Electromagnetic Induction and Alternating Currents MCQs with Answers – Part 7 (NEET Physics)

Explanation: The resistor and capacitor have the same series current. However, their RMS voltage contributions differ in phase by \(90^\circ\). The measured component voltages therefore satisfy a phasor relation rather than an arithmetic sum. The capacitor-voltage reading determines reactance and capacitance. Its peak voltage determines maximum stored energy. \(\textbf{Resistive opposition:}\) \[ V_R=I_{\text{rms}}R \] Thus, \[ R=\frac{V_R}{I_{\text{rms}}} \] \[ R=\frac{80}{2.0} \] \[ R=40\,\Omega \] \(\textbf{Capacitive reactance:}\) \[ V_C=I_{\text{rms}}X_C \] Therefore, \[ X_C=\frac{60}{2.0} \] \[ X_C=30\,\Omega \] \(\textbf{Recover capacitance:}\) \[ X_C=\frac1{\omega C} \] Hence, \[ C=\frac1{\omega X_C} \] \[ C=\frac1{(50)(30)} \] \[ C=\frac1{1500}\,\text{F} \] \(\textbf{Source-voltage consistency:}\) For the series RC circuit, \[ V_s=\sqrt{V_R^2+V_C^2} \] \[ V_s=\sqrt{80^2+60^2} \] \[ V_s=100\,\text{V} \] This agrees with the third voltmeter. \(\textbf{Average real power:}\) Only the resistor dissipates positive average power: \[ P_{\text{avg}}=I_{\text{rms}}^2R \] \[ P_{\text{avg}}=(2.0)^2(40) \] \[ P_{\text{avg}}=160\,\text{W} \] \(\textbf{Recover capacitor peak voltage:}\) \[ V_{Cm}=\sqrt2V_{C,\text{rms}} \] \[ V_{Cm}=60\sqrt2\,\text{V} \] \(\textbf{Maximum capacitor energy:}\) \[ U_{C,\max}=\frac12CV_{Cm}^2 \] Substituting, \[ U_{C,\max}=\frac12\left(\frac1{1500}\right)(60\sqrt2)^2 \] \[ U_{C,\max}=\frac{3600}{1500} \] \[ U_{C,\max}=2.4\,\text{J} \] \(\textbf{Physical interpretation:}\) The capacitor's RMS voltage differs from the source RMS voltage because the resistor and capacitor voltage phasors are perpendicular. The capacitor stores energy according to the square of its instantaneous voltage. Therefore, peak capacitor voltage, not RMS voltage, belongs in the maximum-energy expression. The resistor is responsible for the circuit's positive average real-power consumption. The capacitor affects the impedance and phase while exchanging stored energy reversibly. The combination of voltage and current measurements determines both circuit components and maximum electric-energy storage. \(\textbf{Final answer:}\) \[ \boxed{C=\frac1{1500}\,\text{F},\qquad P_{\text{avg}}=160\,\text{W},\qquad U_{C,\max}=2.4\,\text{J}} \]

Explanation: The first measurement determines inductive reactance and inductance. Maximum magnetic energy depends on the square of peak current. The second experiment therefore requires a specified current-amplitude increase. The voltage must then be adjusted to overcome the greater inductive reactance at the higher frequency. \(\textbf{Initial inductive reactance:}\) For a pure inductor, \[ X_{L1}=\frac{V_1}{I_1} \] Substituting, \[ X_{L1}=\frac{40}{2.0} \] \[ X_{L1}=20\,\Omega \] \(\textbf{Recover inductance:}\) \[ X_{L1}=\omega_1L \] Therefore, \[ L=\frac{20}{100} \] \[ L=0.20\,\text{H} \] \(\textbf{Initial peak current:}\) \[ I_{m1}=\sqrt2I_1 \] \[ I_{m1}=2\sqrt2\,\text{A} \] \(\textbf{Initial maximum magnetic energy:}\) \[ U_{1,\max}=\frac12LI_{m1}^2 \] \[ U_{1,\max}=\frac12(0.20)(2\sqrt2)^2 \] \[ U_{1,\max}=0.80\,\text{J} \] \(\textbf{Required final maximum energy:}\) The new maximum energy must be twice as large: \[ U_{2,\max}=2U_{1,\max} \] \[ U_{2,\max}=1.6\,\text{J} \] \(\textbf{Recover required current:}\) Because inductance is fixed, \[ \frac{U_{2,\max}}{U_{1,\max}}=\left(\frac{I_{m2}}{I_{m1}}\right)^2 \] Therefore, \[ \frac{I_{m2}}{I_{m1}}=\sqrt2 \] The same ratio applies to RMS current: \[ I_2=I_1\sqrt2 \] \[ I_2=2\sqrt2\,\text{A} \] \(\textbf{New inductive reactance:}\) \[ X_{L2}=\omega_2L \] \[ X_{L2}=(300)(0.20) \] \[ X_{L2}=60\,\Omega \] \(\textbf{Required new source voltage:}\) \[ V_2=I_2X_{L2} \] \[ V_2=(2\sqrt2)(60) \] \[ V_2=120\sqrt2\,\text{V} \] \(\textbf{Physical interpretation:}\) The higher frequency increases the inductive reactance threefold. The requirement of twice the stored magnetic energy demands a further factor of \(\sqrt2\) in current amplitude. The source voltage must therefore accommodate both changes. An increase in source voltage does not necessarily imply a corresponding increase in average real power. The ideal inductor continues to return its stored magnetic energy during each cycle and consumes zero average real power. \(\textbf{Final answer:}\) \[ \boxed{L=0.20\,\text{H},\qquad V_2=120\sqrt2\,\text{V},\qquad \frac{I_Q}{I_P}=\sqrt2} \]

Explanation: RMS current measures the heating-equivalent magnitude of an alternating current. It does not independently establish that the connected component consumes positive average real power. An ideal capacitor can carry alternating current while receiving and returning equal amounts of energy during successive portions of a cycle. Statement I is correct. For an ideal inductor, \[ X_L=\omega L \] At unchanged source voltage, \[ I_L=\frac{V_{\text{rms}}}{\omega L} \] Doubling angular frequency halves RMS and peak current. Its maximum magnetic energy depends on peak current squared: \[ U_{L,\max}=\frac12LI_m^2 \] Therefore, maximum magnetic energy becomes one-quarter. Statement II is correct. For an ideal capacitor, \[ I_C=\omega CV_{\text{rms}} \] Doubling frequency doubles current amplitude at unchanged voltage. However, the maximum electric energy is \[ U_{C,\max}=\frac12CV_m^2 \] Because capacitance and voltage amplitude remain fixed, maximum electric energy is unchanged. Statement III is incorrect. A pure resistor has no inductive or capacitive reactance. Its average power is \[ P_R=\frac{V_{\text{rms}}^2}{R} \] At unchanged voltage and resistance, this remains constant regardless of source frequency. Statement IV is correct. The statements distinguish reactive current from real-power consumption and differentiate the energy-storage relations for inductors and capacitors. The correct combination is Statements I, II and IV only.

