AC Concepts: RMS & Frequency - Physics 2

1. What is the ratio of the average power delivered by Generator 2 to that delivered by Generator 1?
   Answer: $\frac{P_2}{P_1} = 2$, because $P \propto V_{rms}^2 \propto V_0^2$ and $V_2^2 = 2V_1^2$; frequency does not appear in the time-averaged power formula for a resistive load. Whenever you see a question comparing power from AC sources, your first instinct should be to reach for the time-averaged power formula for a resistive load: $P = \frac{V_{rms}^2}{R}$, where $V_{rms} = \frac{V_0}{\sqrt{2}}$. Notice what's in that formula — voltage and resistance only. Frequency is nowhere to be found. For Generator 1: $P_1 = \frac{V_{rms,1}^2}{R} = \frac{V_1^2}{2R}$. For Generator 2, with $V_2 = \sqrt{2}\,V_1$: $P_2 = \frac{(\sqrt{2}\,V_1)^2}{2R} = \frac{2V_1^2}{2R}$. Taking the ratio gives $\frac{P_2}{P_1} = 2$, confirming that answer A is correct. The frequency $f_2 = 2f_1$ plays no role whatsoever — a resistor dissipates energy based on the instantaneous voltage across it, and when you time-average a sinusoid squared, the frequency cancels out completely. Answer B is a classic trap: it assumes frequency somehow multiplies power because "more cycles per second means more energy delivery." This confuses frequency with power — a 60 Hz source and a 120 Hz source driving the same resistor at the same $V_{rms}$ deliver identical average power. Answer C builds on the same misconception, explicitly writing $P \propto V_0^2 f$, which is simply not a valid formula for resistive loads. Answer D correctly converts to $V_{rms}$ but then incorrectly treats that as the final ratio rather than squaring it — power depends on $V_{rms}^2$, not $V_{rms}$. Your go-to study tip: for any AC resistive power question, write down $P = V_{rms}^2/R$ immediately and check whether frequency even appears. If it doesn't, eliminate every answer choice that factors it in.
2. Which expression correctly gives the rms value of this current?
   Answer: $I_{rms} = \sqrt{I_{DC}^2 + \dfrac{I_0^2}{2}}$, because the DC and sinusoidal components are orthogonal over one period (their cross term integrates to zero), so their squared rms values add independently. Whenever you see a waveform that mixes a DC offset with an AC component, resist the urge to simply add amplitudes — the rms value demands integration from the definition. Start from first principles: $I_{rms} = \sqrt{\frac{1}{T}\int_0^T i^2(t)\, dt}$. Squaring $i(t) = I_{DC} + I_0\sin(\omega t)$ gives three terms: $I_{DC}^2$, $2I_{DC}I_0\sin(\omega t)$, and $I_0^2\sin^2(\omega t)$. Over exactly one period, the middle cross term — a pure sinusoid — integrates to zero. The DC term averages to $I_{DC}^2$, and the squared sine averages to $I_0^2/2$. Combining: $I_{rms} = \sqrt{I_{DC}^2 + \dfrac{I_0^2}{2}}$, confirming C is correct. The DC and sinusoidal components are mathematically "orthogonal" over one period, so their power contributions simply add. A is wrong because the factor $1/\sqrt{2}$ only converts a single sinusoid's peak to rms — it cannot be applied to an arbitrary peak amplitude of a non-sinusoidal waveform. The expression $\sqrt{I_{DC}^2 + I_0^2}$ is not a meaningful "combined peak." B is wrong because rms values don't add arithmetically. Power (proportional to $I^2$) adds, not current amplitude. Adding $I_{DC} + I_0/\sqrt{2}$ ignores this and overestimates the rms. D is wrong in its reasoning: the cross term $2I_{DC}I_0\sin(\omega t)$ averages to zero over a full period — not a nonzero value — so it never shifts the result above $\sqrt{I_{DC}^2 + I_0^2/2}$. Study tip: For any composite waveform, always expand $i^2(t)$ and integrate term by term. Cross terms between DC and AC vanish over a full cycle — this is the key insight that makes power from independent sources additive.
