Electromagnetic Induction and Alternating Currents MCQs with Answers – Part 4 (NEET Physics)

Explanation: Instantaneous power is negative whenever voltage and current have opposite signs. The current lags voltage by \(60^\circ\). Consequently, the signs of the two waveforms disagree during portions of each cycle. The duration of those portions must be distinguished from the average power calculation. \(\textbf{Write the instantaneous power:}\) \[ p(t)=v(t)i(t) \] Substituting, \[ p(t)=400\sin\omega t\sin\left(\omega t-\frac{\pi}{3}\right) \] Using \[ 2\sin A\sin B=\cos(A-B)-\cos(A+B) \] gives \[ p(t)=200\left[\cos\frac{\pi}{3}-\cos\left(2\omega t-\frac{\pi}{3}\right)\right] \] Thus, \[ p(t)=100-200\cos\left(2\omega t-\frac{\pi}{3}\right) \] \(\textbf{Identify negative-power intervals:}\) The condition \(p\lt 0\) requires \[ 100-200\cos\left(2\omega t-\frac{\pi}{3}\right)\lt 0 \] Therefore, \[ \cos\left(2\omega t-\frac{\pi}{3}\right)\gt \frac12 \] Within one \(2\pi\) interval of the cosine argument, this inequality holds over a total angular width \[ \Delta\alpha=\frac{2\pi}{3} \] Hence, the fraction of that argument's cycle giving negative power is \[ \frac{\Delta\alpha}{2\pi}=\frac13 \] The power waveform repeats twice during one complete voltage cycle. Both repetitions contain the same negative-power fraction. Therefore, instantaneous power is negative for one-third of each complete voltage cycle. \(\textbf{Average absorbed power:}\) The oscillating cosine term averages to zero over a complete cycle. Therefore, \[ P_{\text{avg}}=100\,\text{W} \] Equivalently, \[ P_{\text{avg}}=\frac{V_mI_m}{2}\cos\phi \] \[ P_{\text{avg}}=\frac{(100)(4.0)}2\cos60^\circ \] \[ P_{\text{avg}}=100\,\text{W} \] \(\textbf{Physical interpretation:}\) The load returns energy to the source during one-third of the cycle. Nevertheless, its positive-power portions transfer a greater total amount of energy into the load. Its average real power is therefore positive. A negative instantaneous power interval does not imply that the entire load is wattless. The fraction of negative-power time is controlled by the phase displacement, whereas average power additionally depends on the voltage and current amplitudes. \(\textbf{Final answer:}\) \[ \boxed{\text{Negative-power fraction}=\frac13,\qquad P_{\text{avg}}=100\,\text{W}} \]

Explanation: The source frequency changes, but all quantities appearing in the sinusoidal average-power relation remain fixed. Consequently, average real power remains unchanged. However, a complete cycle takes less time at the higher frequency. The net energy associated with one cycle therefore changes. \(\textbf{Average real power:}\) \[ P_{\text{avg}}=V_{\text{rms}}I_{\text{rms}}\cos\phi \] Substituting, \[ P_{\text{avg}}=(200)(4.0)\cos60^\circ \] \[ P_{\text{avg}}=(800)\left(\frac12\right) \] \[ P_{\text{avg}}=400\,\text{W} \] This value applies at both frequencies under the stated controlled conditions. \(\textbf{Cycle duration after the change:}\) The new frequency is \[ f_2=100\,\text{Hz} \] Therefore, \[ T_2=\frac1{f_2} \] \[ T_2=\frac1{100} \] \[ T_2=0.010\,\text{s} \] \(\textbf{Net energy absorbed per cycle:}\) \[ E_{\text{cycle},2}=P_{\text{avg}}T_2 \] Substituting, \[ E_{\text{cycle},2}=(400)(0.010) \] \[ E_{\text{cycle},2}=4.0\,\text{J} \] \(\textbf{Energy absorbed during ten seconds:}\) For a duration of \(10\,\text{s}\), \[ E_{10}=P_{\text{avg}}\Delta t \] \[ E_{10}=(400)(10) \] \[ E_{10}=4000\,\text{J} \] \(\textbf{Compare with the original frequency:}\) Initially, \[ T_1=\frac1{50}=0.020\,\text{s} \] The original energy per cycle was \[ E_{\text{cycle},1}=(400)(0.020) \] \[ E_{\text{cycle},1}=8.0\,\text{J} \] Thus, doubling frequency halves net energy per cycle while doubling the number of cycles per second. The two effects leave energy transferred per second unchanged. This conclusion relies on the controller maintaining RMS current and phase difference; it would not generally follow for a fixed passive reactive circuit supplied at different frequencies. \(\textbf{Final answer:}\) \[ \boxed{P_{\text{avg}}=400\,\text{W},\qquad E_{\text{cycle},2}=4.0\,\text{J},\qquad E_{10}=4000\,\text{J}} \]

Explanation: Both parallel branches experience the same sinusoidal voltage. However, their currents have different phases. The resistor current is in phase with voltage, while the ideal inductor current lags voltage by \(90^\circ\). Their RMS magnitudes must therefore be combined as perpendicular phasor components. \(\textbf{Resistor-branch current:}\) \[ I_R=\frac{V_{\text{rms}}}{R} \] Substituting, \[ I_R=\frac{240}{120} \] \[ I_R=2.0\,\text{A} \] This current is in phase with the source voltage. \(\textbf{Inductor-branch current:}\) \[ I_L=\frac{V_{\text{rms}}}{X_L} \] \[ I_L=\frac{240}{160} \] \[ I_L=1.5\,\text{A} \] This current lags the source voltage by \(90^\circ\). \(\textbf{Total source current:}\) The active and quadrature branch currents are perpendicular in phase. Therefore, \[ I_s=\sqrt{I_R^2+I_L^2} \] \[ I_s=\sqrt{2.0^2+1.5^2} \] \[ I_s=\sqrt{4.0+2.25} \] \[ I_s=2.5\,\text{A} \] \(\textbf{Overall power factor:}\) The source current has an in-phase component of \(2.0\,\text{A}\). Therefore, \[ \cos\phi=\frac{I_R}{I_s} \] \[ \cos\phi=\frac{2.0}{2.5} \] \[ \cos\phi=\frac45 \] The overall source current lags voltage because of the inductive branch. \(\textbf{Average real power:}\) \[ P=V_{\text{rms}}I_s\cos\phi \] Substituting, \[ P=(240)(2.5)\left(\frac45\right) \] \[ P=480\,\text{W} \] \(\textbf{Independent branch-power check:}\) The resistor dissipates \[ P_R=\frac{V_{\text{rms}}^2}{R} \] \[ P_R=\frac{240^2}{120} \] \[ P_R=480\,\text{W} \] The ideal inductor consumes zero average real power. Thus, the total average power is also \(480\,\text{W}\). The source current exceeds the resistor-branch current because it also supplies the wattless inductive current. Adding the two RMS branch currents arithmetically would give an incorrect source-current magnitude. \(\textbf{Final answer:}\) \[ \boxed{I_s=2.5\,\text{A},\qquad \cos\phi=\frac45,\qquad P=480\,\text{W}} \]

