Electromagnetic Induction and Alternating Currents MCQs with Answers – Part 3 (NEET Physics)

Explanation: The steady response of an electrical component depends on whether the applied voltage is constant or varies with time. Inductors and capacitors respond to rates of change of current and voltage, respectively. Resistors instead relate instantaneous voltage and current through Ohm's law. \(\textbf{Statement I:}\) For an ideal inductor, \[ v_L=L\frac{dI}{dt} \] Under steady DC conditions, current is constant. Therefore, its time derivative is zero and the voltage across the ideal inductor is zero. The current itself need not be zero. Statement I is correct. \(\textbf{Statement II:}\) For an ideal capacitor, \[ i_C=C\frac{dv_C}{dt} \] Under steady DC conditions, capacitor voltage is constant. Thus, the capacitor current is zero. The capacitor can nevertheless remain charged and maintain a nonzero voltage. Statement II is correct. \(\textbf{Statement III:}\) For an ideal inductor, \[ X_L=\omega L \] Increasing frequency increases inductive reactance, reducing RMS current at fixed RMS voltage. For an ideal capacitor, \[ X_C=\frac1{\omega C} \] Increasing frequency decreases capacitive reactance, increasing RMS current at fixed RMS voltage. Statement III reverses both trends. It is incorrect. \(\textbf{Statement IV:}\) For an ideal ohmic resistor, \[ I_{\text{rms}}=\frac{V_{\text{rms}}}{R} \] The resistance is assumed independent of frequency. Therefore, changing frequency alone does not change RMS current when RMS voltage remains fixed. Statement IV is correct. \(\textbf{Combined interpretation:}\) A steady magnetic field can persist in an inductor without continuing self-induced emf. A capacitor can retain electric charge without continuous conduction current through its dielectric. Under alternating excitation, the different frequency dependences of inductive and capacitive reactance produce opposite current trends. The resistor provides a frequency-independent comparison. The correct combination is Statements I, II and IV only.

Explanation: The resistor and inductor are connected in series, so they carry the same RMS current. Increasing frequency increases the inductive reactance. This changes both the impedance magnitude and the current-voltage phase difference. \(\textbf{Initial inductive reactance:}\) \[ X_{L1}=\omega_1L \] Substituting, \[ X_{L1}=(100)(0.30) \] \[ X_{L1}=30\,\Omega \] \(\textbf{Initial impedance:}\) \[ Z_1=\sqrt{R^2+X_{L1}^2} \] \[ Z_1=\sqrt{40^2+30^2} \] \[ Z_1=50\,\Omega \] \(\textbf{New inductive reactance:}\) The angular frequency doubles: \[ \omega_2=200\,\text{rad s}^{-1} \] Therefore, \[ X_{L2}=(200)(0.30) \] \[ X_{L2}=60\,\Omega \] \(\textbf{New impedance:}\) \[ Z_2=\sqrt{R^2+X_{L2}^2} \] \[ Z_2=\sqrt{40^2+60^2} \] \[ Z_2=\sqrt{1600+3600} \] \[ Z_2=\sqrt{5200}\,\Omega \] \(\textbf{Compare average resistor powers:}\) The average resistor heating is \[ P=I_{\text{rms}}^2R \] Since \[ I_{\text{rms}}=\frac{V_{\text{rms}}}{Z} \] we obtain \[ P=\frac{V_{\text{rms}}^2R}{Z^2} \] Both RMS source voltage and resistance remain unchanged. Hence, \[ \frac{P_2}{P_1}=\frac{Z_1^2}{Z_2^2} \] Substituting, \[ \frac{P_2}{P_1}=\frac{2500}{5200} \] \[ \frac{P_2}{P_1}=\frac{25}{52} \] \(\textbf{Compare phase angles:}\) For a series resistor-inductor circuit, \[ \tan\phi=\frac{X_L}{R} \] Initially, \[ \tan\phi_1=\frac{30}{40} \] \[ \tan\phi_1=\frac34 \] After doubling frequency, \[ \tan\phi_2=\frac{60}{40} \] \[ \tan\phi_2=\frac32 \] Therefore, \[ \phi_2\gt \phi_1 \] The current lags the source voltage by a greater angle at the higher frequency. \(\textbf{Physical interpretation:}\) The increasing inductive reactance raises the total impedance and reduces RMS current. Resistive heating decreases because it depends on current squared. At the same time, the inductive contribution becomes more significant relative to the resistive contribution. Consequently, the current-voltage lag angle increases. The result connects frequency dependence, impedance, phase and resistor heating without assuming that inductive reactance itself dissipates power. \(\textbf{Final answer:}\) \[ \frac{P_2}{P_1}=\frac{25}{52} \] The current lag angle increases.

