Physics 2 Quiz: Ac Concepts Rms And Frequency

What this quiz covers

This quiz focuses on Ac Concepts Rms And Frequency, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 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 $P = \frac{\mathcal{E}_{rms}^2}{R} = \frac{\mathcal{E}_0^2}{2R}$. (correct answer)
2. The claim is always correct because the average power depends only on the peak EMF and the resistance; the reactive elements store and release energy with zero net dissipation per cycle, so $L$ and $C$ never appear in the power formula.
3. The claim is correct only when $R \gg |Z_L - Z_C|$, because in that limit the reactive impedance is negligible and the circuit behaves as a purely resistive load with $P \approx \frac{\mathcal{E}_0^2}{2R}$.
4. The claim is never correct because the average power must be written as $P = \frac{\mathcal{E}_0^2}{2Z}$ where $Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$, and $Z = R$ only in the special case where both $L = 0$ and $C \to \infty$.

Question 2

1. The student conflated the instantaneous value at $t = 0$ with the peak amplitude. Because $i(0) = I_0\cos(\phi)$, the student mistook this instantaneous value for the waveform's amplitude; the true peak is $I_0$, so $I_{rms} = I_0/\sqrt{2}$.
2. The student made a trigonometric error; the correct substitution should use $\sin(\phi)$ rather than $\cos(\phi)$, giving $I_{rms} = I_0 \sin(\phi)/\sqrt{2}$, which reduces to $I_0/\sqrt{2}$ only when $\phi = \pi/2$.
3. The student's formula is dimensionally inconsistent; because $\cos(\phi)$ is dimensionless and $I_0$ carries units of amperes, the product $I_0\cos(\phi)$ must be normalized by the period $T$ before dividing by $\sqrt{2}$, giving $I_{rms} = I_0\cos(\phi)/(T\sqrt{2})$.
4. 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 $\phi$. (correct answer)

Question 3

1. $\frac{P_2}{P_1} = 2$, 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. (correct answer)
2. $\frac{P_2}{P_1} = 2\sqrt{2}$, because the higher frequency doubles the number of cycles per second, each carrying energy proportional to $V_{rms}^2$, so both the voltage factor $\sqrt{2}$ and the frequency factor 2 multiply the power.
3. $\frac{P_2}{P_1} = 4$, because average power is proportional to $V_0^2 f$, giving a combined factor of $(\sqrt{2})^2 \times 2 = 4$.
4. $\frac{P_2}{P_1} = \sqrt{2}$, because rms voltage is $V_0/\sqrt{2}$ and the $\sqrt{2}$ factors partially cancel, leaving only the ratio of peak voltages $V_2/V_1 = \sqrt{2}$ as the relevant quantity.

Question 4

1. 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. (correct answer)
2. The rms current doubles because capacitive reactance halves with doubled frequency; the average power dissipated also doubles because more charge is moved per unit time through a larger rms current, increasing the energy transferred to the capacitor.
3. The rms current halves because doubling frequency doubles the rate of charge/discharge, effectively halving the time available to charge the capacitor each half-cycle; the average power dissipated remains zero for a pure capacitor.
4. The rms current remains unchanged because $V_0$ is constant and the capacitor's energy storage depends only on voltage, not frequency; the average power dissipated is zero because no resistance is present in the circuit.

Question 5

1. 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. (correct answer)
2. The secondary rms voltage increases above 12 V because higher frequency increases the rate of flux change, inducing a larger EMF per turn; the magnetizing current also increases because the core saturates more rapidly at higher frequency, drawing additional primary current to sustain the flux.
3. The secondary rms voltage remains 12 V because the turns ratio is unchanged, and the magnetizing current increases because higher frequency raises hysteresis losses, which manifest as an increased real-power draw on the primary that adds to the magnetizing current.
4. The secondary rms voltage decreases below 12 V because the primary leakage reactance grows with frequency and drops additional voltage before ideal coupling occurs, reducing the flux linkage; the magnetizing current also decreases because the higher total primary impedance limits current flow.

Question 6

1. 2400 W, because using the peak voltage inflates the calculated resistance by $\sqrt{2}$, which halves the resistance relative to the correct value, and halving resistance doubles the power for a fixed rms voltage.
2. 1200 W, because the rms voltage is defined precisely so that a resistive load dissipates the same power as it would from a DC voltage equal to the rms value; the method of calculation does not affect the actual physical power.
3. 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 $P_{actual} = V_{rms}^2/R = (120)^2/24.1 \approx 600\,\text{W}$. (correct answer)
4. 848 W, because the peak voltage is $\sqrt{2}$ times the rms, so the resistance is inflated by a factor of $\sqrt{2}$, and since $P \propto 1/R$, the actual power is $1200/\sqrt{2} \approx 848\,\text{W}$.