Explanation: The ideal inductor and capacitor are connected across the same alternating voltage. Their reactance magnitudes are equal, so their branch current amplitudes are equal. However, the inductor current lags voltage by \(90^\circ\), whereas the capacitor current leads voltage by \(90^\circ\). The two branch currents consequently differ in phase by \(180^\circ\). \(\textbf{Inductor-branch current:}\) \[ I_L=\frac{100}{50} \] \[ I_L=2.0\,\text{A} \] \(\textbf{Capacitor-branch current:}\) \[ I_C=\frac{100}{50} \] \[ I_C=2.0\,\text{A} \] Since both currents are purely sinusoidal with opposite phases, their instantaneous sum is zero at every instant. Therefore, the source current is zero in the stated ideal steady state. This does not mean that the branch currents disappear. The inductor stores magnetic energy according to \[ U_L=\frac12Li_L^2 \] The capacitor stores electric energy according to \[ U_C=\frac12Cv^2 \] Their individual stored energies vary during the cycle. The inductor and capacitor can exchange energy even though the source supplies no net current in the ideal cancellation condition. An ideal inductor and an ideal capacitor each absorb zero average real power over a full cycle. Their average powers are not equal and opposite nonzero wattages; each is individually zero. The result therefore distinguishes absence of source current from absence of internal current or energy storage.

Explanation: The load receives the same real power at the same RMS terminal voltage in both conditions. However, current is inversely proportional to power factor when those two quantities remain fixed. The leads dissipate heat according to the square of their RMS current. \(\textbf{Current at the lower power factor:}\) For sinusoidal AC, \[ P=V_{\text{rms}}I_{\text{rms}}\cos\phi \] Therefore, \[ I_P=\frac{P}{V_{\text{rms}}\cos\phi_P} \] Substituting, \[ I_P=\frac{2400}{(200)(0.60)} \] \[ I_P=20\,\text{A} \] \(\textbf{Current after power-factor improvement:}\) \[ I_Q=\frac{2400}{(200)(0.80)} \] \[ I_Q=15\,\text{A} \] \(\textbf{Lead heating in condition P:}\) \[ P_{\text{loss},P}=I_P^2R_{\text{lead}} \] \[ P_{\text{loss},P}=(20)^2(1.0) \] \[ P_{\text{loss},P}=400\,\text{W} \] \(\textbf{Lead heating in condition Q:}\) \[ P_{\text{loss},Q}=I_Q^2R_{\text{lead}} \] \[ P_{\text{loss},Q}=(15)^2(1.0) \] \[ P_{\text{loss},Q}=225\,\text{W} \] For equal operating durations, heat losses are proportional to these powers. Therefore, \[ \frac{H_Q}{H_P}=\frac{225}{400} \] \[ \frac{H_Q}{H_P}=\frac9{16} \] \(\textbf{Physical interpretation:}\) Improving power factor reduces the current needed to deliver a specified real power. Because line heating depends on current squared, the fractional reduction in loss is greater than the fractional reduction in current. The power-factor improvement does not create additional useful electrical energy. It reduces the current associated with the same useful power transfer. The specified maintained load-terminal voltage is important because otherwise changing the line drop could also change the load operating condition. \(\textbf{Final answer:}\) \[ \boxed{I_P=20\,\text{A},\qquad I_Q=15\,\text{A},\qquad \frac{H_Q}{H_P}=\frac9{16}} \]

Explanation: The appliance current has an active component in phase with voltage and a lagging wattless component. The parallel capacitor contributes a leading wattless current. These quadrature components partially cancel at the source terminals. The capacitor does not consume average real power under the stated ideal conditions. \(\textbf{Original active current:}\) \[ I_{\text{active}}=I_{\text{load}}\cos\phi \] Substituting, \[ I_{\text{active}}=(10)(0.60) \] \[ I_{\text{active}}=6.0\,\text{A} \] \(\textbf{Original lagging wattless component:}\) \[ |\sin\phi|=\sqrt{1-0.60^2} \] \[ |\sin\phi|=0.80 \] Therefore, \[ I_{\text{reactive,load}}=(10)(0.80) \] \[ I_{\text{reactive,load}}=8.0\,\text{A} \] This component lags voltage by \(90^\circ\). \(\textbf{Effect of the parallel capacitor:}\) The capacitor provides \(6.0\,\text{A}\) of leading current. Therefore, the resultant quadrature component is \[ I_{\text{reactive,source}}=8.0-6.0 \] \[ I_{\text{reactive,source}}=2.0\,\text{A} \] It remains lagging. The active component remains \(6.0\,\text{A}\). \(\textbf{New source RMS current:}\) \[ I_s=\sqrt{I_{\text{active}}^2+I_{\text{reactive,source}}^2} \] \[ I_s=\sqrt{6.0^2+2.0^2} \] \[ I_s=\sqrt{40} \] \[ I_s=2\sqrt{10}\,\text{A} \] \(\textbf{New overall power factor:}\) \[ \cos\phi_s=\frac{I_{\text{active}}}{I_s} \] \[ \cos\phi_s=\frac6{2\sqrt{10}} \] \[ \cos\phi_s=\frac3{\sqrt{10}} \] The overall power factor remains lagging. \(\textbf{Average real power:}\) \[ P=V_{\text{rms}}I_{\text{active}} \] \[ P=(240)(6.0) \] \[ P=1440\,\text{W} \] \(\textbf{Physical interpretation:}\) The appliance still receives the same real power and draws the same branch current. The parallel capacitor reduces the lagging quadrature current that must be supplied externally. The source-current magnitude consequently falls from \(10\,\text{A}\) to \(2\sqrt{10}\,\text{A}\). The power factor improves without changing the appliance's real power. The new power factor is not exactly unity because the capacitor cancels only part of the original reactive component. \(\textbf{Final answer:}\) \[ \boxed{I_s=2\sqrt{10}\,\text{A},\qquad \cos\phi_s=\frac3{\sqrt{10}},\qquad P=1440\,\text{W}} \]