Explanation: The RMS ammeter gives the heating-equivalent current, not its instantaneous maximum. The oscilloscope frequency determines the time available for the positive half-cycle. Heat and transferred charge must then be calculated differently because power depends on current squared while charge depends on current itself. \(\textbf{Recover peak current:}\) \[ I_m=\sqrt2I_{\text{rms}} \] Substituting, \[ I_m=2\sqrt2\,\text{A} \] \(\textbf{Determine angular frequency:}\) \[ \omega=2\pi f \] \[ \omega=2\pi(25) \] \[ \omega=50\pi\,\text{rad s}^{-1} \] \(\textbf{Current waveform:}\) \[ i(t)=2\sqrt2\sin(50\pi t) \] \(\textbf{Maximum instantaneous resistor power:}\) \[ p(t)=i^2(t)R \] The maximum occurs when \(|i|=I_m\): \[ p_{\max}=I_m^2R \] \[ p_{\max}=(2\sqrt2)^2(10) \] \[ p_{\max}=80\,\text{W} \] \(\textbf{Determine half-cycle duration:}\) \[ T=\frac1{25} \] \[ T=0.040\,\text{s} \] Therefore, \[ \frac T2=0.020\,\text{s} \] \(\textbf{Average resistor heating:}\) \[ P_{\text{avg}}=I_{\text{rms}}^2R \] \[ P_{\text{avg}}=(2.0)^2(10) \] \[ P_{\text{avg}}=40\,\text{W} \] \(\textbf{Heat generated during the positive half-cycle:}\) The mean of \(\sin^2\) over a half-cycle is one-half, equal to its complete-cycle mean. Thus, \[ H_{\text{half}}=P_{\text{avg}}\frac T2 \] \[ H_{\text{half}}=(40)(0.020) \] \[ H_{\text{half}}=0.80\,\text{J} \] \(\textbf{Transferred charge during the same interval:}\) \[ q_{\text{half}}=\int_0^{T/2}i(t)\,dt \] \[ q_{\text{half}}=2\sqrt2\int_0^{1/50}\sin(50\pi t)\,dt \] Integrating, \[ q_{\text{half}}=-\frac{2\sqrt2}{50\pi}\left[\cos(50\pi t)\right]_0^{1/50} \] \[ q_{\text{half}}=\frac{4\sqrt2}{50\pi} \] \[ q_{\text{half}}=\frac{2\sqrt2}{25\pi}\,\text{C} \] \(\textbf{Physical interpretation:}\) The maximum instantaneous power is twice the average power for a sinusoidal current through a pure resistor. The positive half-cycle produces half the complete-cycle heat. Its transferred charge is determined by the area under the positive current waveform. The charge cannot be calculated by assuming the current remains at its maximum throughout the interval. The result distinguishes peak current, RMS current, heating and charge accumulation. \(\textbf{Final answer:}\) \[ \boxed{p_{\max}=80\,\text{W},\qquad H_{\text{half}}=0.80\,\text{J},\qquad q_{\text{half}}=\frac{2\sqrt2}{25\pi}\,\text{C}} \]

Explanation: The two frequency measurements produce equal current magnitudes but opposite phase characters. Equal current at fixed source RMS voltage implies equal impedance magnitudes. The known resistance allows the magnitude of net reactance to be determined. The lead-lag information supplies the signs needed to form two independent equations for \(L\) and \(C\). \(\textbf{Impedance from either current measurement:}\) \[ Z=\frac{V_{\text{rms}}}{I_{\text{rms}}} \] \[ Z=\frac{100}{2.0} \] \[ Z=50\,\Omega \] \(\textbf{Magnitude of net reactance:}\) For a series LCR circuit, \[ Z^2=R^2+(X_L-X_C)^2 \] Therefore, \[ |X_L-X_C|=\sqrt{50^2-40^2} \] \[ |X_L-X_C|=\sqrt{2500-1600} \] \[ |X_L-X_C|=30\,\Omega \] \(\textbf{Low-frequency condition:}\) At \[ \omega_1=100\,\text{rad s}^{-1} \] the current leads voltage. The circuit is therefore capacitive: \[ X_{L1}-X_{C1}=-30\,\Omega \] Substituting reactance expressions, \[ 100L-\frac1{100C}=-30 \] \(\textbf{High-frequency condition:}\) At \[ \omega_2=400\,\text{rad s}^{-1} \] the current lags voltage. The circuit is inductive: \[ X_{L2}-X_{C2}=+30\,\Omega \] Therefore, \[ 400L-\frac1{400C}=30 \] \(\textbf{Solve for inductance and capacitance:}\) Multiply the first equation by \(100\): \[ 10000L-\frac1C=-3000 \] Multiply the second equation by \(400\): \[ 160000L-\frac1C=12000 \] Subtracting, \[ 150000L=15000 \] \[ L=0.10\,\text{H} \] \(\textbf{Recover capacitance:}\) Using the first equation, \[ 100(0.10)-\frac1{100C}=-30 \] \[ 10-\frac1{100C}=-30 \] \[ \frac1{100C}=40 \] \[ C=\frac1{4000}\,\text{F} \] \[ C=250\,\mu\text{F} \] \(\textbf{Resonance angular frequency:}\) At resonance, \[ \omega_0=\frac1{\sqrt{LC}} \] Substituting, \[ \omega_0=\frac1{\sqrt{(0.10)(250\times10^{-6})}} \] \[ \omega_0=\frac1{0.005} \] \[ \omega_0=200\,\text{rad s}^{-1} \] \(\textbf{Phase consistency:}\) The first measurement lies below resonance and is capacitive. The second lies above resonance and is inductive. Their equal impedance magnitudes are consistent with equal and opposite net reactances. \(\textbf{Physical interpretation:}\) Equal current magnitudes do not necessarily mean that an LCR circuit has the same phase character at two frequencies. The same impedance magnitude can occur on opposite sides of resonance. The lead-lag observations identify which reactance dominates in each case. Together, the two current measurements determine both unknown reactive components. The recovered resonance frequency lies between the two test frequencies, as expected. \(\textbf{Final answer:}\) \[ \boxed{L=0.10\,\text{H},\qquad C=250\,\mu\text{F},\qquad \omega_0=200\,\text{rad s}^{-1}} \]