Question 7

1. The average power decreases by a factor of 4, because power is inversely proportional to frequency in an inductive circuit, and the frequency increases by a factor of 4, giving $P_{new} = P_{old}/4$ regardless of the specific values of $R$ and $L$.
2. The average power decreases by exactly a factor of 16, because the inductive reactance $X_L = \omega L$ quadruples and appears squared in the impedance, so $Z^2$ increases by a factor of 16 and $P \propto 1/Z^2$ decreases by 16.
3. The average power remains unchanged because $V_{rms}$ is held constant, and average power in any AC circuit is determined solely by $V_{rms}$ and the resistance $R$ through $P = V_{rms}^2/R$.
4. 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. (correct answer)

Question 8

1. Option 1 reduces losses by a factor of 4; Option 2 reduces losses by a factor of 2. Doubling voltage halves the current and quarters the losses, but doubling the power factor only improves the effective current by $\sqrt{2}$, halving the losses.
2. 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 $I_{rms}$). (correct answer)
3. Option 1 reduces losses by a factor of 2; Option 2 reduces losses by a factor of 4. Doubling voltage reduces current in proportion (factor of 2 in current, factor of 2 in losses), whereas doubling the power factor reduces the reactive component of current more dramatically.
4. Option 1 reduces losses by a factor of 4; Option 2 leaves losses unchanged. Improving the power factor redistributes energy between real and reactive power but does not change $I_{rms}$, which depends only on the apparent power $S = V_{rms} I_{rms}$.

Question 9

1. $I_{rms} = \dfrac{1}{\sqrt{2}}\sqrt{I_{DC}^2 + I_0^2}$, because the combined peak amplitude of the waveform is $\sqrt{I_{DC}^2 + I_0^2}$, and the standard sinusoidal factor $1/\sqrt{2}$ converts any peak value to its rms equivalent.
2. $I_{rms} = I_{DC} + \dfrac{I_0}{\sqrt{2}}$, because the total rms is the arithmetic sum of the DC component's rms value ($I_{DC}$) and the AC component's rms value ($I_0/\sqrt{2}$).
3. $I_{rms} = \sqrt{I_{DC}^2 + \dfrac{I_0^2}{2}}$, because the DC and sinusoidal components are orthogonal over one period (their cross term integrates to zero), so their squared rms values add independently. (correct answer)
4. $I_{rms} = I_{DC} + \dfrac{I_0}{\sqrt{2}}$, because squaring $(I_{DC} + I_0\sin\omega t)$ produces a cross term $2I_{DC}I_0\sin\omega t$ that has a nonzero time average, shifting the rms above $\sqrt{I_{DC}^2 + I_0^2/2}$.

Question 10

1. The argument is partially correct. While $\cos(\pi/4) < \cos(0)$, the reactive power $Q = \frac{V_0 I_0}{2}\sin\phi$ reaches a local maximum near $\phi = \pi/4$, and since both $P$ and $Q$ contribute to the apparent power $S = \sqrt{P^2 + Q^2}$, the total energy delivered per cycle — real plus reactive — is greatest at an intermediate phase angle rather than at $\phi = 0$.
2. 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. (correct answer)
3. The argument is incorrect for this specific circuit but would be valid in a source-limited system. When apparent power $S = V_0 I_0/2$ is the constrained quantity, maximizing real power still requires $\phi = 0$; however, if the rms current $I_0/\sqrt{2}$ is instead held fixed and voltage is allowed to vary, the optimal phase angle shifts away from zero toward $\phi = \pi/4$, making the student's intuition applicable under different constraints.
4. The argument is correct. Expanding $p(t)$ via a product-to-sum identity yields both a constant term $\frac{V_0I_0}{2}\cos\phi$ and an oscillating term of amplitude $\frac{V_0I_0}{2}$. At $\phi = \pi/4$ the oscillating term has a smaller amplitude than at $\phi = 0$, which means less energy is returned to the source each cycle; this reduction in back-flow effectively increases the net forward energy delivery beyond what $\cos\phi$ alone would suggest.