An inductor and a capacitor are connected in an AC circuit. Which statement best explains why neither element dissipates average power?
AInductors and capacitors have zero resistance, so no power can flow through them
BThey store energy in fields during one half-cycle and return it to the circuit during the next, so net energy transfer averages to zero
CTheir reactance exactly cancels the applied voltage, preventing any current from flowing
DAC power is imaginary, so reactive elements don't interact with it
Inductors store energy in a magnetic field and capacitors in an electric field. On each half-cycle, energy flows into the element as the field builds; on the next half-cycle, the same energy flows back out as the field collapses. The average over a full cycle is zero net energy transfer. This is why reactive elements appear in the imaginary part of impedance — they cause phase shifts without real power dissipation. Only resistance converts electrical energy irreversibly into heat.
Question 2 Multiple Choice
At the resonant frequency ω₀ = 1/√(LC), what is the total impedance of a series RLC circuit?
AZero — the circuit has no opposition to current at resonance
BEqual to R — the inductive and capacitive reactances cancel each other exactly
CEqual to √(R² + (X_L − X_C)²), which is minimized but not zero
DInfinite — energy stored in L and C blocks current at this frequency
At resonance, X_L = X_C, so they cancel: the imaginary part of impedance is zero. The remaining impedance is purely resistive: Z = R. Impedance is at its minimum and current is at its maximum. Note that option A ('zero') would require R = 0, which only applies to an ideal superconducting circuit. In any real circuit R > 0, so impedance at resonance equals R, not zero. Option C is the general formula but at resonance it correctly reduces to R.
Question 3 True / False
A series RLC circuit with a high quality factor Q has a sharper, narrower resonance peak than the same circuit with a low Q factor.
TTrue
FFalse
Answer: True
Q = ω₀/Δω, where Δω is the bandwidth — the range of frequencies where power exceeds half its resonance maximum. High Q means small Δω: the circuit responds strongly only to frequencies very close to ω₀. Physically, high Q corresponds to low R (low damping): energy dissipates slowly, so near-resonant oscillations persist at nearly full amplitude. This selectivity is why high-Q circuits are used in radio tuners to isolate a single station from its neighbors.
Question 4 True / False
An ideal transformer can step up DC voltage just as effectively as it steps up AC voltage, since the turns ratio N₂/N₁ applies in either case.
TTrue
FFalse
Answer: False
Transformers operate on mutual inductance: a changing magnetic flux in the primary coil induces a voltage in the secondary. DC current produces a constant (not changing) magnetic field, generating no changing flux and therefore no induced voltage in the secondary. A DC source produces no transformer output. This is one of the historic reasons AC was chosen for the electrical grid: it can be efficiently transformed, while DC cannot (without modern solid-state power electronics).
Question 5 Short Answer
Explain why high-voltage AC transmission reduces energy losses compared to transmitting the same power at low voltage.
Think about your answer, then reveal below.
Model answer: Power loss in a transmission line is P_loss = I²R, where R is the line's fixed resistance. For a given power P = VI, increasing voltage V means decreasing current I proportionally. Since losses scale as I², doubling voltage halves current and cuts losses by 75%; multiplying voltage by 10 reduces losses by 99%. Transformers make this lossless voltage conversion possible for AC.
This is the dominant economic argument for AC power transmission. A step-up transformer near the generator raises voltage (and lowers current) for long-distance transmission; a step-down transformer near the end user restores safe voltage levels. The I²R loss formula directly connects transformer operation to resistive dissipation: you want to minimize I for a given power delivered, which means maximizing V.