Explanation: The resistor and capacitor carry the same current because they are connected in series. However, the current leads the applied source voltage by a phase angle. Therefore, the current does not generally reach its maximum when the source voltage reaches its maximum. \(\textbf{Determine circuit impedance:}\) For a series resistor-capacitor circuit, \[ Z=\sqrt{R^2+X_C^2} \] Substituting, \[ Z=\sqrt{30^2+40^2} \] \[ Z=50\,\Omega \] \(\textbf{Peak current:}\) The source peak voltage is \[ V_m=100\,\text{V} \] Therefore, \[ I_m=\frac{V_m}{Z} \] \[ I_m=\frac{100}{50} \] \[ I_m=2.0\,\text{A} \] \(\textbf{Determine the phase angle:}\) Current leads the source voltage by an angle \(\phi\) satisfying \[ \tan\phi=\frac{X_C}{R} \] \[ \tan\phi=\frac{40}{30} \] \[ \tan\phi=\frac43 \] From the impedance triangle, \[ \cos\phi=\frac RZ \] \[ \cos\phi=\frac{30}{50} \] \[ \cos\phi=\frac35 \] \(\textbf{Instantaneous current at source-voltage maximum:}\) The source voltage reaches its positive maximum when \[ \omega t=\frac{\pi}{2} \] Because current leads voltage, \[ i(t)=I_m\sin(\omega t+\phi) \] At the specified instant, \[ i=I_m\sin\left(\frac{\pi}{2}+\phi\right) \] Using \[ \sin\left(\frac{\pi}{2}+\phi\right)=\cos\phi \] we obtain \[ i=I_m\cos\phi \] \[ i=(2.0)\left(\frac35\right) \] \[ i=1.2\,\text{A} \] \(\textbf{Instantaneous resistor voltage:}\) \[ v_R=iR \] Substituting, \[ v_R=(1.2)(30) \] \[ v_R=36\,\text{V} \] \(\textbf{Circuit-voltage interpretation:}\) At this instant, the source voltage is \(100\,\text{V}\). The resistor voltage is only \(36\,\text{V}\). The remaining instantaneous voltage is associated with the capacitor: \[ v_C=100-36 \] \[ v_C=64\,\text{V} \] The component voltages add algebraically at every instant even though their RMS values must be combined as phasors. The current is positive but below its peak because it reached its positive maximum earlier than the source voltage. This is consistent with the leading-current behaviour of a capacitive circuit. \(\textbf{Final answer:}\) \[ \boxed{i=1.2\,\text{A},\qquad v_R=36\,\text{V}} \]

Explanation: The resistor and inductor are in series, but their voltage contributions differ in phase. The current lags the applied voltage because of the inductive reactance. Therefore, a zero instantaneous source voltage does not necessarily mean that the circuit current is zero. \(\textbf{Series impedance:}\) \[ Z=\sqrt{R^2+X_L^2} \] Substituting, \[ Z=\sqrt{30^2+40^2} \] \[ Z=50\,\Omega \] \(\textbf{Peak current:}\) The source peak voltage is \[ V_m=100\,\text{V} \] Therefore, \[ I_m=\frac{V_m}{Z} \] \[ I_m=\frac{100}{50} \] \[ I_m=2.0\,\text{A} \] \(\textbf{Determine current phase:}\) For a series resistor-inductor circuit, \[ \tan\phi=\frac{X_L}{R} \] Thus, \[ \tan\phi=\frac{40}{30} \] \[ \tan\phi=\frac43 \] From the impedance triangle, \[ \sin\phi=\frac{X_L}{Z} \] \[ \sin\phi=\frac{40}{50} \] \[ \sin\phi=\frac45 \] The current lags source voltage: \[ i(t)=I_m\sin(\omega t-\phi) \] \(\textbf{Current at the upward source-voltage zero