3. How do the transmission line losses ($P_{loss} = I_{rms}^2 R_{line}$) change under each option?
   Answer: Option 1 reduces losses by a factor of 4; Option 2 reduces losses by a factor of 4. Both options achieve the same goal through different mechanisms: Option 1 reduces current by halving it (since $P = V_{rms} I_{rms} \cos\phi$ with constant $P$ and $\cos\phi$), and Option 2 reduces current by halving it (since doubling $\cos\phi$ with constant $P$ and $V_{rms}$ halves $I_{rms}$). Whenever you see a question about transmission line losses, anchor your thinking to two equations simultaneously: the power equation $P = V_{rms} I_{rms} \cos\phi$ and the loss equation $P_{loss} = I_{rms}^2 R_{line}$. The key variable connecting them is $I_{rms}$ — anything that reduces current dramatically cuts losses even more dramatically, since losses go as current squared. Option 1 doubles $V_{rms}$ while holding real power $P$ and $\cos\phi$ constant. Solving for current: $I_{rms} = \frac{P}{V_{rms}\cos\phi}$. Doubling $V_{rms}$ halves $I_{rms}$, so losses scale as $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ — a factor-of-4 reduction. Option 2 doubles $\cos\phi$ while holding $P$ and $V_{rms}$ constant. Using the same formula, doubling $\cos\phi$ also halves $I_{rms}$, giving the same $\frac{1}{4}$ reduction in losses. Both options achieve identical loss reduction through different physical mechanisms. This makes B correct. Choice A is wrong because it incorrectly claims Option 2 only halves losses. Doubling $\cos\phi$ halves $I_{rms}$ fully — not partially — producing a factor-of-4 reduction, not 2. Choice C is wrong about Option 1: doubling voltage halves current, and since losses depend on $I_{rms}^2$, losses drop by four, not two. Choice D is wrong because it treats power factor improvement as purely redistributing power without changing $I_{rms}$. In reality, adding capacitors reduces the total current drawn from the line, directly lowering $I_{rms}$ and transmission losses. Study tip: Always convert everything to $I_{rms}$ first, then square it for losses — the squaring step is where students most often lose points.
4. If the frequency is increased from $\omega$ to $4\omega$ while $V_{rms}$ and all component values are held fixed, by what factor does the average power delivered to the circuit change?
   Answer: The average power decreases by a factor less than 16, because $P = V_{rms}^2 R / Z^2$ where $Z^2 = R^2 + (4\omega L)^2$; the exact factor depends on the ratio $\omega L / R$, but $P$ decreases and is not simply related to the frequency by a fixed numerical factor. When analyzing AC circuits with frequency changes, your starting point should always be the average power formula: $P = \frac{V_{rms}^2 R}{Z^2}$, where $Z^2 = R^2 + X_L^2$. This formula reveals that power depends on both the resistance and the impedance together — not one in isolation. When frequency increases from $\omega$ to $4\omega$, the inductive reactance grows from $X_L = \omega L$ to $X_L' = 4\omega L$. The new impedance becomes $Z_{new}^2 = R^2 + (4\omega L)^2 = R^2 + 16\omega^2 L^2$. Notice that $Z_{new}^2$ is not simply 16 times $Z_{old}^2 = R^2 + \omega^2 L^2$, because the $R^2$ term doesn't scale with frequency. Therefore, the ratio $P_{new}/P_{old}$ depends on the specific values of $R$ and $\omega L$ — making D the correct answer. Power definitely decreases, but by a factor strictly between 1 and 16. Choice A is wrong because there is no simple inverse-proportional relationship between power and frequency in a series RL circuit — that's a fabricated rule. Choice B contains a tempting but flawed step: while $X_L$ does quadruple, $Z^2$ only equals $16Z_{old}^2$ if $R = 0$, which is never physically true for a resistor. Choice C incorrectly applies $P = V_{rms}^2/R$, which only holds when the entire source voltage drops across $R$ — in a series RL circuit, the inductor shares the voltage. Study tip: Whenever a question changes frequency in an RL or RC circuit, always write out the full impedance expression first. Never assume a component's change propagates uniformly to $Z^2$ — the constant $R^2$ term always "resists" clean scaling.