Explanation: The three component voltages are measured in the same series circuit, so they correspond to the same RMS current. The inductor and capacitor voltage contributions oppose each other in the phasor representation. Changing capacitance changes capacitive reactance and therefore the total impedance. The current must be recalculated after the change. \(\textbf{Initial circuit current:}\) For the resistor, \[ V_R=I_1R \] Therefore, \[ I_1=\frac{V_R}{R} \] \[ I_1=\frac{30}{15} \] \[ I_1=2.0\,\text{A} \] \(\textbf{Initial inductive reactance:}\) \[ X_L=\frac{V_L}{I_1} \] \[ X_L=\frac{80}{2.0} \] \[ X_L=40\,\Omega \] \(\textbf{Initial capacitive reactance:}\) \[ X_{C1}=\frac{V_C}{I_1} \] \[ X_{C1}=\frac{40}{2.0} \] \[ X_{C1}=20\,\Omega \] \(\textbf{Initial source-voltage check:}\) The net reactance is \[ X_{L}-X_{C1}=40-20 \] \[ X_L-X_{C1}=20\,\Omega \] The initial impedance is \[ Z_1=\sqrt{15^2+20^2} \] \[ Z_1=25\,\Omega \] Thus, \[ V_s=I_1Z_1 \] \[ V_s=(2.0)(25) \] \[ V_s=50\,\text{V} \] This agrees with the source. \(\textbf{Capacitance-doubling effect:}\) At unchanged angular frequency, \[ X_C=\frac1{\omega C} \] Doubling capacitance halves capacitive reactance: \[ X_{C2}=\frac{20}{2} \] \[ X_{C2}=10\,\Omega \] \(\textbf{New net reactance:}\) \[ X_{\text{net},2}=X_L-X_{C2} \] \[ X_{\text{net},2}=40-10 \] \[ X_{\text{net},2}=30\,\Omega \] \(\textbf{New impedance:}\) \[ Z_2=\sqrt{R^2+X_{\text{net},2}^2} \] \[ Z_2=\sqrt{15^2+30^2} \] \[ Z_2=15\sqrt5\,\Omega \] \(\textbf{Final current:}\) \[ I_2=\frac{V_s}{Z_2} \] \[ I_2=\frac{50}{15\sqrt5} \] \[ I_2=\frac{2\sqrt5}{3}\,\text{A} \] \(\textbf{Final phase character:}\) Since \[ X_L-X_{C2}\gt 0 \] the circuit remains inductive. Therefore, current lags source voltage. \(\textbf{Physical interpretation:}\) Doubling capacitance weakens the capacitive reactance. Because the inductor initially dominated, the reduction in capacitive opposition increases the magnitude of net inductive reactance. The complete impedance increases, so the RMS current decreases. The resistor, inductor and capacitor voltage readings before the modification cannot simply be reused afterwards because their common circuit current has changed. The phase remains lagging, with a greater inductive contribution relative to resistance. \(\textbf{Final answer:}\) \[ \boxed{I_1=2.0\,\text{A},\qquad I_2=\frac{2\sqrt5}{3}\,\text{A}} \] The final current lags the source voltage.

Explanation: At series resonance, inductive and capacitive reactances have equal magnitudes. Their instantaneous voltage contributions oppose each other in the source-voltage phasor equation. The individual reactive voltage magnitudes can therefore exceed the applied source voltage. The measured current and component voltages determine the component values and maximum stored energies. \(\textbf{Recover resistance:}\) At resonance, the source RMS voltage equals the resistor RMS voltage: \[ V_s=V_R=50\,\text{V} \] The series current is \[ I_{\text{rms}}=2.5\,\text{A} \] Therefore, \[ R=\frac{V_R}{I_{\text{rms}}} \] \[ R=\frac{50}{2.5} \] \[ R=20\,\Omega \] \(\textbf{Inductive reactance at resonance:}\) \[ X_L=\frac{V_L}{I_{\text{rms}}} \] \[ X_L=\frac{200}{2.5} \] \[ X_L=80\,\Omega \] \(\textbf{Recover inductance:}\) \[ X_L=\omega L \] Therefore, \[ L=\frac{80}{200} \] \[ L=0.40\,\text{H} \] \(\textbf{Capacitive reactance at resonance:}\) \[ X_C=X_L \] \[ X_C=80\,\Omega \] \(\textbf{Recover capacitance:}\) \[ X_C=\frac1{\omega C} \] Therefore, \[ C=\frac1{\omega X_C} \] \[ C=\frac1{(200)(80)} \] \[ C=62.5\,\mu\text{F} \] \(\textbf{Peak circuit current:}\) \[ I_m=\sqrt2I_{\text{rms}} \] \[ I_m=2.5\sqrt2\,\text{A} \] \(\textbf{Maximum inductor energy:}\) \[ U_{L,\max}=\frac12LI_m^2 \] \[ U_{L,\max}=\frac12(0.40)(2.5\sqrt2)^2 \] \[ U_{L,\max}=2.5\,\text{J} \] \(\textbf{Peak capacitor voltage:}\) \[ V_{Cm}=\sqrt2V_{C,\text{rms}} \] \[ V_{Cm}=200\sqrt2\,\text{V} \] \(\textbf{Maximum capacitor energy:}\) \[ U_{C,\max}=\frac12CV_{Cm}^2 \] Substituting, \[ U_{C,\max}=\frac12(62.5\times10^{-6})(200\sqrt2)^2 \] \[ U_{C,\max}=2.5\,\text{J} \] \(\textbf{Energy interpretation:}\) The inductor and capacitor exchange energy during the AC cycle. The maximum magnetic and electric energies are equal at resonance in the stated ideal reactive elements. The resistor simultaneously dissipates positive average power, which the source must replace. Thus, equal maximum reactive energies do not imply zero source-power consumption. \(\textbf{Physical interpretation:}\) Resonance minimizes the net reactive opposition in a series LCR circuit. Large individual inductor and capacitor voltages can cancel in the source-voltage phasor sum. The same resonance condition also permits energy to be exchanged between magnetic and electric storage. The measured reactive component voltages determine maximum stored energies when their peak values are used. The resistor remains responsible for irreversible energy dissipation. \(\textbf{Final answer:}\) \[ \boxed{R=20\,\Omega,\qquad U_{L,\max}=2.5\,\text{J},\qquad U_{C,\max}=2.5\,\text{J}} \]

Explanation: The source RMS voltage remains unchanged. Equal RMS currents therefore imply equal impedance magnitudes. Statement I is correct. For a series LCR circuit, \[ Z=\sqrt{R^2+(X_L-X_C)^2} \] The resistance is fixed. Consequently, equal impedance magnitudes require equal magnitudes of net reactance. However, the phase observations identify opposite reactive characters. Current leads at the lower frequency, so capacitive reactance dominates. Current lags at the higher frequency, so inductive reactance dominates. The net reactances therefore have equal magnitudes but opposite signs. Statement II is correct. The phase angle is determined by \[ \tan\phi=\frac{X_L-X_C}{R} \] Equal phase-angle magnitudes do not imply a fixed arithmetic ratio between the corresponding source frequencies. The frequency dependence combines a term proportional to \(\omega\) and another inversely proportional to \(\omega\). Statement III is incorrect. At the lower frequency, net reactance is capacitive. At the higher frequency, it is inductive. For fixed positive \(L\) and \(C\), the net reactance passes continuously through zero between these conditions. That zero-reactance condition is series resonance. Statement IV is correct. The identical wattmeter readings are consistent with the equal currents and unchanged resistance because \[ P_{\text{avg}}=I_{\text{rms}}^2R \] The experiment demonstrates that identical impedance magnitude and real power can occur at two different frequencies on opposite sides of resonance. Phase information is essential for distinguishing those conditions. The correct combination is Statements I, II and IV only.