crossing:}\) At \(t=0\), \[ i(0)=I_m\sin(-\phi) \] \[ i(0)=-I_m\sin\phi \] Substituting, \[ i(0)=-(2.0)\left(\frac45\right) \] \[ i(0)=-1.6\,\text{A} \] \(\textbf{Resistor voltage:}\) Using the passive sign convention, \[ v_R=iR \] Therefore, \[ v_R(0)=(-1.6)(30) \] \[ v_R(0)=-48\,\text{V} \] \(\textbf{Inductor voltage:}\) Kirchhoff's voltage law holds instantaneously: \[ v_{\text{source}}=v_R+v_L \] At \(t=0\), \[ v_{\text{source}}=0 \] Therefore, \[ 0=-48+v_L \] \[ v_L=+48\,\text{V} \] \(\textbf{Energy-transfer interpretation:}\) The current is negative while the inductor voltage is positive. Consequently, the instantaneous power absorbed by the inductor is negative. \[ p_L=v_Li \] \[ p_L=(+48)(-1.6) \] \[ p_L=-76.8\,\text{W} \] The resistor heating is \[ p_R=i^2R \] \[ p_R=(-1.6)^2(30) \] \[ p_R=76.8\,\text{W} \] The inductor releases magnetic energy at exactly the rate at which the resistor dissipates it at this instant. The source supplies zero instantaneous power because its instantaneous voltage is zero. This energy exchange explains why nonzero current can exist even when the source voltage is zero. The resistor and inductor voltages cancel algebraically rather than each becoming zero. \(\textbf{Final answer:}\) \[ \boxed{i(0)=-1.6\,\text{A},\qquad v_R(0)=-48\,\text{V},\qquad v_L(0)=+48\,\text{V}} \]

Explanation: Inductive reactance increases linearly with angular frequency, whereas capacitive reactance decreases inversely with angular frequency. The graph of inductive reactance can therefore be used to determine the inductance. The separate capacitor measurement determines its capacitance. The two relations can then be combined to find the frequency at which the reactances have equal magnitudes. \(\textbf{Extract inductance from the graph:}\) For an ideal inductor, \[ X_L=\omega L \] Therefore, the slope of the graph of \(X_L\) against \(\omega\) equals \(L\). \[ L=\frac{\Delta X_L}{\Delta\omega} \] Substituting the graph coordinates, \[ L=\frac{90-30}{300-100} \] \[ L=\frac{60}{200} \] \[ L=0.30\,\text{H} \] \(\textbf{Determine the capacitance:}\) For an ideal capacitor, \[ X_C=\frac1{\omega C} \] The measured values are \[ X_C=120\,\Omega,\qquad \omega=100\,\text{rad s}^{-1} \] Therefore, \[ C=\frac1{\omega X_C} \] \[ C=\frac1{(100)(120)} \] \[ C=\frac1{12000}\,\text{F} \] \(\textbf{Find the equal-reactance frequency:}\) The equality condition is \[ X_L=X_C \] Therefore, \[ \omega L=\frac1{\omega C} \] Rearranging, \[ \omega^2=\frac1{LC} \] Substituting, \[ \omega^2=\frac1{(0.30)(1/12000)} \] \[ \omega^2=40000 \] \[ \omega=200\,\text{rad s}^{-1} \] \(\textbf{Reactance at the common frequency:}\) \[ X_L=(200)(0.30) \] \[ X_L=60\,\Omega \] Since the reactances are equal, \[ X_C=60\,\Omega \] \(\textbf{RMS currents:}\) Both elements are connected separately to a source with \[ V_{\text{rms}}=120\,\text{V} \] Hence, \[ I_{L,\text{rms}}=\frac{120}{60} \] \[ I_{L,\text{rms}}=2.0\,\text{A} \] Similarly, \[ I_{C,\text{rms}}=\frac{120}{60} \] \[ I_{C,\text{rms}}=2.0\,\text{A} \] \(\textbf{Physical