Explanation: Parallel appliances experience the same source voltage, but their currents can have different phase angles. Their active current components add in phase. Their wattless components must be combined with attention to whether they lead or lag voltage. \(\textbf{Current drawn by appliance P:}\) \[ P_P=VI_P\cos\phi_P \] Therefore, \[ I_P=\frac{800}{(200)(0.80)} \] \[ I_P=5.0\,\text{A} \] Its active component is \[ I_{P,\text{active}}=5.0(0.80) \] \[ I_{P,\text{active}}=4.0\,\text{A} \] Since \[ \sin\phi_P=\sqrt{1-0.80^2}=0.60 \] its lagging quadrature component is \[ I_{P,\text{reactive}}=5.0(0.60) \] \[ I_{P,\text{reactive}}=3.0\,\text{A} \] \(\textbf{Current drawn by appliance Q:}\) \[ I_Q=\frac{600}{(200)(0.60)} \] \[ I_Q=5.0\,\text{A} \] Its active component is \[ I_{Q,\text{active}}=5.0(0.60) \] \[ I_{Q,\text{active}}=3.0\,\text{A} \] The magnitude of its leading quadrature component is \[ I_{Q,\text{reactive}}=5.0\sqrt{1-0.60^2} \] \[ I_{Q,\text{reactive}}=5.0(0.80) \] \[ I_{Q,\text{reactive}}=4.0\,\text{A} \] \(\textbf{Combine the components:}\) The total active current is \[ I_{\text{active}}=4.0+3.0 \] \[ I_{\text{active}}=7.0\,\text{A} \] The lagging component from P and leading component from Q oppose: \[ I_{\text{reactive}}=4.0-3.0 \] \[ I_{\text{reactive}}=1.0\,\text{A}\quad\text{leading} \] \(\textbf{Total source current:}\) \[ I_s=\sqrt{I_{\text{active}}^2+I_{\text{reactive}}^2} \] \[ I_s=\sqrt{7.0^2+1.0^2} \] \[ I_s=5\sqrt2\,\text{A} \] \(\textbf{Overall power factor:}\) \[ \cos\phi_s=\frac{I_{\text{active}}}{I_s} \] \[ \cos\phi_s=\frac7{5\sqrt2} \] The source current leads voltage because the remaining quadrature component is leading. \(\textbf{Total real power:}\) \[ P_{\text{total}}=800+600 \] \[ P_{\text{total}}=1400\,\text{W} \] This is consistent with \[ P_{\text{total}}=(200)(7.0) \] \[ P_{\text{total}}=1400\,\text{W} \] \(\textbf{Physical interpretation:}\) Each appliance draws \(5.0\,\text{A}\), but the source current is not \(10\,\text{A}\). Their differently phased current components undergo partial cancellation. The combined power factor is close to unity even though neither appliance operates at unity power factor. The overall current is leading because Q supplies a larger leading quadrature component than P's lagging component. Real powers add directly, while current phasors must be combined according to phase. \(\textbf{Final answer:}\) \[ \boxed{I_s=5\sqrt2\,\text{A},\qquad \cos\phi_s=\frac7{5\sqrt2},\qquad P_{\text{total}}=1400\,\text{W}} \] The overall current leads the supply voltage.

Explanation: The original load draws active current together with lagging wattless current. A capacitor connected in parallel contributes leading wattless current. Unity overall power factor requires complete cancellation of the net quadrature current at the supply terminals. \(\textbf{Original load current:}\) \[ P=VI\cos\phi \] Therefore, \[ I_{\text{initial}}=\frac{1200}{(200)(0.60)} \] \[ I_{\text{initial}}=10\,\text{A} \] \(\textbf{Active current required by the load:}\) \[ I_{\text{active}}=I_{\text{initial}}\cos\phi \] \[ I_{\text{active}}=(10)(0.60) \] \[ I_{\text{active}}=6.0\,\text{A} \] \(\textbf{Lagging component to be cancelled:}\) \[ I_{\text{wattless}}=I_{\text{initial}}\sin\phi \] Using \[ \sin\phi=\sqrt{1-0.60^2}=0.80 \] gives \[ I_{\text{wattless}}=(10)(0.80) \] \[ I_{\text{wattless}}=8.0\,\text{A} \] The capacitor must therefore draw \(8.0\,\text{A}\) RMS of leading current. \(\textbf{Determine angular frequency:}\) \[ \omega=2\pi f \] \[ \omega=2\pi(50) \] \[ \omega=100\pi\,\text{rad s}^{-1} \] \(\textbf{Required capacitance:}\) For an ideal capacitor under sinusoidal voltage, \[ I_C=\omega CV \] Therefore, \[ C=\frac{I_C}{\omega V} \] Substituting, \[ C=\frac8{(100\pi)(200)} \] \[ C=\frac1{2500\pi}\,\text{F} \] \[ C=\frac{400}{\pi}\,\mu\text{F} \] \(\textbf{Corrected supply current:}\) The quadrature components now cancel. Only the active current remains: \[ I_{\text{corrected}}=6.0\,\text{A} \] The real power remains \[ P=(200)(6.0) \] \[ P=1200\,\text{W} \] \(\textbf{Compare resistive line losses:}\) For the same line resistance, \[ P_{\text{line}}=I_s^2R_{\text{line}} \] Therefore, \[ \frac{P_{\text{line,new}}}{P_{\text{line,old}}} =\left(\frac{6.0}{10}\right)^2 \] \[ \frac{P_{\text{line,new}}}{P_{\text{line,old}}}=\frac9{25} \] \(\textbf{Physical interpretation:}\) The capacitor does not reduce the appliance's required real power. Instead, it provides a local leading current that cancels the load's lagging wattless component. This reduces the current supplied through the external conductors. The corresponding line heating decreases according to the square of that current. Power-factor correction therefore improves the supply-current requirement without changing useful load power. \(\textbf{Final answer:}\) \[ \boxed{C=\frac{400}{\pi}\,\mu\text{F},\qquad I_{\text{corrected}}=6.0\,\text{A},\qquad \frac{P_{\text{line,new}}}{P_{\text{line,old}}}=\frac9{25}} \]