Explanation: The circuit contains both magnetic and electric energy-storage elements. At the specified operating frequency, their reactances are equal. The circuit is therefore at series resonance. An important consequence is that the instantaneous inductor and capacitor voltages cancel, even though each can be nonzero. \(\textbf{Check the resonance condition:}\) The inductive reactance is \[ X_L=\omega L \] Substituting, \[ X_L=(100)(0.30) \] \[ X_L=30\,\Omega \] The capacitive reactance is \[ X_C=\frac1{\omega C} \] \[ X_C=\frac1{100(1/3000)} \] \[ X_C=30\,\Omega \] Therefore, \[ X_L=X_C \] \(\textbf{Instantaneous inductor voltage:}\) For the ideal inductor, \[ v_L=L\frac{di}{dt} \] Differentiating the measured current, \[ \frac{di}{dt}=200\cos(100t) \] Thus, \[ v_L(t)=(0.30)(200)\cos(100t) \] \[ v_L(t)=60\cos(100t)\,\text{V} \] At \(t=0\), \[ v_L(0)=+60\,\text{V} \] \(\textbf{Instantaneous capacitor voltage:}\) The capacitor voltage satisfies \[ i=C\frac{dv_C}{dt} \] The sinusoidal steady-state solution without a DC offset is \[ v_C(t)=-\frac{I_m}{\omega C}\cos(100t) \] Substituting, \[ v_C(t)=-\frac{2.0}{100(1/3000)}\cos(100t) \] \[ v_C(t)=-60\cos(100t)\,\text{V} \] Therefore, \[ v_C(0)=-60\,\text{V} \] \(\textbf{Stored energies at the initial instant:}\) At \(t=0\), current is zero: \[ i(0)=0 \] Hence, magnetic energy is \[ U_L(0)=\frac12Li^2(0)=0 \] The capacitor stores \[ U_C(0)=\frac12Cv_C^2(0) \] \[ U_C(0)=\frac12\left(\frac1{3000}\right)(60)^2 \] \[ U_C(0)=0.60\,\text{J} \] \(\textbf{Check energy during the cycle:}\) The magnetic energy at any instant is \[ U_L(t)=\frac12(0.30)[2\sin(100t)]^2 \] \[ U_L(t)=0.60\sin^2(100t)\,\text{J} \] Similarly, \[ U_C(t)=\frac12\left(\frac1{3000}\right)[60\cos(100t)]^2 \] \[ U_C(t)=0.60\cos^2(100t)\,\text{J} \] Adding, \[ U_L(t)+U_C(t)=0.60[\sin^2(100t)+\cos^2(100t)] \] \[ U_L(t)+U_C(t)=0.60\,\text{J} \] \(\textbf{Energy-conservation interpretation:}\) The inductor and capacitor exchange energy as their instantaneous current and voltage change. At resonance, the source voltage is in phase with current. The instantaneous source power is therefore equal to instantaneous resistor heating. Consequently, the source continuously replaces resistive losses while the total energy stored in the ideal reactive elements remains constant in this sinusoidal steady state. \(\textbf{Physical interpretation:}\) Zero current does not imply zero voltage across an inductor or capacitor. At the initial instant, the capacitor stores all the reactive energy. One-quarter cycle later, the inductor stores all of it. The separate energies oscillate, but their sum remains constant. This distinguishes energy exchange between reactive elements from irreversible heating in the resistor. \(\textbf{Final answer:}\) \[ \boxed{v_L(0)=+60\,\text{V},\qquad v_C(0)=-60\,\text{V},\qquad U_L+U_C=0.60\,\text{J}\ \text{(constant)}} \]

Explanation: Initially, the inductive and capacitive reactances are equal. The circuit is therefore at resonance and its impedance equals the resistance. Closing the switch removes the capacitor's contribution from the conducting path. The new circuit has resistance and inductive reactance, so its current and phase must be recalculated. \(\textbf{Initial resonance condition:}\) \[ X_L=40\,\Omega \] \[ X_C=40\,\Omega \] Therefore, \[ X_L-X_C=0 \] \(\textbf{Initial impedance and current:}\) \[ Z_1=\sqrt{R^2+(X_L-X_C)^2} \] \[ Z_1=\sqrt{30^2+0^2} \] \[ Z_1=30\,\Omega \] Thus, \[ I_1=\frac{V_{\text{rms}}}{Z_1} \] \[ I_1=\frac{150}{30} \] \[ I_1=5.0\,\text{A} \] \(\textbf{Initial resistor heating:}\) \[ P_1=I_1^2R \] \[ P_1=(5.0)^2(30) \] \[ P_1=750\,\text{W} \] \(\textbf{Circuit after the capacitor is bypassed:}\) The ideal closed switch makes the capacitor terminal voltage zero in steady operation. Current now flows through the series resistor and inductor. Their impedance is \[ Z_2=\sqrt{R^2+X_L^2} \] \[ Z_2=\sqrt{30^2+40^2} \] \[ Z_2=50\,\Omega \] \(\textbf{Final RMS current:}\) \[ I_2=\frac{V_{\text{rms}}}{Z_2} \] \[ I_2=\frac{150}{50} \] \[ I_2=3.0\,\text{A} \] \(\textbf{Final resistor power:}\) \[ P_2=I_2^2R \] \[ P_2=(3.0)^2(30) \] \[ P_2=270\,\text{W} \] \(\textbf{Determine final phase:}\) The remaining reactance is inductive: \[ X_{\text{net},2}=+40\,\Omega \] Therefore, current lags the applied source voltage. Its power factor is \[ \cos\phi_2=\frac R{Z_2} \] \[ \cos\phi_2=\frac{30}{50} \] \[ \cos\phi_2=\frac35 \] \(\textbf{Physical interpretation:}\) Before switching, inductive and capacitive opposition cancel in the series circuit. Bypassing the capacitor destroys this cancellation. The circuit impedance increases from \(30\,\Omega\) to \(50\,\Omega\), reducing current. The resistor therefore dissipates less average power, even though its own resistance is unchanged. The current also changes from being in phase with source voltage to lagging it. The experiment illustrates why changing circuit connectivity can be more significant than merely changing a component value. \(\textbf{Final answer:}\) \[ \boxed{I_1=5.0\,\text{A},\qquad P_2=270\,\text{W},\qquad \phi_2=\tan^{-1}\!\left(\frac43\right)\ \text{(lagging)}} \]