interpretation:}\) Equal reactance magnitudes produce equal RMS current magnitudes when the applied RMS voltages are equal. However, the currents do not have the same phase relationship with the applied voltage. The inductor current lags its applied voltage by \(90^\circ\), whereas the capacitor current leads by \(90^\circ\). Thus, equality of reactances does not imply identical instantaneous currents. The graph and capacitor measurement together reveal a frequency-dependent equality that cannot be determined from either measurement alone. \(\textbf{Final answer:}\) \[ \boxed{\omega=200\,\text{rad s}^{-1},\qquad I_{L,\text{rms}}=I_{C,\text{rms}}=2.0\,\text{A}} \]

Explanation: In a series resistor-capacitor circuit, increasing frequency reduces capacitive reactance. The resulting reduction in impedance increases the RMS current at fixed source RMS voltage. However, the changing ratio of capacitive reactance to resistance also changes the current-leading angle. \(\textbf{Initial capacitive reactance:}\) \[ X_{C1}=\frac1{\omega_1C} \] The given values are \[ \omega_1=100\,\text{rad s}^{-1} \] \[ C=100\times10^{-6}\,\text{F} \] Therefore, \[ X_{C1}=\frac1{(100)(100\times10^{-6})} \] \[ X_{C1}=100\,\Omega \] \(\textbf{Initial impedance:}\) For a series resistor-capacitor circuit, \[ Z_1=\sqrt{R^2+X_{C1}^2} \] \[ Z_1=\sqrt{50^2+100^2} \] \[ Z_1=50\sqrt5\,\Omega \] \(\textbf{Initial resistor heating:}\) The average power dissipated in the resistor is \[ P_1=I_1^2R \] Since \[ I_1=\frac{V_{\text{rms}}}{Z_1} \] we obtain \[ P_1=\frac{V_{\text{rms}}^2R}{Z_1^2} \] \(\textbf{Changed capacitive reactance:}\) The angular frequency doubles: \[ \omega_2=200\,\text{rad s}^{-1} \] Therefore, \[ X_{C2}=\frac1{(200)(100\times10^{-6})} \] \[ X_{C2}=50\,\Omega \] The new impedance is \[ Z_2=\sqrt{50^2+50^2} \] \[ Z_2=50\sqrt2\,\Omega \] \(\textbf{Compare heating powers:}\) Because source RMS voltage and resistance are unchanged, \[ \frac{P_2}{P_1}=\frac{Z_1^2}{Z_2^2} \] Substituting, \[ \frac{P_2}{P_1} =\frac{(50\sqrt5)^2}{(50\sqrt2)^2} \] \[ \frac{P_2}{P_1}=\frac52 \] \(\textbf{Compare phase angles:}\) In a series resistor-capacitor circuit, current leads the source voltage. The tangent of the leading angle is \[ \tan\phi=\frac{X_C}{R} \] Initially, \[ \tan\phi_1=\frac{100}{50} \] \[ \tan\phi_1=2 \] After doubling frequency, \[ \tan\phi_2=\frac{50}{50} \] \[ \tan\phi_2=1 \] Therefore, \[ \phi_2\lt \phi_1 \] \(\textbf{Physical interpretation:}\) The capacitor offers less opposition to alternating current at the higher frequency. The increased current produces greater resistive heating. However, the reactive contribution becomes smaller relative to the fixed resistance. The current therefore becomes more nearly in phase with the applied voltage. Increasing frequency in a series resistor-capacitor circuit can simultaneously increase current and decrease its phase lead. \(\textbf{Final answer:}\) \[ \frac{P_2}{P_1}=\frac52,\qquad \tan\phi_1=2,\qquad \tan\phi_2=1 \] The current-leading angle decreases.