5. Under what condition, if any, is the student's claim correct, and what physical principle justifies this?
   Answer: The claim is correct only when $\omega L = \frac{1}{\omega C}$ (resonance), because at resonance the impedance equals $R$ and the rms voltage across the source equals the rms voltage across the resistor, giving $P = \frac{\mathcal{E}_{rms}^2}{R} = \frac{\mathcal{E}_0^2}{2R}$. Whenever you see a question about power in a series RLC circuit, your instinct should be to reach for the general power formula and ask: when does it simplify? Average power delivered to a series RLC circuit is $P = I_{rms}^2 R$, where the rms current is $I_{rms} = \frac{\mathcal{E}_{rms}}{Z}$ and the total impedance is $Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}$. Substituting gives $P = \frac{\mathcal{E}_{rms}^2 \cdot R}{Z^2}$. Since $\mathcal{E}_{rms} = \frac{\mathcal{E}_0}{\sqrt{2}}$, this becomes $P = \frac{\mathcal{E}_0^2 R}{2Z^2}$. For this to equal $\frac{\mathcal{E}_0^2}{2R}$, you need $Z^2 = R^2$, which requires $\omega L = \frac{1}{\omega C}$ — the resonance condition. At resonance, the inductive and capacitive reactances cancel exactly, $Z = R$, and the full source voltage drives the resistor. Answer A is correct. Answer B is tempting but dangerously wrong. While reactive elements dissipate zero net power themselves, they do affect the current magnitude, which directly controls how much power reaches $R$. Ignoring $L$ and $C$ in the power formula is a classic misconception. Answer C describes an approximation, not an exact condition. The student's claim uses an equals sign, not an approximation — so "negligible" reactance doesn't satisfy it rigorously. Answer D is almost right in its formula but wrong in its conclusion. It correctly identifies that $Z = R$ requires $\omega L = \frac{1}{\omega C}$, but incorrectly claims this only happens when $L = 0$ and $C \to \infty$. Resonance is a frequency condition, not a requirement that components vanish. Study tip: Always write out the full impedance $Z$ first in RLC power problems, then ask what condition collapses it to $R$. That question leads directly to resonance every time.
6. A sinusoidal current $i(t) = I_0 \cos(\omega t + \phi)$ flows through a resistor $R$. A student computes the rms current as $I_{rms} = I_0 \cos(\phi)/\sqrt{2}$ by substituting the phase angle into the peak value. Which of the following correctly identifies the error and gives the true rms current?
   Answer: The student incorrectly treated the phase constant $\phi$ as if it reduced the effective amplitude. The rms value of a sinusoid depends only on its amplitude, not its phase, so $I_{rms} = I_0/\sqrt{2}$ regardless of $\phi$. When you see a question about rms values of sinusoidal signals, anchor yourself to the definition: the rms value is computed by squaring the function, averaging over a full period, and taking the square root — a process that has nothing to do with where the waveform "starts" in time. For $i(t) = I_0\cos(\omega t + \phi)$, the square is $i^2(t) = I_0^2\cos^2(\omega t + \phi)$. Because the average of $\cos^2$ over any complete cycle is always $\frac{1}{2}$ — regardless of the phase shift $\phi$ — the mean-square value is $I_0^2/2$, and therefore $I_{rms} = I_0/\sqrt{2}$. The phase constant simply slides the waveform left or right in time; it never changes the amplitude or the shape of the oscillation. Answer D correctly captures this: the rms value depends only on the amplitude $I_0$, not on $\phi$. Answer A correctly diagnoses the student's numerical mistake (confusing $i(0) = I_0\cos(\phi)$ with the peak amplitude), but then claims this is the full explanation. Actually, D is the more complete and precise statement of principle — the phase is irrelevant, full stop. Answer B is a fabricated correction; replacing $\cos(\phi)$ with $\sin(\phi)$ has no mathematical justification and just produces another wrong expression. Answer C invents a dimensional argument that doesn't hold up: $I_0\cos(\phi)$ already has units of amperes, and dividing by $T$ would give amperes per second — which is dimensionally wrong for current. Study tip: Whenever you compute rms, go back to the definition. The rms of any sinusoid with amplitude $A$ is always $A/\sqrt{2}$, regardless of phase, frequency, or when you start your clock.