Explanation: A leading power factor indicates that the source current has a net leading quadrature component. The aim of power-factor correction is to reduce that component without unnecessarily changing the active component. An ideal inductor contributes lagging wattless current. Therefore, a suitably chosen parallel inductor can oppose the original leading component. Statement I is correct. An ideal capacitor contributes leading wattless current. Adding another capacitor to an already leading load generally increases the leading quadrature component rather than cancelling it. Its zero average real power does not guarantee an improvement in overall power factor. Statement II is incorrect. If the parallel inductor draws exactly the required lagging wattless current, the leading and lagging components cancel. If it draws more than required, the remaining quadrature current becomes lagging. The overall phase relationship can therefore reverse without changing the original load's real-power consumption. Statement III is correct. At unity overall power factor, the total source current is in phase with voltage. Individual reactive branches may nevertheless carry nonzero currents whose quadrature contributions cancel. These branch currents do not have to vanish merely because the resultant wattless component is zero. Statement IV is correct. The distinction is between wattless current in each branch and the net wattless current demanded from the source. The correct combination is Statements I, III and IV only.

Explanation: An ideal inductor can return previously stored magnetic energy to its source. The net electrical energy absorbed during a selected interval is therefore not necessarily positive. The given interval lasts one-quarter of the voltage cycle, beginning at an upward voltage zero crossing. \(\textbf{Determine inductive reactance:}\) \[ X_L=\omega L \] Substituting, \[ X_L=(100)(0.20) \] \[ X_L=20\,\Omega \] \(\textbf{Steady sinusoidal current:}\) The peak voltage is \[ V_m=60\,\text{V} \] Therefore, \[ I_m=\frac{V_m}{X_L} \] \[ I_m=\frac{60}{20} \] \[ I_m=3.0\,\text{A} \] The current lags voltage by \(90^\circ\): \[ i(t)=3.0\sin\left(100t-\frac{\pi}{2}\right) \] \[ i(t)=-3.0\cos(100t)\,\text{A} \] \(\textbf{Initial magnetic energy:}\) At \(t=0\), \[ i(0)=-3.0\,\text{A} \] Therefore, \[ U_i=\frac12Li^2(0) \] \[ U_i=\frac12(0.20)(3.0)^2 \] \[ U_i=0.90\,\text{J} \] \(\textbf{Final magnetic energy:}\) At \[ t_f=\frac{\pi}{200}\,\text{s} \] the current phase is \[ 100t_f=\frac{\pi}{2} \] Hence, \[ i(t_f)=-3.0\cos\frac{\pi}{2} \] \[ i(t_f)=0 \] Therefore, \[ U_f=\frac12L(0)^2 \] \[ U_f=0 \] \(\textbf{Net energy absorbed:}\) For an ideal inductor, electrical work absorbed equals its magnetic-energy change: \[ E_{\text{absorbed}}=\Delta U=U_f-U_i \] \[ E_{\text{absorbed}}=0-0.90 \] \[ E_{\text{absorbed}}=-0.90\,\text{J} \] \(\textbf{Instantaneous-power check:}\) During the interval, source voltage is positive. The current is negative until the end of the interval. Therefore, \[ p(t)=v(t)i(t)\lt 0 \] throughout the interior of the interval. The negative power confirms that magnetic energy is being returned to the source. \(\textbf{Physical interpretation:}\) The inductor initially carries its maximum negative current and stores its maximum magnetic energy. Its current magnitude decreases as the applied positive voltage drives the current toward zero. The magnetic energy consequently decreases to zero. This energy returns to the source rather than being dissipated as Joule heat. Over a complete steady cycle, the inductor receives the same amount of energy that it returns. \(\textbf{Final answer:}\) \[ \boxed{E_{\text{absorbed}}=-0.90\,\text{J},\qquad U_i=0.90\,\text{J},\qquad U_f=0} \]

Explanation: The voltage and current contain constant components together with sinusoidal components. The standard sinusoidal power formula applies to the purely sinusoidal parts, but the complete average power must also include the constant parts. The mixed products of a constant term and a sinusoidal term average to zero over one complete cycle. \(\textbf{Instantaneous electrical power:}\) \[ p(t)=v(t)i(t) \] Writing \(\theta=\omega t\), \[ p=(20+100\sin\theta) \left[1.0+4.0\sin\left(\theta-\frac{\pi}{3}\right)\right] \] Expanding, \[ p=20+80\sin\left(\theta-\frac{\pi}{3}\right) +100\sin\theta +400\sin\theta\sin\left(\theta-\frac{\pi}{3}\right) \] \(\textbf{Average of the constant contribution:}\) The constant term is \[ P_{\text{constant}}=(20)(1.0) \] \[ P_{\text{constant}}=20\,\text{W} \] \(\textbf{Average of the mixed terms:}\) Over one complete period, \[ \langle\sin\theta\rangle=0 \] and \[ \left\langle\sin\left(\theta-\frac{\pi}{3}\right)\right\rangle=0 \] Therefore, the two terms containing only one sinusoidal factor contribute zero to average power. \(\textbf{Average of the sinusoidal product:}\) For sinusoidal terms of equal frequency, \[ \left\langle\sin\theta\sin(\theta-\phi)\right\rangle =\frac12\cos\phi \] Hence, \[ P_{\text{sinusoidal}} =\frac{(100)(4.0)}2\cos\frac{\pi}{3} \] \[ P_{\text{sinusoidal}}=(200)\left(\frac12\right) \] \[ P_{\text{sinusoidal}}=100\,\text{W} \] \(\textbf{Total average power:}\) \[ P_{\text{avg}}=P_{\text{constant}}+P_{\text{sinusoidal}} \] \[ P_{\text{avg}}=20+100 \] \[ P_{\text{avg}}=120\,\text{W} \] \(\textbf{Physical interpretation:}\) A sinusoidal voltage or current can have zero algebraic average and still contribute nonzero average power when multiplied by another correlated sinusoidal quantity. The steady voltage and current components also transfer energy continuously. Their contribution must not be omitted merely because the complete waveform contains an alternating part. Using a single peak-value power formula for the entire offset waveform would be incorrect. The average electrical power is the sum of the constant-component power and the phase-dependent sinusoidal power. \(\textbf{Final answer:}\) \[ \boxed{P_{\text{avg}}=120\,\text{W},\qquad P_{\text{constant}}=20\,\text{W},\qquad P_{\text{sinusoidal}}=100\,\text{W}} \]