Explanation: The inductive reactance remains unchanged because frequency and inductance are fixed. Changing the capacitor changes capacitive reactance. The two capacitances are chosen so that the net reactances have equal magnitudes but opposite signs. This gives equal current magnitudes while reversing the circuit's reactive phase character. \(\textbf{Fixed inductive reactance:}\) \[ X_L=\omega L \] \[ X_L=(100)(0.30) \] \[ X_L=30\,\Omega \] \(\textbf{Experiment P — Smaller capacitor:}\) \[ X_{CP}=\frac1{\omega C_P} \] \[ X_{CP}=\frac1{(100)(200\times10^{-6})} \] \[ X_{CP}=50\,\Omega \] The net reactance is \[ X_P=X_L-X_{CP} \] \[ X_P=30-50 \] \[ X_P=-20\,\Omega \] Thus, P is capacitive. \(\textbf{Impedance and current in P:}\) \[ Z_P=\sqrt{20^2+(-20)^2} \] \[ Z_P=20\sqrt2\,\Omega \] Therefore, \[ I_P=\frac{20\sqrt2}{20\sqrt2} \] \[ I_P=1.0\,\text{A} \] \(\textbf{Experiment Q — Larger capacitor:}\) \[ X_{CQ}=\frac1{(100)(1000\times10^{-6})} \] \[ X_{CQ}=10\,\Omega \] Hence, \[ X_Q=30-10 \] \[ X_Q=+20\,\Omega \] The circuit is now inductive. \(\textbf{Impedance and current in Q:}\) \[ Z_Q=\sqrt{20^2+20^2} \] \[ Z_Q=20\sqrt2\,\Omega \] Therefore, \[ I_Q=\frac{20\sqrt2}{20\sqrt2} \] \[ I_Q=1.0\,\text{A} \] Thus, \[ \frac{I_Q}{I_P}=1 \] \(\textbf{Compare capacitor RMS voltages:}\) \[ V_C=I_{\text{rms}}X_C \] For P, \[ V_{CP}=(1.0)(50)=50\,\text{V} \] For Q, \[ V_{CQ}=(1.0)(10)=10\,\text{V} \] \(\textbf{Maximum capacitor energy:}\) Since \[ V_{Cm}=\sqrt2V_{C,\text{rms}} \] the maximum electric energy is \[ U_{C,\max}=\frac12C(2V_{C,\text{rms}}^2) \] \[ U_{C,\max}=CV_{C,\text{rms}}^2 \] For P, \[ U_{CP,\max}=(200\times10^{-6})(50)^2 \] \[ U_{CP,\max}=0.50\,\text{J} \] For Q, \[ U_{CQ,\max}=(1000\times10^{-6})(10)^2 \] \[ U_{CQ,\max}=0.10\,\text{J} \] Therefore, \[ \frac{U_{CQ,\max}}{U_{CP,\max}}=\frac{0.10}{0.50} \] \[ \frac{U_{CQ,\max}}{U_{CP,\max}}=\frac15 \] \(\textbf{Physical interpretation:}\) Both configurations carry the same RMS current and dissipate the same average resistor power. Nevertheless, their capacitor voltages and maximum stored electric energies differ. The larger capacitor has smaller reactance at the same frequency. Its voltage amplitude is sufficiently reduced that its maximum stored energy is lower. The net circuit phase also reverses from leading to lagging. Equal current and impedance magnitudes therefore do not establish identical energy-storage behaviour. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_Q}{I_P}=1,\qquad \frac{U_{CQ,\max}}{U_{CP,\max}}=\frac15,\qquad X_{\text{net},Q}=+20\,\Omega\ \text{(lagging current)}} \]

Explanation: The generator's angular speed controls both its output frequency and its induced-emf amplitude. Doubling the speed therefore changes the circuit in two ways simultaneously. The source voltage increases, but the LCR impedance also changes because both reactances depend on frequency. The final power must be calculated by combining these effects. \(\textbf{Initial resonance condition:}\) The circuit's resonance angular frequency is \[ \omega_0=\frac1{\sqrt{LC}} \] Substituting, \[ \omega_0=\frac1{\sqrt{(0.20)(500\times10^{-6})}} \] \[ \omega_0=\frac1{0.010} \] \[ \omega_0=100\,\text{rad s}^{-1} \] Thus, the circuit initially operates at resonance. \(\textbf{Initial current and resistor power:}\) At resonance, \[ Z_1=R=20\,\Omega \] Therefore, \[ I_1=\frac{100}{20} \] \[ I_1=5.0\,\text{A} \] The average power is \[ P_1=I_1^2R \] \[ P_1=(5.0)^2(20) \] \[ P_1=500\,\text{W} \] \(\textbf{Generator output after doubling speed:}\) For fixed generator geometry and magnetic field, \[ E_m=NBA\omega \] Therefore, RMS induced emf also varies directly with angular speed. Hence, \[ E_{2,\text{rms}}=2E_{1,\text{rms}} \] \[ E_{2,\text{rms}}=200\,\text{V} \] The new angular frequency is \[ \omega_2=200\,\text{rad s}^{-1} \] \(\textbf{New reactive opposition:}\) \[ X_{L2}=\omega_2L \] \[ X_{L2}=(200)(0.20) \] \[ X_{L2}=40\,\Omega \] For the capacitor, \[ X_{C2}=\frac1{\omega_2C} \] \[ X_{C2}=\frac1{(200)(500\times10^{-6})} \] \[ X_{C2}=10\,\Omega \] Thus, \[ X_{\text{net},2}=40-10 \] \[ X_{\text{net},2}=+30\,\Omega \] \(\textbf{New impedance and current:}\) \[ Z_2=\sqrt{20^2+30^2} \] \[ Z_2=10\sqrt{13}\,\Omega \] Therefore, \[ I_2=\frac{200}{10\sqrt{13}} \] \[ I_2=\frac{20}{\sqrt{13}}\,\text{A} \] \(\textbf{Final average resistor power:}\) \[ P_2=I_2^2R \] \[ P_2=\left(\frac{20}{\sqrt{13}}\right)^2(20) \] \[ P_2=\frac{8000}{13}\,\text{W} \] \(\textbf{Compare average powers:}\) \[ \frac{P_2}{P_1}=\frac{8000/13}{500} \] \[ \frac{P_2}{P_1}=\frac{16}{13} \] \(\textbf{Determine phase character:}\) Since \[ X_{L2}\gt X_{C2} \] the modified circuit is inductive. Current therefore lags the source voltage. \(\textbf{Physical interpretation:}\) At fixed source voltage, moving away from resonance would reduce average resistor power. However, this source is an AC generator whose voltage amplitude itself increases with rotational speed. The doubled emf more than compensates for the increased impedance in this particular experiment. As a result, average resistor power increases even though the circuit is no longer at resonance. This demonstrates why resonance comparisons must identify whether source voltage is held constant. \(\textbf{Final answer:}\) \[ \boxed{\frac{P_2}{P_1}=\frac{16}{13},\qquad X_{\text{net},2}=+30\,\Omega\ \text{(lagging current)}} \]