7. How do the rms current and the average power dissipated in the capacitor each change?
   Answer: The rms current doubles because $I_{rms} = V_{rms}/(1/\omega C)$ and doubling $\omega$ halves the capacitive reactance; the average power dissipated remains zero because a pure capacitor stores and returns energy with no net dissipation per cycle. When analyzing AC circuits with reactive components, your first instinct should be to ask: how does frequency affect the component's opposition to current, and does this component dissipate real power? For a capacitor, the opposition to current flow is called capacitive reactance: $X_C = \frac{1}{\omega C}$. Notice that $X_C$ is inversely proportional to frequency. When $\omega$ doubles, $X_C$ halves — the capacitor offers less resistance to current flow. Since $I_{rms} = \frac{V_{rms}}{X_C}$, and $V_{rms} = V_0/\sqrt{2}$ remains fixed, halving $X_C$ doubles $I_{rms}$. Now for power: a pure capacitor shifts voltage and current by exactly 90°, making the power factor $\cos(90°) = 0$. Average power is $P_{avg} = V_{rms} I_{rms} \cos\phi = 0$ regardless of how large the current gets. The capacitor simply stores energy in its electric field during one half-cycle and returns it during the next — no net dissipation. This confirms A is correct. Choice B correctly identifies the current doubling but incorrectly concludes power doubles. Moving more charge per cycle doesn't create dissipation — only resistance does. Choice C has the physics backwards: higher frequency reduces $X_C$, increasing current, not decreasing it. The reasoning about "less time to charge" conflates a transient DC picture with steady-state AC behavior. Choice D incorrectly claims current is unchanged — energy storage depends on voltage, yes, but current through a capacitor depends on both voltage and frequency. A reliable tip: whenever a question pairs current and power for a pure reactive element (capacitor or inductor), the average power is always zero — no matter what happens to the current.
8. After applying a product-to-sum trigonometric identity, the average power over one full cycle is found to be $P_{avg} = \frac{V_0 I_0}{2}\cos\phi$. A student argues that this expression shows the average power is maximized not when $\phi = 0$ but when $\phi = \pi/4$, because that is when the product $\cos\phi$ is still large but the reactive energy exchange is also significant, enhancing net energy transfer. Which response best evaluates this argument?
   Answer: The argument is incorrect. $\cos\phi$ is a strictly decreasing function on $[0, \pi/2]$, so $P_{avg}$ is maximized at $\phi = 0$ (power factor = 1). Reactive energy exchange does not enhance net energy transfer; it represents energy oscillating between source and reactive element with zero net contribution to average power. When analyzing average power in AC circuits, your anchor should always be the formula itself: $P_{avg} = \frac{V_0 I_0}{2}\cos\phi$. The math tells the complete story about net energy delivery — everything else is interpretation. Since $\cos\phi$ is strictly decreasing on $[0, \pi/2]$, the function reaches its maximum at $\phi = 0$, giving $P_{avg} = \frac{V_0 I_0}{2}$. At $\phi = \pi/4$, you get $\cos(\pi/4) = \frac{\sqrt{2}}{2} \approx 0.707$, which is unambiguously smaller. Answer B is correct: maximizing average power means minimizing the phase angle, and reactive energy exchange never enhances net transfer — it simply oscillates between source and reactive element (inductor or capacitor), contributing exactly zero to the cycle average. Answer A contains a subtle but critical error. Reactive power $Q$ and real power $P$ combine to form apparent power $S$, but $Q$ represents energy that bounces back and forth — it does not add to energy actually consumed by the load. Conflating apparent power with "total energy delivered" is a classic misconception. Answer C invents a scenario where shifting constraints changes the optimal phase angle. This is fabricated reasoning — under standard fixed-amplitude conditions, $\phi = 0$ always maximizes real power regardless of which quantity you hold fixed. Answer D misreads the product-to-sum expansion. A smaller oscillating-term amplitude at $\phi = \pi/4$ doesn't mean more net forward energy; it just reflects lower peak instantaneous power throughout the cycle. Study tip: Whenever a distractor appeals to a real quantity (like reactive power $Q$) to explain a maximum, ask yourself: does that quantity actually contribute to the average being optimized? If not, it's a red herring.
9. If the engineer mistakenly uses the peak voltage (170 V) instead of the rms voltage to calculate the required resistance, and then manufactures the heater with that resistance value, what power will the heater actually dissipate when connected to the 120 V rms supply?