Explanation: Both circuits contain the same resistance and inductive reactance. However, connecting the elements in parallel or in series changes how the source voltage is distributed and how current flows. The two arrangements can have equal power factors without drawing equal current or absorbing equal real power. \(\textbf{Circuit P — Parallel branches:}\) The resistor current is \[ I_R=\frac{100}{50} \] \[ I_R=2.0\,\text{A} \] The inductor current is \[ I_L=\frac{100}{50} \] \[ I_L=2.0\,\text{A} \] The two branch currents differ in phase by \(90^\circ\). Therefore, \[ I_P=\sqrt{I_R^2+I_L^2} \] \[ I_P=\sqrt{2.0^2+2.0^2} \] \[ I_P=2\sqrt2\,\text{A} \] The active current component equals \(2.0\,\text{A}\). Hence, \[ \cos\phi_P=\frac{2.0}{2\sqrt2} \] \[ \cos\phi_P=\frac1{\sqrt2} \] The average real power is dissipated in the resistor: \[ P_P=\frac{100^2}{50} \] \[ P_P=200\,\text{W} \] \(\textbf{Circuit Q — Series elements:}\) The impedance magnitude is \[ Z_Q=\sqrt{R^2+X_L^2} \] \[ Z_Q=\sqrt{50^2+50^2} \] \[ Z_Q=50\sqrt2\,\Omega \] Therefore, \[ I_Q=\frac{100}{50\sqrt2} \] \[ I_Q=\sqrt2\,\text{A} \] Its power factor is \[ \cos\phi_Q=\frac R{Z_Q} \] \[ \cos\phi_Q=\frac{50}{50\sqrt2} \] \[ \cos\phi_Q=\frac1{\sqrt2} \] The average real power is \[ P_Q=I_Q^2R \] \[ P_Q=(\sqrt2)^2(50) \] \[ P_Q=100\,\text{W} \] \(\textbf{Compare the results:}\) \[ \frac{I_P}{I_Q}=\frac{2\sqrt2}{\sqrt2} \] \[ \frac{I_P}{I_Q}=2 \] Similarly, \[ \frac{P_P}{P_Q}=\frac{200}{100} \] \[ \frac{P_P}{P_Q}=2 \] \(\textbf{Physical interpretation:}\) Both arrangements have equal magnitudes of active and reactive contributions in their respective current or voltage phasor relations. Their overall power factors are therefore equal. Nevertheless, the parallel resistor receives the entire source RMS voltage, while the resistor in the series circuit receives only part of it. The parallel arrangement consequently draws more current and dissipates more real power. Equal power factors do not imply equal apparent power or equal real power. \(\textbf{Final answer:}\) \[ \boxed{\cos\phi_P=\cos\phi_Q=\frac1{\sqrt2},\qquad \frac{I_P}{I_Q}=2,\qquad \frac{P_P}{P_Q}=2} \]

Explanation: For sinusoidal voltage and current, \[ P=V_{\text{rms}}I_{\text{rms}}\cos\phi \] The power factor is the cosine of a phase difference. Therefore, its magnitude cannot exceed one. For the stated positive RMS values, a necessary condition is \[ |P|\leq V_{\text{rms}}I_{\text{rms}} \] Each record can be checked against this physical limit. \(\textbf{Experiment P:}\) Its apparent power is \[ S_P=(120)(5.0) \] \[ S_P=600\,\text{VA} \] The implied power factor is \[ \cos\phi_P=\frac{480}{600} \] \[ \cos\phi_P=0.80 \] This is physically possible. \(\textbf{Experiment Q:}\) Its apparent power is \[ S_Q=(100)(3.0) \] \[ S_Q=300\,\text{VA} \] The implied power factor is \[ \cos\phi_Q=\frac{330}{300} \] \[ \cos\phi_Q=1.10 \] This exceeds the maximum possible cosine magnitude. Therefore, Q is inconsistent. \(\textbf{Experiment R:}\) The measured power is zero despite nonzero RMS current. This is possible for an ideal purely reactive load, where voltage and current differ in phase by \(90^\circ\). \(\textbf{Experiment S:}\) Its apparent power is \[ S_S=(200)(2.0) \] \[ S_S=400\,\text{VA} \] The measured real power equals this value. A unity power factor is possible for a purely resistive load. Thus, S is consistent. \(\textbf{Physical interpretation:}\) Real power cannot exceed the product of RMS voltage and RMS current for two sinusoidal quantities of the same frequency. Zero real power can coexist with nonzero current when the load is ideal and reactive. Likewise, equality of real and apparent power corresponds to unity power factor. The impossible result in Q is not explained by whether current leads or lags; neither phase direction can produce a cosine greater than one. The inconsistent measurement set is Experiment Q.