Explanation: The same resistor is used in every arrangement. Therefore, average resistor power depends on the square of the RMS current. At a fixed source voltage, current is determined by the complete impedance of each arrangement. The three impedances must be calculated using their own reactive contributions. \(\textbf{Arrangement P — Series RL circuit:}\) \[ Z_P=\sqrt{R^2+X_L^2} \] Substituting, \[ Z_P=\sqrt{40^2+30^2} \] \[ Z_P=50\,\Omega \] The RMS current is \[ I_P=\frac{100}{50} \] \[ I_P=2.0\,\text{A} \] Therefore, \[ P_P=I_P^2R \] \[ P_P=(2.0)^2(40) \] \[ P_P=160\,\text{W} \] \(\textbf{Arrangement Q — Series RC circuit:}\) \[ Z_Q=\sqrt{R^2+X_C^2} \] \[ Z_Q=\sqrt{40^2+40^2} \] \[ Z_Q=40\sqrt2\,\Omega \] The RMS current is \[ I_Q=\frac{100}{40\sqrt2} \] \[ I_Q=\frac5{2\sqrt2}\,\text{A} \] Thus, \[ P_Q=\frac{V_{\text{rms}}^2R}{Z_Q^2} \] \[ P_Q=\frac{(100)^2(40)}{(40\sqrt2)^2} \] \[ P_Q=125\,\text{W} \] \(\textbf{Arrangement S — Complete series LCR circuit:}\) The net reactance is \[ X_S=X_L-X_C \] \[ X_S=30-40 \] \[ X_S=-10\,\Omega \] The impedance is \[ Z_S=\sqrt{40^2+(-10)^2} \] \[ Z_S=10\sqrt{17}\,\Omega \] \(\textbf{Average resistor power in S:}\) \[ P_S=\frac{V_{\text{rms}}^2R}{Z_S^2} \] \[ P_S=\frac{(100)^2(40)}{(10\sqrt{17})^2} \] \[ P_S=\frac{4000}{17}\,\text{W} \] \[ P_S\approx235.3\,\text{W} \] \(\textbf{Compare the three results:}\) \[ P_S\approx235.3\,\text{W} \] \[ P_P=160\,\text{W} \] \[ P_Q=125\,\text{W} \] Therefore, \[ P_S\gt P_P\gt P_Q \] \(\textbf{Phase character in S:}\) The net reactance is negative: \[ X_S=-10\,\Omega \] Thus, the complete series circuit is capacitive and its current leads source voltage. \(\textbf{Physical interpretation:}\) Adding a capacitor to an RL circuit can reduce the net reactive opposition. In arrangement S, the capacitor partly compensates the inductive reactance. Consequently, the complete LCR arrangement has lower impedance than either two-component arrangement. It carries greater RMS current and dissipates more resistor power. The circuit remains slightly capacitive because capacitive reactance exceeds inductive reactance. The result illustrates that adding a reactive component does not necessarily increase impedance or reduce real power. \(\textbf{Final answer:}\) \[ P_S=\frac{4000}{17}\,\text{W}\gt P_P=160\,\text{W}\gt P_Q=125\,\text{W},\qquad X_S=-10\,\Omega\ \text{(leading current)} \]

Explanation: The original load current contains a component in phase with the applied voltage and a component perpendicular to it. The in-phase component accounts for average real-power consumption. The quadrature component represents the lagging reactive-current contribution. A suitable parallel capacitor supplies an opposing leading current component. \(\textbf{Original apparent power and current:}\) Average power is \[ P=VI\cos\phi \] Therefore, \[ I=\frac{P}{V\cos\phi} \] Substituting, \[ I=\frac{1200}{(200)(0.60)} \] \[ I=10\,\text{A} \] \(\textbf{In-phase current component:}\) \[ I_{\text{active}}=I\cos\phi \] \[ I_{\text{active}}=(10)(0.60) \] \[ I_{\text{active}}=6.0\,\text{A} \] \(\textbf{Lagging quadrature component:}\) Using \[ \sin\phi=\sqrt{1-\cos^2\phi} \] we obtain \[ \sin\phi=\sqrt{1-(0.60)^2} \] \[ \sin\phi=0.80 \] Therefore, \[ I_{\text{reactive}}=I\sin\phi \] \[ I_{\text{reactive}}=(10)(0.80) \] \[ I_{\text{reactive}}=8.0\,\text{A} \] \(\textbf{Required capacitor current:}\) The capacitor current leads voltage by \(90^\circ\). To cancel the lagging quadrature component, \[ I_C=8.0\,\text{A} \] \(\textbf{Recover the compensating capacitance:}\) For a pure capacitor, \[ I_C=\omega CV \] Therefore, \[ C=\frac{I_C}{\omega V} \] Substituting, \[ C=\frac{8.0}{(100)(200)} \] \[ C=4.0\times10^{-4}\,\text{F} \] \[ C=400\,\mu\text{F} \] \(\textbf{Corrected source current:}\) After the quadrature components cancel, the source supplies only the in-phase current: \[ I_{\text{source,new}}=I_{\text{active}} \] \[ I_{\text{source,new}}=6.0\,\text{A} \] \(\textbf{Real-power consistency:}\) At unity power factor, \[ P_{\text{new}}=VI_{\text{source,new}} \] \[ P_{\text{new}}=(200)(6.0) \] \[ P_{\text{new}}=1200\,\text{W} \] The real power remains unchanged. \(\textbf{Physical interpretation:}\) The capacitor does not remove the load's useful real-power requirement. It provides a local leading reactive-current component that cancels the load's lagging component at the source. The source current consequently decreases from \(10\,\text{A}\) to \(6.0\,\text{A}\). The original load current need not decrease; the reduction concerns the combined supply current. The compensation improves power factor without creating additional useful energy. \(\textbf{Final answer:}\) \[ \boxed{I_{\text{original}}=10\,\text{A},\qquad C=400\,\mu\text{F},\qquad I_{\text{corrected}}=6.0\,\text{A}} \]

Explanation: A capacitor connected in parallel with an inductive load can reduce the source's lagging reactive-current component. However, an excessively large compensating current can reverse the net phase character. In that case, the combined source current can become leading rather than lagging. Its magnitude need not be smaller than the original current. \(\textbf{Original active current:}\) \[ I_{\text{active}}=I\cos\phi \] Substituting, \[ I_{\text{active}}=(10)(0.60) \] \[ I_{\text{active}}=6.0\,\text{A} \] \(\textbf{Original lagging reactive current:}\) \[ \sin\phi=\sqrt{1-(0.60)^2} \] \[ \sin\phi=0.80 \] Therefore, \[ I_{\text{lagging}}=I\sin\phi \] \[ I_{\text{lagging}}=(10)(0.80) \] \[ I_{\text{lagging}}=8.0\,\text{A} \] \(\textbf{Combine the reactive components:}\) The capacitor contributes \(16\,\text{A}\) in the leading direction. Taking leading quadrature current as positive, \[ I_{\text{reactive,new}}=16-8.0 \] \[ I_{\text{reactive,new}}=+8.0\,\text{A} \] Thus, the final reactive contribution is leading. \(\textbf{Final source-current magnitude:}\) The in-phase component remains \(6.0\,\text{A}\). The quadrature component is \(8.0\,\text{A}\). Therefore, \[ I_{\text{source,new}}=\sqrt{I_{\text{active}}^2+I_{\text{reactive,new}}^2} \] \[ I_{\text{source,new}}=\sqrt{6^2+8^2} \] \[ I_{\text{source,new}}=10\,\text{A} \] \(\textbf{Final power factor:}\) \[ \cos\phi_{\text{new}}=\frac{I_{\text{active}}}{I_{\text{source,new}}} \] \[ \cos\phi_{\text{new}}=\frac6{10} \] \[ \cos\phi_{\text{new}}=0.60 \] The phase character is leading. \(\textbf{Average real power:}\) The ideal capacitor consumes zero average real power. Therefore, \[ P_{\text{new}}=VI_{\text{active}} \] \[ P_{\text{new}}=(100)(6.0) \] \[ P_{\text{new}}=600\,\text{W} \] \(\textbf{Compare source-current heating effects:}\) The original source current was \(10\,\text{A}\). The compensated source current is also \(10\,\text{A}\). Hence, for unchanged supply-wire resistance, \[ I_{\text{new}}^2R_{\text{wire}}=I_{\text{original}}^2R_{\text{wire}} \] The supply-wire heating is unchanged. \(\textbf{Physical interpretation:}\) The capacitor is too large for unity-power-factor correction. It first cancels the original lagging reactive current and then introduces an additional leading component. The final current has the same magnitude as initially, but its phase character is reversed. Average real power remains unchanged because the load's active current remains \(6.0\,\text{A}\). This demonstrates that adding more capacitance does not necessarily improve the magnitude of the power factor or reduce supply current. \(\textbf{Final answer:}\) \[ \boxed{I_{\text{new}}=10\,\text{A},\qquad \cos\phi_{\text{new}}=0.60\ \text{(leading)},\qquad P_{\text{new}}=600\,\text{W}} \]