Explanation: The maximum charge and maximum voltage determine the capacitance. The instantaneous current depends on the rate at which voltage changes, not merely on its instantaneous value. The information that the voltage is decreasing determines the sign of the current. \(\textbf{Determine capacitance:}\) For an ideal capacitor, \[ q=Cv \] Therefore, \[ C=\frac{Q_m}{V_m} \] Substituting, \[ C=\frac{8.0\times10^{-6}}{40} \] \[ C=2.0\times10^{-7}\,\text{F} \] \[ C=0.20\,\mu\text{F} \] \(\textbf{Represent the sinusoidal voltage:}\) Choose a phase variable \(\theta\) such that \[ v=V_m\sin\theta \] where \[ V_m=40\,\text{V} \] At the specified instant, \[ 20=40\sin\theta \] Therefore, \[ \sin\theta=\frac12 \] The voltage is decreasing, so its derivative is negative. Consequently, the cosine of the phase must be negative. \[ \cos\theta=-\frac{\sqrt3}{2} \] \(\textbf{Determine the voltage-change rate:}\) Differentiating, \[ \frac{dv}{dt}=\omega V_m\cos\theta \] Substituting, \[ \frac{dv}{dt} =(500)(40)\left(-\frac{\sqrt3}{2}\right) \] \[ \frac{dv}{dt}=-10000\sqrt3\,\text{V s}^{-1} \] \(\textbf{Calculate instantaneous current:}\) \[ i=C\frac{dv}{dt} \] \[ i=(2.0\times10^{-7})(-10000\sqrt3) \] \[ i=-2.0\times10^{-3}\sqrt3\,\text{A} \] \[ i=-2\sqrt3\,\text{mA} \] \(\textbf{Calculate stored electric energy:}\) The energy stored at instantaneous voltage \(v\) is \[ U=\frac12Cv^2 \] Substituting, \[ U=\frac12(2.0\times10^{-7})(20)^2 \] \[ U=4.0\times10^{-5}\,\text{J} \] \[ U=40\,\mu\text{J} \] \(\textbf{Physical interpretation:}\) The capacitor voltage is positive, but its magnitude is decreasing. The charge on the designated positive plate is therefore decreasing. The capacitor current is negative under the stated reference convention. The stored electric energy is still positive because it depends on the square of voltage. The capacitor is releasing energy at that instant, even though the voltage has not yet reached zero. The sign of current and the sign of stored energy therefore convey different physical information. \(\textbf{Final answer:}\) \[ \boxed{i=-2\sqrt3\,\text{mA},\qquad U=40\,\mu\text{J}} \]

Explanation: An ideal inductor stores magnetic energy when the magnitude of current increases and releases it when the magnitude decreases. Its instantaneous electrical power can therefore be negative even though its average power over a complete sinusoidal cycle is zero. The sign depends on the simultaneous values of voltage and current. \(\textbf{Determine inductive reactance:}\) \[ X_L=\omega L \] With \[ \omega=100\,\text{rad s}^{-1} \] and \[ L=0.50\,\text{H} \] we obtain \[ X_L=(100)(0.50) \] \[ X_L=50\,\Omega \] \(\textbf{Peak inductor current:}\) \[ I_m=\frac{V_m}{X_L} \] \[ I_m=\frac{100}{50} \] \[ I_m=2.0\,\text{A} \] \(\textbf{Write the steady current waveform:}\) Current lags voltage by \(\pi/2\) in an ideal inductor. Therefore, \[ i(t)=2.0\sin\left(100t-\frac{\pi}{2}\right) \] Equivalently, \[ i(t)=-2.0\cos(100t) \] \(\textbf{Instantaneous voltage:}\) At \[ 100t=\frac{\pi}{4} \] the voltage is \[ v=100\sin\frac{\pi}{4} \] \[ v=50\sqrt2\,\text{V} \] \(\textbf{Instantaneous current:}\) \[ i=-2.0\cos\frac{\pi}{4} \] \[ i=-\sqrt2\,\text{A} \] \(\textbf{Instantaneous power:}\) Under the passive sign convention, \[ p=vi \] Substituting, \[ p=(50\sqrt2)(-\sqrt2) \] \[ p=-100\,\text{W} \] The negative sign means that the inductor is returning energy to the source. \(\textbf{Stored magnetic energy:}\) \[ U=\frac12Li^2 \] \[ U=\frac12(0.50)(-\sqrt2)^2 \] \[ U=\frac12(0.50)(2) \] \[ U=0.50\,\text{J} \] \(\textbf{Physical check:}\) The voltage is positive while current is negative. The instantaneous power is therefore negative. The current magnitude is decreasing from its negative peak toward zero, so the magnetic energy is also decreasing. This agrees with energy flowing out of the inductor. The stored magnetic energy remains positive despite the negative current and negative instantaneous power. The distinction between stored energy and energy-transfer rate is essential when analysing ideal reactive elements. \(\textbf{Final answer:}\) \[ \boxed{p=-100\,\text{W},\qquad U=0.50\,\text{J}} \]