   Answer: 600 W, because the engineer calculates $R = V_{peak}^2/P = (170)^2/1200 \approx 24.1\,\Omega$, but the heater operates on $V_{rms} = 120\,\text{V}$, giving $P_{actual} = V_{rms}^2/R = (120)^2/24.1 \approx 600\,\text{W}$. Whenever you see a question mixing peak and rms voltages, your instinct should be to track which voltage was used at each step — the mistake happens at the design stage, but the physics happens at the operating stage. Here's what goes wrong: the engineer wants $R = V^2/P$, but plugs in the peak voltage instead of rms. This gives $R = (170)^2/1200 \approx 24{,}100/1200 \approx 24.1\,\Omega$. That resistance is now physically built into the heater. When the heater is actually plugged in, it sees $V_{rms} = 120\,\text{V}$ — because rms voltage is what drives real power dissipation in a resistor. So the actual power is $P_{actual} = V_{rms}^2/R = (120)^2/24.1 \approx 14{,}400/24.1 \approx 600\,\text{W}$. The heater delivers exactly half the intended power, confirming C. Choice A claims the heater dissipates 2400 W by arguing resistance is halved — but the error inflates resistance, not halves it. Using a larger voltage in $V^2/P$ produces a larger R, which means less power, not more. Choice B is a conceptual trap: it correctly states that rms voltage governs power, but ignores that the wrong resistance was manufactured. The rms definition doesn't save you if your component is already wrong. Choice D arrives at $1200/\sqrt{2}$ by applying a factor of $\sqrt{2}$ to the power directly, but the relationship is $P \propto 1/R \propto 1/V^2$, so the power ratio scales as $(V_{rms}/V_{peak})^2 = 1/2$, not $1/\sqrt{2}$. Study tip: In AC circuits, always confirm you're using rms values for power calculations. Peak voltage is for waveform analysis — the moment you write $P = V^2/R$, V must be $V_{rms}$.
10. A transformer operates from a 120 V rms, 60 Hz source. Its secondary coil delivers power to a load at 12 V rms. A technician replaces the 60 Hz source with a 120 V rms, 400 Hz source while keeping all other components identical. Which of the following best describes the effect on the secondary rms voltage and the magnetizing current drawn from the primary?
    Answer: The secondary rms voltage remains 12 V because the turns ratio is unchanged, and the magnetizing current decreases because the inductive reactance of the primary $X_L = \omega L$ increases with frequency, reducing the no-load primary current. Transformer questions become much clearer when you anchor everything to two foundational equations: the turns-ratio voltage relationship $\frac{V_s}{V_p} = \frac{N_s}{N_p}$ and the inductive reactance of the primary winding $X_L = \omega L = 2\pi f L$. The turns ratio depends only on geometry — the number of wire loops — not on frequency. As long as the primary still receives 120 V rms, the secondary delivers exactly $V_s = 120 \times \frac{N_s}{N_p} = 12 \text{ V rms}$, regardless of whether the source runs at 60 Hz or 400 Hz. Meanwhile, the magnetizing current — the small no-load current the primary draws to establish core flux — is governed by $I_{mag} \approx \frac{V_p}{X_L} = \frac{V_p}{2\pi f L}$. Increasing frequency from 60 Hz to 400 Hz raises $X_L$ by a factor of $\frac{400}{60} \approx 6.7$, so the magnetizing current drops dramatically. This makes A correct. B is wrong on both counts: the secondary voltage doesn't rise with frequency (the turns ratio still governs it), and higher frequency actually reduces magnetizing current rather than causing saturation. C correctly identifies that the secondary voltage is unchanged, but then incorrectly claims magnetizing current increases due to hysteresis losses — hysteresis does increase with frequency, but it's a small real-power effect, not the dominant driver of magnetizing current, which falls because $X_L$ rises. D misattributes the voltage drop to leakage reactance; in an ideal (or near-ideal) transformer model, leakage reactance is neglected, and the turns ratio controls the secondary voltage. A useful rule of thumb: on transformer problems, always separate what the turns ratio controls (voltage transformation) from what frequency controls (reactive impedance and core losses) — they're independent knobs.

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