Explanation: The RMS currents of different parallel AC loads cannot generally be added arithmetically. Their phase differences relative to the common supply voltage must be considered. Each load current can be resolved into an in-phase component and a quadrature component. The corresponding components can then be added with their signs. \(\textbf{Load P — Active component:}\) \[ I_{P,\parallel}=I_P\cos\phi_P \] \[ I_{P,\parallel}=(5.0)(0.60) \] \[ I_{P,\parallel}=3.0\,\text{A} \] \(\textbf{Load P — Reactive component:}\) \[ \sin\phi_P=\sqrt{1-(0.60)^2} \] \[ \sin\phi_P=0.80 \] Thus, \[ I_{P,\perp}=(5.0)(0.80) \] \[ I_{P,\perp}=4.0\,\text{A}\ \text{lagging} \] \(\textbf{Load Q — Active component:}\) \[ I_{Q,\parallel}=(5.0)(0.80) \] \[ I_{Q,\parallel}=4.0\,\text{A} \] \(\textbf{Load Q — Reactive component:}\) \[ \sin\phi_Q=\sqrt{1-(0.80)^2} \] \[ \sin\phi_Q=0.60 \] Hence, \[ I_{Q,\perp}=(5.0)(0.60) \] \[ I_{Q,\perp}=3.0\,\text{A}\ \text{leading} \] \(\textbf{Combine active components:}\) \[ I_{\text{active,total}}=3.0+4.0 \] \[ I_{\text{active,total}}=7.0\,\text{A} \] \(\textbf{Combine quadrature components:}\) Taking lagging as positive, \[ I_{\text{reactive,total}}=4.0-3.0 \] \[ I_{\text{reactive,total}}=1.0\,\text{A}\ \text{lagging} \] \(\textbf{Source RMS current:}\) \[ I_s=\sqrt{I_{\text{active,total}}^2+I_{\text{reactive,total}}^2} \] \[ I_s=\sqrt{7^2+1^2} \] \[ I_s=\sqrt{50} \] \[ I_s=5\sqrt2\,\text{A} \] \(\textbf{Combined power factor:}\) \[ \cos\phi_s=\frac{I_{\text{active,total}}}{I_s} \] \[ \cos\phi_s=\frac7{5\sqrt2} \] The remaining quadrature contribution is lagging. \(\textbf{Total real power:}\) \[ P_{\text{total}}=VI_{\text{active,total}} \] \[ P_{\text{total}}=(100)(7.0) \] \[ P_{\text{total}}=700\,\text{W} \] \(\textbf{Individual-power check:}\) \[ P_P=(100)(5.0)(0.60)=300\,\text{W} \] \[ P_Q=(100)(5.0)(0.80)=400\,\text{W} \] \[ P_P+P_Q=700\,\text{W} \] \(\textbf{Physical interpretation:}\) The two loads draw equal RMS currents but have different power factors and opposite reactive phase characters. Their active current components add. Their leading and lagging reactive components partly cancel. The combined source current is therefore less than the arithmetic sum of the branch currents. The combined power factor is also closer to unity than either individual power factor. The remaining small lagging component determines the final phase character. \(\textbf{Final answer:}\) \[ \boxed{I_s=5\sqrt2\,\text{A},\qquad \cos\phi_s=\frac7{5\sqrt2}\ \text{(lagging)},\qquad P_{\text{total}}=700\,\text{W}} \]

Explanation: The receiving load requires the same real power before and after correction. However, lower power factor requires a larger RMS current for the same real power and voltage. That current also passes through the supply wires. Their resistive heating therefore depends on power factor indirectly through current magnitude. \(\textbf{Initial line current:}\) For the load, \[ P=VI\cos\phi \] Therefore, \[ I_1=\frac{P}{V\cos\phi_1} \] Substituting, \[ I_1=\frac{1600}{(200)(0.80)} \] \[ I_1=10\,\text{A} \] \(\textbf{Initial wire-heating power:}\) \[ P_{\text{loss},1}=I_1^2R_{\text{line}} \] \[ P_{\text{loss},1}=(10)^2(2.0) \] \[ P_{\text{loss},1}=200\,\text{W} \] \(\textbf{Current after unity-power-factor correction:}\) The load's real power remains \[ P=1600\,\text{W} \] At unity power factor, \[ I_2=\frac{P}{V} \] \[ I_2=\frac{1600}{200} \] \[ I_2=8.0\,\text{A} \] \(\textbf{New wire-heating power:}\) \[ P_{\text{loss},2}=I_2^2R_{\text{line}} \] \[ P_{\text{loss},2}=(8.0)^2(2.0) \] \[ P_{\text{loss},2}=128\,\text{W} \] \(\textbf{Reduction in transmission heating:}\) \[ \Delta P_{\text{loss}}=P_{\text{loss},1}-P_{\text{loss},2} \] \[ \Delta P_{\text{loss}}=200-128 \] \[ \Delta P_{\text{loss}}=72\,\text{W} \] \(\textbf{Sending-end real-power requirement before correction:}\) Neglecting all other losses, \[ P_{\text{send},1}=P_{\text{load}}+P_{\text{loss},1} \] \[ P_{\text{send},1}=1600+200 \] \[ P_{\text{send},1}=1800\,\text{W} \] \(\textbf{Sending-end real-power requirement after correction:}\) \[ P_{\text{send},2}=1600+128 \] \[ P_{\text{send},2}=1728\,\text{W} \] Therefore, \[ P_{\text{send},1}-P_{\text{send},2}=72\,\text{W} \] \(\textbf{Physical interpretation:}\) Power-factor correction does not reduce the useful real power required by the load. It reduces the additional current associated with the load's reactive demand. The smaller line current produces less resistive heating. Consequently, the sending end needs to provide less real power while the receiving load continues using \(1600\,\text{W}\). The saving comes from reduced transmission loss rather than the creation of extra electrical energy. \(\textbf{Final answer:}\) \[ \boxed{P_{\text{loss},1}=200\,\text{W},\qquad P_{\text{loss},2}=128\,\text{W},\qquad \Delta P_{\text{send}}=72\,\text{W}} \]