Explanation: In a series circuit containing resistance, inductance and capacitance, the total impedance depends on the difference between inductive and capacitive reactances. Two different frequencies can produce net reactances of equal magnitude but opposite signs. The impedance magnitudes and RMS currents may then be equal even though the phase relationships are opposite. \(\textbf{Circuit impedance:}\) For a series RLC circuit, \[ Z=\sqrt{R^2+(X_L-X_C)^2} \] where \[ X_L=\omega L,\qquad X_C=\frac1{\omega C} \] \(\textbf{At }100\,\text{rad s}^{-1}\textbf{:}\) The inductive reactance is \[ X_{L1}=(100)(0.20) \] \[ X_{L1}=20\,\Omega \] The capacitive reactance is \[ X_{C1}=\frac1{(100)(250\times10^{-6})} \] \[ X_{C1}=40\,\Omega \] The net reactance is \[ X_{L1}-X_{C1}=20-40 \] \[ X_{L1}-X_{C1}=-20\,\Omega \] The circuit therefore behaves capacitively, and current leads the source voltage. The impedance magnitude is \[ Z_1=\sqrt{15^2+(-20)^2} \] \[ Z_1=\sqrt{225+400} \] \[ Z_1=25\,\Omega \] Consequently, \[ I_1=\frac{100}{25} \] \[ I_1=4.0\,\text{A} \] \(\textbf{At }200\,\text{rad s}^{-1}\textbf{:}\) The inductive reactance becomes \[ X_{L2}=(200)(0.20) \] \[ X_{L2}=40\,\Omega \] The capacitive reactance becomes \[ X_{C2}=\frac1{(200)(250\times10^{-6})} \] \[ X_{C2}=20\,\Omega \] The net reactance is \[ X_{L2}-X_{C2}=40-20 \] \[ X_{L2}-X_{C2}=+20\,\Omega \] The circuit now behaves inductively, so current lags the source voltage. Its impedance magnitude is \[ Z_2=\sqrt{15^2+20^2} \] \[ Z_2=25\,\Omega \] Therefore, \[ I_2=\frac{100}{25} \] \[ I_2=4.0\,\text{A} \] \(\textbf{Physical interpretation:}\) At the lower frequency, capacitive reactance exceeds inductive reactance. At the higher frequency, the reverse is true. The net reactances have equal magnitudes but opposite signs. Therefore, the impedance magnitudes and RMS currents are equal while the current-voltage phase relationship reverses. The result demonstrates why impedance magnitude alone cannot identify whether a circuit behaves inductively or capacitively. The sign of the net reactance provides that additional information. \(\textbf{Final answer:}\) \[ I_1=I_2=4.0\,\text{A} \] Current leads the source voltage at \(100\,\text{rad s}^{-1}\) and lags at \(200\,\text{rad s}^{-1}\).

Explanation: The resistor and reactive element carry the same instantaneous current because they are connected in series. However, an ideal inductor and an ideal capacitor store different forms of energy. Their voltages also depend on different aspects of current variation. \(\textbf{Statement I:}\) For the ideal inductor, \[ v_L=L\frac{di}{dt} \] At a positive maximum of sinusoidal current, its rate of change is zero. The inductor voltage is therefore zero. For an ideal capacitor, current leads capacitor voltage by \(90^\circ\). At a positive current maximum, the sinusoidal capacitor voltage passes through zero. Thus, the instantaneous voltage across the ideal reactive element is zero in both cases. Statement I is correct. \(\textbf{Statement II:}\) The inductor stores magnetic energy according to \[ U_L=\frac12Li^2 \] Current has maximum magnitude at the instant considered, so the magnetic energy is maximum. The capacitor stores electric energy according to \[ U_C=\frac12Cv_C^2 \] At the positive current maximum, its instantaneous voltage is zero. Therefore, its stored electric energy is zero. Statement II is correct. \(\textbf{Statement III:}\) The source voltage equals the instantaneous sum of resistor and reactive-element voltages. At the current maximum, the reactive-element voltage is zero. However, the resistor still carries nonzero current. Its instantaneous voltage is \[ v_R=iR \] which is nonzero for nonzero resistance. Thus, the source voltage equals the resistor voltage at that instant rather than necessarily vanishing. Statement III is incorrect. \(\textbf{Physical significance:}\) The same current condition produces maximum magnetic energy in an inductor but zero electric energy in a capacitor. A vanishing reactive-element voltage does not imply that the entire series circuit has zero voltage. The resistor provides an instantaneous voltage contribution because current remains nonzero. The correct combination is Statements I and II only.