Explanation: The instantaneous power of a sinusoidal voltage-current pair contains a constant mean term and a term oscillating at twice the supply frequency. The mean and oscillating amplitude can be recovered from the maximum and minimum power readings. These quantities then determine real power, apparent power and power factor. \(\textbf{Power waveform relation:}\) Let the voltage and current be \[ v=V_m\sin\omega t \] \[ i=I_m\sin(\omega t-\phi) \] Their product is \[ p=V_mI_m\sin\omega t\sin(\omega t-\phi) \] Using a trigonometric identity, \[ p=\frac{V_mI_m}{2} \left[\cos\phi-\cos(2\omega t-\phi)\right] \] Since \[ \frac{V_mI_m}{2}=V_{\text{rms}}I_{\text{rms}} \] we may write \[ p=V_{\text{rms}}I_{\text{rms}}\cos\phi -V_{\text{rms}}I_{\text{rms}}\cos(2\omega t-\phi) \] \(\textbf{Recover average real power:}\) The mean of the maximum and minimum instantaneous powers is the constant term: \[ P_{\text{avg}}=\frac{p_{\max}+p_{\min}}2 \] Substituting, \[ P_{\text{avg}}=\frac{360+(-120)}2 \] \[ P_{\text{avg}}=120\,\text{W} \] \(\textbf{Recover apparent power from the oscillation amplitude:}\) The amplitude of the oscillating power term is \[ S=\frac{p_{\max}-p_{\min}}2 \] Substituting, \[ S=\frac{360-(-120)}2 \] \[ S=240\,\text{VA} \] Thus, \[ V_{\text{rms}}I_{\text{rms}}=240\,\text{VA} \] \(\textbf{Determine RMS current:}\) The source RMS voltage is \[ V_{\text{rms}}=120\,\text{V} \] Therefore, \[ I_{\text{rms}}=\frac{S}{V_{\text{rms}}} \] \[ I_{\text{rms}}=\frac{240}{120} \] \[ I_{\text{rms}}=2.0\,\text{A} \] \(\textbf{Determine power factor:}\) \[ \cos\phi=\frac{P_{\text{avg}}}{S} \] \[ \cos\phi=\frac{120}{240} \] \[ \cos\phi=0.50 \] The power measurements alone do not reveal whether current leads or lags. \(\textbf{Recover source frequency:}\) The instantaneous-power waveform repeats at twice the source-voltage frequency. Its measured period is \[ T_p=0.010\,\text{s} \] Therefore, \[ f_p=\frac1{T_p} \] \[ f_p=100\,\text{Hz} \] Hence, \[ f_{\text{source}}=\frac{f_p}{2} \] \[ f_{\text{source}}=50\,\text{Hz} \] \(\textbf{Physical interpretation:}\) The signed power graph contains more information than an ordinary average-power reading. Its mean gives real power, while the amplitude of its varying part gives apparent power for the stated sinusoidal conditions. The negative minimum shows that energy is returned to the source during part of each cycle. The power variation occurs at twice the voltage frequency because power involves the product of two sinusoids of the same frequency. The graph therefore permits independent recovery of current magnitude, power factor and source frequency without directly measuring the current waveform. \(\textbf{Final answer:}\) \[ \boxed{P_{\text{avg}}=120\,\text{W},\qquad I_{\text{rms}}=2.0\,\text{A},\qquad \cos\phi=0.50,\qquad f_{\text{source}}=50\,\text{Hz}} \]

Explanation: A wattmeter measures average real power, which is dissipated only in the resistor under the stated ideal-inductor assumption. Increasing frequency changes the inductive reactance and therefore changes the circuit current. The two power readings can be used to recover the unknown resistance and inductance without directly measuring current. \(\textbf{Relate measured power to circuit parameters:}\) For a series resistor-inductor circuit, \[ Z^2=R^2+\omega^2L^2 \] The average power is \[ P=I_{\text{rms}}^2R \] Using \(I_{\text{rms}}=V_{\text{rms}}/Z\), \[ P=\frac{V_{\text{rms}}^2R}{R^2+\omega^2L^2} \] \(\textbf{Use the first measurement:}\) At \(\omega_1=100\,\text{rad s}^{-1}\), \[ 800=\frac{10000R}{R^2+10000L^2} \] Rearranging, \[ R^2+10000L^2=12.5R \] \(\textbf{Use the second measurement:}\) At \(\omega_2=200\,\text{rad s}^{-1}\), \[ 500=\frac{10000R}{R^2+40000L^2} \] Therefore, \[ R^2+40000L^2=20R \] \(\textbf{Eliminate the inductance:}\) Subtracting the two equations, \[ 30000L^2=7.5R \] \[ L^2=\frac{R}{4000} \] Substituting into the first equation, \[ R^2+10000\left(\frac{R}{4000}\right)=12.5R \] \[ R^2+2.5R=12.5R \] Since \(R\gt 0\), \[ R=10\,\Omega \] \(\textbf{Recover inductance:}\) \[ L^2=\frac{10}{4000} \] \[ L^2=0.0025\,\text{H}^2 \] \[ L=0.050\,\text{H} \] \(\textbf{Resolve the current at the higher frequency:}\) At \(\omega_2=200\,\text{rad s}^{-1}\), \[ X_{L2}=\omega_2L \] \[ X_{L2}=(200)(0.050) \] \[ X_{L2}=10\,\Omega \] Therefore, \[ Z_2=\sqrt{10^2+10^2} \] \[ Z_2=10\sqrt2\,\Omega \] The RMS current is \[ I_2=\frac{100}{10\sqrt2} \] \[ I_2=5\sqrt2\,\text{A} \] The lag angle satisfies \[ \sin\phi_2=\frac{X_{L2}}{Z_2} \] \[ \sin\phi_2=\frac1{\sqrt2} \] Hence, the wattless current magnitude is \[ I_{\text{wattless}}=I_2\sin\phi_2 \] \[ I_{\text{wattless}}=(5\sqrt2)\left(\frac1{\sqrt2}\right) \] \[ I_{\text{wattless}}=5.0\,\text{A} \] \(\textbf{Physical check:}\) At the higher frequency, the active current component is also \(5.0\,\text{A}\). The corresponding real power is \[ P=(100)(5.0) \] \[ P=500\,\text{W} \] This agrees with the second wattmeter reading. The experiment demonstrates that frequency-dependent real-power measurements can reveal both resistance and inductance. The decrease in measured power does not mean that the ideal inductor has begun dissipating power; it results from the reduced RMS current. \(\textbf{Final answer:}\) \[ \boxed{R=10\,\Omega,\qquad L=0.050\,\text{H},\qquad I_{\text{wattless}}=5.0\,\text{A}} \]