Explanation: The voltmeter determines the RMS applied voltage, while the ammeter determines the RMS current. Their product is the apparent power: \[ S=VI \] For the supplied readings, \[ S=(100)(5.0)=500\,\text{VA} \] The wattmeter indicates the average real power: \[ P=300\,\text{W} \] Therefore, the magnitude of the power factor is \[ \cos\phi=\frac{P}{S} \] \[ \cos\phi=\frac{300}{500}=0.60 \] This determines the magnitude of the phase angle but not its sign. A circuit whose current lags by an angle \(\phi\) and one whose current leads by the same angle have identical values of \(\cos\phi\). Both can therefore produce the same RMS voltage, RMS current and average-power readings. The readings also establish the magnitude of the active current component: \[ I\cos\phi=(5.0)(0.60)=3.0\,\text{A} \] The magnitude of the quadrature component is \[ I|\sin\phi|=(5.0)(0.80)=4.0\,\text{A} \] However, these measurements do not identify whether that quadrature component is leading or lagging. A phase-sensitive observation, such as comparing corresponding voltage and current zero crossings on an oscilloscope, would distinguish the two possibilities. The Assertion is therefore true. The Reason correctly explains the limitation because cosine has the same value for opposite phase angles. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Explanation: The initial RL circuit has nonzero inductive reactance and therefore draws less current than a pure resistor connected to the same voltage. Adding a suitable series capacitor cancels the inductor's reactance. This improves the power factor to unity, but it also changes the complete circuit impedance. At fixed voltage, the changed impedance affects current and real power. \(\textbf{Initial RL impedance:}\) \[ Z_P=\sqrt{R^2+X_L^2} \] Substituting, \[ Z_P=\sqrt{20^2+40^2} \] \[ Z_P=20\sqrt5\,\Omega \] \(\textbf{Initial RMS current:}\) \[ I_P=\frac{V}{Z_P} \] \[ I_P=\frac{100}{20\sqrt5} \] \[ I_P=\sqrt5\,\text{A} \] \(\textbf{Initial resistor power:}\) \[ P_P=I_P^2R \] \[ P_P=(\sqrt5)^2(20) \] \[ P_P=100\,\text{W} \] \(\textbf{New reactance after adding the capacitor:}\) The capacitive reactance is \[ X_C=40\,\Omega \] Therefore, \[ X_L-X_C=40-40 \] \[ X_L-X_C=0 \] The series LCR circuit is at resonance. \(\textbf{New impedance and current:}\) At resonance, \[ Z_Q=R \] \[ Z_Q=20\,\Omega \] Hence, \[ I_Q=\frac{100}{20} \] \[ I_Q=5.0\,\text{A} \] \(\textbf{New average resistor power:}\) \[ P_Q=I_Q^2R \] \[ P_Q=(5.0)^2(20) \] \[ P_Q=500\,\text{W} \] \(\textbf{Power ratio:}\) \[ \frac{P_Q}{P_P}=\frac{500}{100} \] \[ \frac{P_Q}{P_P}=5 \] \(\textbf{Final power factor:}\) At resonance, \[ \cos\phi_Q=\frac{R}{Z_Q} \] \[ \cos\phi_Q=1 \] The current is in phase with the source voltage. \(\textbf{Compare source currents:}\) Initially, \[ I_P=\sqrt5\,\text{A} \] Finally, \[ I_Q=5.0\,\text{A} \] Thus, source RMS current increases. \(\textbf{Physical interpretation:}\) The added capacitor is ideal and does not itself consume average real power. Nevertheless, by cancelling the inductor's reactance, it increases the current through the series resistor. The resistor consequently dissipates more real power. This is different from parallel power-factor correction, where a capacitor can reduce supply current while leaving the original load voltage and real power unchanged. Improving power factor must therefore be interpreted together with circuit topology and what the source holds constant. \(\textbf{Final answer:}\) \[ \boxed{\frac{P_Q}{P_P}=5,\qquad \cos\phi_Q=1,\qquad I_Q=5.0\,\text{A}\gt I_P=\sqrt5\,\text{A}} \]

Explanation: The resistor and inductor receive the same source RMS voltage because they are connected in parallel. The resistor current is in phase with source voltage. The ideal inductor current lags voltage by \(90^\circ\). The supply current is the phasor sum of these two perpendicular branch currents. \(\textbf{Determine the resistive branch current:}\) Only the resistor consumes positive average real power. Therefore, \[ P=VI_R \] where \(I_R\) is in phase with source voltage. Thus, \[ I_R=\frac PV \] \[ I_R=\frac{800}{100} \] \[ I_R=8.0\,\text{A} \] \(\textbf{Recover resistance:}\) \[ R=\frac{V}{I_R} \] \[ R=\frac{100}{8.0} \] \[ R=12.5\,\Omega \] \(\textbf{Determine the inductive branch current:}\) Because the branch currents differ in phase by \(90^\circ\), \[ I_s^2=I_R^2+I_L^2 \] Therefore, \[ I_L=\sqrt{I_s^2-I_R^2} \] \[ I_L=\sqrt{10^2-8^2} \] \[ I_L=6.0\,\text{A} \] \(\textbf{Recover inductive reactance:}\) \[ X_L=\frac{V}{I_L} \] \[ X_L=\frac{100}{6.0} \] \[ X_L=\frac{50}{3}\,\Omega \] \(\textbf{Recover inductance:}\) \[ X_L=\omega L \] Therefore, \[ L=\frac{X_L}{\omega} \] \[ L=\frac{50/3}{100} \] \[ L=\frac16\,\text{H} \] \(\textbf{Determine combined power factor:}\) \[ \cos\phi=\frac{P}{VI_s} \] Substituting, \[ \cos\phi=\frac{800}{(100)(10)} \] \[ \cos\phi=0.80 \] The reactive branch is inductive, so the combined supply current lags voltage. \(\textbf{Branch-current consistency:}\) \[ I_s=\sqrt{8^2+6^2} \] \[ I_s=10\,\text{A} \] This matches the ammeter. \(\textbf{Physical interpretation:}\) The wattmeter does not measure the full apparent power of the parallel combination. It measures the real power consumed by the resistor. The inductor draws an additional quadrature current that increases the supply RMS current without contributing positive average real power. The current components must be combined vectorially rather than arithmetically. The measured current and wattmeter readings together determine both unknown component values. \(\textbf{Final answer:}\) \[ \boxed{R=12.5\,\Omega,\qquad L=\frac16\,\text{H},\qquad \cos\phi=0.80\ \text{(lagging)}} \]