I^2 = I^2$
$E_P/E_Q = 2I^2/I^2 = \mathbf{2}$
Answer: 2
Impedance: $Z = \sqrt{R^2+(X_L-X_C)^2}$ where $X_L=\omega L$, $X_C=1/\omega C$. Current: $I = V/Z$. Phase angle: $\tan\phi = (X_L-X_C)/R$. Power factor: $\cos\phi = R/Z$. At resonance ($X_L=X_C$): $Z=R$ (minimum), $I$ maximum, $\phi=0$ (purely resistive). Resonant frequency: $f_0 = 1/(2\pi\sqrt{LC})$.
Series (no mutual inductance): $L_{eq} = L_1+L_2+...$ (same current, voltages add). Parallel (no mutual inductance): $1/L_{eq} = 1/L_1+1/L_2+...$ (same voltage, currents add). With mutual inductance (series aiding): $L_{eq} = L_1+L_2+2M$. With mutual inductance (series opposing): $L_{eq} = L_1+L_2-2M$. Analogous to resistors for DC; valid for AC at any frequency.
Transformer uses mutual inductance. Ideal transformer: $V_s/V_p = N_s/N_p$; $I_p/I_s = N_s/N_p$; $P_{in}=P_{out}$. Power transmission: high voltage (low current) to reduce $I^2 R$ losses in transmission lines. India: 765 kV AC transmission. Step up at power station, step down near consumers. Without transformers, AC power transmission over long distances would be impractical (DC transmission is used for very long distances with HVDC technology).
Low-pass filter: passes low frequencies, blocks high. LC circuit: $f < f_c$ passes. High-pass: blocks low frequencies, passes high. Band-pass (resonant circuit): passes only frequencies near $f_0$, blocks others. Used in: radio tuning (variable capacitor selects resonant frequency = station frequency), audio crossover networks (separate bass/treble for speakers), signal processing, communication systems. Bandwidth = $R/L = f_0/Q$.
At resonance $X_L = X_C$, so the two reactances cancel and the impedance falls to its minimum value, $Z = R$. The current is then maximum and in phase with the voltage, giving a power factor of 1. The resonant frequency is $f_0 = \frac{1}{2\pi\sqrt{LC}}$, independent of R — resistance controls how sharp the resonance is, not where it occurs. This is the principle behind tuning a radio to one station out of many.
The quality factor $Q = \frac{1}{R}\sqrt{L/C}$ measures how sharply the circuit selects its resonant frequency. A large Q means a narrow bandwidth and a tall, thin resonance curve; a large R lowers Q and flattens the curve. Bandwidth is $\Delta f = f_0/Q$. A circuit designed to pick out one frequency needs high Q, while one designed to pass a range of frequencies needs low Q.
Average power in an AC circuit is $P = V_{rms}I_{rms}\cos\phi$, not simply VI. A purely inductive or purely capacitive circuit has $\cos\phi = 0$, so it consumes no average power at all — energy flows into the component for half a cycle and back out the next. Only resistance dissipates energy. Industrial installations improve a poor power factor by adding capacitors, because a low factor means large currents for the same useful power, and therefore larger $I^2R$ losses in the supply lines.
Adding reactances directly. $X_L$ and $X_C$ act in opposition, so they subtract before entering the Pythagorean expression for Z.
Using peak values where rms is required. Power formulas use rms values, and $V_{rms} = V_0/\sqrt{2}$. Mixing the two gives an answer wrong by a factor of 2.