Explanation: The appliance initially draws an active current component together with a larger lagging wattless component. The parallel capacitor supplies leading wattless current. The overall power factor depends on the magnitude of the resultant quadrature current, not simply on whether the capacitor has been added. \(\textbf{Original active component:}\) \[ I_{\text{active}}=I\cos\phi \] Substituting, \[ I_{\text{active}}=(10)(0.60) \] \[ I_{\text{active}}=6.0\,\text{A} \] \(\textbf{Original lagging component:}\) \[ \sin\phi=\sqrt{1-0.60^2} \] \[ \sin\phi=0.80 \] Therefore, \[ I_{\text{lagging}}=(10)(0.80) \] \[ I_{\text{lagging}}=8.0\,\text{A} \] \(\textbf{Quadrature current required by the target power factor:}\) The required overall power-factor magnitude is \[ |\cos\phi_s|=0.80 \] For the passive appliance and ideal capacitor, the active component remains positive. Hence, \[ I_s=\frac{I_{\text{active}}}{0.80} \] \[ I_s=\frac{6.0}{0.80} \] \[ I_s=7.5\,\text{A} \] The required magnitude of the net quadrature component is \[ |I_{\text{reactive},s}|=\sqrt{I_s^2-I_{\text{active}}^2} \] \[ |I_{\text{reactive},s}|=\sqrt{7.5^2-6.0^2} \] \[ |I_{\text{reactive},s}|=\sqrt{56.25-36} \] \[ |I_{\text{reactive},s}|=4.5\,\text{A} \] \(\textbf{First capacitor setting — Remaining lag:}\) Let the capacitor RMS current be \(I_C\). For an overall lagging current, \[ 8.0-I_C=+4.5 \] Therefore, \[ I_{C1}=3.5\,\text{A} \] \(\textbf{Second capacitor setting — Excessive compensation:}\) For an overall leading current, \[ 8.0-I_C=-4.5 \] Therefore, \[ I_{C2}=12.5\,\text{A} \] \(\textbf{Convert branch currents to capacitances:}\) For an ideal capacitor, \[ I_C=\omega CV_{\text{rms}} \] The supply angular frequency is \[ \omega=2\pi(50) \] \[ \omega=100\pi\,\text{rad s}^{-1} \] For the first setting, \[ C_1=\frac{3.5}{(100\pi)(100)} \] \[ C_1=\frac{350}{\pi}\,\mu\text{F} \] For the second setting, \[ C_2=\frac{12.5}{(100\pi)(100)} \] \[ C_2=\frac{1250}{\pi}\,\mu\text{F} \] \(\textbf{Physical interpretation:}\) The smaller capacitor cancels part of the appliance's lagging quadrature current. The larger capacitor exceeds the amount needed for unity power factor and leaves a leading quadrature current. Both cases produce the same magnitude of overall power factor because their remaining reactive current magnitudes are equal. The source RMS current is \(7.5\,\text{A}\) in both cases. Therefore, a power-factor magnitude measurement alone cannot determine whether correction is insufficient or excessive. An additional phase-direction measurement would distinguish the two operating conditions. The correct option is the one allowing both capacitances and assigning lagging behaviour to the smaller capacitor.

Explanation: For two sinusoids of the same frequency, instantaneous power contains a constant term and an oscillating term. The constant term equals average real power. The amplitude of the oscillating term equals apparent power. \(\textbf{Instantaneous-power relation:}\) Write \[ v=V_m\sin\omega t \] and \[ i=I_m\sin(\omega t-\phi) \] Then \[ p=vi \] Using the product identity, \[ p=\frac{V_mI_m}{2} \left[\cos\phi-\cos(2\omega t-\phi)\right] \] Since \[ S=\frac{V_mI_m}{2}=V_{\text{rms}}I_{\text{rms}} \] the power waveform becomes \[ p=P-S\cos(2\omega t-\phi) \] \(\textbf{Maximum instantaneous power:}\) The cosine term reaches \(-1\). Therefore, \[ p_{\max}=P+S \] Substituting, \[ p_{\max}=300+500 \] \[ p_{\max}=800\,\text{W} \] \(\textbf{Minimum instantaneous power:}\) The cosine term reaches \(+1\). Therefore, \[ p_{\min}=P-S \] \[ p_{\min}=300-500 \] \[ p_{\min}=-200\,\text{W} \] \(\textbf{Interpret the phase information:}\) The power factor is \[ \cos\phi=\frac PS \] \[ \cos\phi=\frac{300}{500} \] \[ \cos\phi=0.60 \] This determines the magnitude of the phase displacement for a passive load. However, the same cosine value corresponds to a leading or lagging current. The instantaneous-power extrema are unchanged when the phase sign reverses. A time-referenced voltage-current waveform would be needed to determine the direction. \(\textbf{Physical interpretation:}\) The negative minimum means that the load returns energy to the source during part of the cycle. Nevertheless, average absorbed power is positive because net energy transfer over a full cycle is into the load. A purely resistive load with unity power factor would have no negative instantaneous-power interval. The wattmeter and apparent-power measurements establish the power waveform's vertical range but do not establish its phase orientation relative to the voltage zero crossings. \(\textbf{Final answer:}\) \[ p_{\max}=800\,\text{W},\qquad p_{\min}=-200\,\text{W},\qquad |\cos\phi|=0.60 \] The measurements do not establish whether current leads or lags.

Explanation: The two capacitor placements perform different physical functions. A capacitor inserted in series changes the impedance of the appliance circuit itself. A capacitor connected in parallel can cancel part or all of the appliance's wattless current without changing the voltage across the original appliance. \(\textbf{Original appliance current:}\) Its impedance magnitude is \[ Z_0=\sqrt{6.0^2+8.0^2} \] \[ Z_0=10\,\Omega \] Therefore, \[ I_0=\frac{120}{10} \] \[ I_0=12\,\text{A} \] The original average real power is \[ P_0=I_0^2R \] \[ P_0=(12)^2(6.0) \] \[ P_0=864\,\text{W} \] \(\textbf{Experiment P — Series capacitor:}\) The added capacitor has \[ X_C=8.0\,\Omega \] The inductor also has \[ X_L=8.0\,\Omega \] The reactive contributions cancel: \[ X_L-X_C=0 \] Thus, the new series impedance is \[ Z_P=R=6.0\,\Omega \] The RMS source current becomes \[ I_P=\frac{120}{6.0} \] \[ I_P=20\,\text{A} \] The average power dissipated in the appliance resistor is \[ P_P=I_P^2R \] \[ P_P=(20)^2(6.0) \] \[ P_P=2400\,\text{W} \] \(\textbf{Experiment Q — Parallel capacitor:}\) The original appliance remains directly across the \(120\,\text{V}\) supply. Therefore, its RMS current remains \[ I_{\text{load}}=12\,\text{A} \] and its average real power remains \[ P_Q=864\,\text{W} \] Its power factor is \[ \cos\phi_0=\frac R{Z_0} \] \[ \cos\phi_0=\frac6{10}=0.60 \] The active current component is \[ I_{\text{active}}=(12)(0.60) \] \[ I_{\text{active}}=7.2\,\text{A} \] An appropriately chosen parallel capacitor cancels the lagging wattless component. The supply therefore carries only the active component: \[ I_Q=7.2\,\text{A} \] \(\textbf{Compare the source powers:}\) For P, \[ P_P=(120)(20)=2400\,\text{W} \] For Q, \[ P_Q=(120)(7.2)=864\,\text{W} \] Both corrected arrangements have unity overall power factor, but their currents and real powers are different. \(\textbf{Physical interpretation:}\) Series compensation reduces the appliance's total impedance and increases the current through its resistor. This greatly increases real power dissipation. Parallel compensation leaves the appliance voltage and current unchanged and instead reduces the wattless current supplied externally. A unity power factor therefore does not, by itself, guarantee reduced electrical consumption. The arrangement of the capacitor and the controlled operating condition must also be considered. \(\textbf{Final answer:}\) \[ \boxed{I_P=20\,\text{A},\quad P_P=2400\,\text{W},\quad I_Q=7.2\,\text{A},\quad P_Q=864\,\text{W}} \]

Explanation: The load current lags the applied voltage by \(60^\circ\). Voltage and current therefore have opposite signs during part of the positive-voltage half-cycle. Nevertheless, the signed energy transferred over the complete half-cycle can be positive. The result must be calculated from instantaneous power rather than from voltage polarity alone. \(\textbf{Instantaneous absorbed power:}\) \[ p(t)=v(t)i(t) \] Substituting, \[ p(t)=400\sin(100t)\sin\left(100t-\frac{\pi}{3}\right) \] Using \[ 2\sin A\sin B=\cos(A-B)-\cos(A+B) \] we obtain \[ p(t)=200\left[\cos\frac{\pi}{3}-\cos\left(200t-\frac{\pi}{3}\right)\right] \] Therefore, \[ p(t)=100-200\cos\left(200t-\frac{\pi}{3}\right) \] \(\textbf{Net energy in the first half-cycle:}\) \[ E_1=\int_0^{\pi/100}p(t)\,dt \] Substituting, \[ E_1=\int_0^{\pi/100} \left[100-200\cos\left(200t-\frac{\pi}{3}\right)\right]dt \] Integrating, \[ E_1=\left[100t-\sin\left(200t-\frac{\pi}{3}\right)\right]_0^{\pi/100} \] At the upper limit, \[ 200t-\frac{\pi}{3}=\frac{5\pi}{3} \] At the lower limit, the argument is \(-\pi/3\). Since \[ \sin\frac{5\pi}{3}=\sin\left(-\frac{\pi}{3}\right) \] the oscillating term contributes zero to the integral. Therefore, \[ E_1=100\left(\frac{\pi}{100}\right) \] \[ E_1=\pi\,\text{J} \] \(\textbf{Energy during the next half-cycle:}\) For sinusoidal voltage and current of the same frequency, \[ v(t+T/2)=-v(t) \] and \[ i(t+T/2)=-i(t) \] Thus, \[ p(t+T/2)=p(t) \] The instantaneous-power waveform repeats after half a voltage cycle. Consequently, \[ E_2=E_1 \] \[ E_2=\pi\,\text{J} \] \(\textbf{Average real power:}\) The complete-cycle energy is \[ E_{\text{cycle}}=E_1+E_2 \] \[ E_{\text{cycle}}=2\pi\,\text{J} \] The complete period is \[ T=\frac{2\pi}{100} \] \[ T=\frac{\pi}{50}\,\text{s} \] Therefore, \[ P_{\text{avg}}=\frac{E_{\text{cycle}}}{T} \] \[ P_{\text{avg}}=\frac{2\pi}{\pi/50} \] \[ P_{\text{avg}}=100\,\text{W} \] \(\textbf{Physical interpretation:}\) The load returns energy at some instants within the positive-voltage half-cycle because its current is then negative. Later in that same interval, positive instantaneous power more than compensates for the returned energy. The resulting net energy absorption is positive. The following negative-voltage half-cycle gives the same net absorption because both voltage and current reverse together. This shows why a negative source voltage does not imply negative average power transfer. \(\textbf{Final answer:}\) \[ \boxed{E_1=\pi\,\text{J},\qquad E_2=\pi\,\text{J},\qquad P_{\text{avg}}=100\,\text{W}} \]

Explanation: In a series RC circuit, average real power is dissipated in the resistor. The capacitor changes the RMS current by contributing a frequency-dependent reactance. At very high frequency, the ideal capacitor's reactance approaches zero, so the power approaches the value for the resistor alone. \(\textbf{Recover resistance from the high-frequency limit:}\) The limiting power is \[ P_{\infty}=\frac{V_{\text{rms}}^2}{R} \] Therefore, \[ R=\frac{V_{\text{rms}}^2}{P_{\infty}} \] Substituting, \[ R=\frac{100^2}{500} \] \[ R=20\,\Omega \] \(\textbf{Use the measured power at the finite frequency:}\) For a series RC circuit, \[ Z^2=R^2+X_C^2 \] The average power is \[ P=\frac{V_{\text{rms}}^2R}{R^2+X_C^2} \] At \[ \omega_1=100\,\text{rad s}^{-1} \] the measured power is \(100\,\text{W}\). Therefore, \[ 100=\frac{(100)^2(20)}{20^2+X_{C1}^2} \] Rearranging, \[ 20^2+X_{C1}^2=2000 \] \[ X_{C1}^2=2000-400 \] \[ X_{C1}=40\,\Omega \] \(\textbf{Recover capacitance:}\) \[ X_C=\frac1{\omega C} \] Therefore, \[ C=\frac1{\omega_1X_{C1}} \] Substituting, \[ C=\frac1{(100)(40)} \] \[ C=2.5\times10^{-4}\,\text{F} \] \[ C=250\,\mu\text{F} \] \(\textbf{Predict power after doubling frequency:}\) At \[ \omega_2=200\,\text{rad s}^{-1} \] the capacitive reactance becomes \[ X_{C2}=\frac1{(200)(250\times10^{-6})} \] \[ X_{C2}=20\,\Omega \] The new impedance magnitude is \[ Z_2=\sqrt{20^2+20^2} \] \[ Z_2=20\sqrt2\,\Omega \] The average power is \[ P_2=\frac{V_{\text{rms}}^2R}{Z_2^2} \] \[ P_2=\frac{10000(20)}{(20\sqrt2)^2} \] \[ P_2=250\,\text{W} \] \(\textbf{Physical interpretation:}\) The high-frequency limiting power reveals the fixed resistance without requiring a separate resistance measurement. The lower-frequency wattmeter reading then reveals the capacitor's reactance. Increasing frequency reduces that reactance, allowing more current and increasing resistor heating. At the second frequency, resistance and capacitive reactance have equal magnitudes. The current therefore leads voltage by \(45^\circ\). The power remains below the \(500\,\text{W}\) high-frequency limit because the capacitive reactance is still nonzero. \(\textbf{Final answer:}\) \[ \boxed{R=20\,\Omega,\qquad C=250\,\mu\text{F},\qquad P_2=250\,\text{W}} \]