Explanation: The two circuits have equal resistance and equal reactance magnitudes. Consequently, their impedance magnitudes and current amplitudes are equal. However, the inductor causes current to lag the source voltage, while the capacitor causes it to lead. This produces opposite instantaneous current signs at an upward source-voltage zero crossing. \(\textbf{Common impedance magnitude:}\) \[ Z=\sqrt{R^2+X^2} \] Substituting, \[ Z=\sqrt{3.0^2+4.0^2} \] \[ Z=5.0\,\Omega \] \(\textbf{Peak current in each circuit:}\) \[ I_m=\frac{V_m}{Z} \] \[ I_m=\frac{50}{5.0} \] \[ I_m=10\,\text{A} \] \(\textbf{Magnitude of the phase angle:}\) \[ \tan\phi=\frac XR \] \[ \tan\phi=\frac43 \] From the impedance triangle, \[ \sin\phi=\frac XZ \] \[ \sin\phi=\frac45 \] \(\textbf{Current in the inductive circuit P:}\) Current lags voltage. Therefore, \[ i_P(t)=10\sin(\omega t-\phi) \] At \(t=0\), \[ i_P(0)=10\sin(-\phi) \] \[ i_P(0)=-10\left(\frac45\right) \] \[ i_P(0)=-8.0\,\text{A} \] \(\textbf{Current in the capacitive circuit Q:}\) Current leads voltage. Therefore, \[ i_Q(t)=10\sin(\omega t+\phi) \] At \(t=0\), \[ i_Q(0)=10\sin\phi \] \[ i_Q(0)=10\left(\frac45\right) \] \[ i_Q(0)=+8.0\,\text{A} \] \(\textbf{Average resistor heating:}\) Both currents are sinusoidal with the same peak magnitude. Their common RMS current is \[ I_{\text{rms}}=\frac{10}{\sqrt2} \] The resistor heating power in either circuit is \[ P_{\text{avg}}=I_{\text{rms}}^2R \] Substituting, \[ P_{\text{avg}}=\left(\frac{10}{\sqrt2}\right)^2(3.0) \] \[ P_{\text{avg}}=50(3.0) \] \[ P_{\text{avg}}=150\,\text{W} \] \(\textbf{Physical interpretation:}\) The two circuits have equal current amplitudes but opposite phase shifts. Their instantaneous currents at the specified voltage zero crossing therefore have equal magnitudes and opposite signs. The average resistor heating remains equal because it depends on the square of RMS current rather than phase sign alone. An inductive lag and a capacitive lead can produce identical impedance magnitudes while giving different instantaneous current histories. \(\textbf{Final answer:}\) \[ \boxed{i_P(0)=-8.0\,\text{A},\qquad i_Q(0)=+8.0\,\text{A},\qquad P_{\text{avg}}=150\,\text{W}} \]

Explanation: The maximum magnetic energy determines the peak current once inductance is known. However, the inductance is initially unknown. The applied voltage and frequency provide a second relation through inductive reactance. These relations can be combined to determine both inductance and current amplitude. \(\textbf{Peak current relation:}\) For an ideal inductor under sinusoidal voltage, \[ I_m=\frac{V_m}{X_L} \] Since \[ X_L=\omega L \] we obtain \[ I_m=\frac{V_m}{\omega L} \] \(\textbf{Use maximum stored energy:}\) The maximum magnetic energy is \[ U_{\max}=\frac12LI_m^2 \] Substituting the expression for peak current, \[ U_{\max} =\frac12L\left(\frac{V_m}{\omega L}\right)^2 \] Simplifying, \[ U_{\max}=\frac{V_m^2}{2\omega^2L} \] \(\textbf{Recover the inductance:}\) Rearranging, \[ L=\frac{V_m^2}{2\omega^2U_{\max}} \] The given values are \[ V_m=80\,\text{V} \] \[ \omega=200\,\text{rad s}^{-1} \] \[ U_{\max}=0.80\,\text{J} \] Substituting, \[ L=\frac{80^2}{2(200)^2(0.80)} \] \[ L=\frac{6400}{64000} \] \[ L=0.10\,\text{H} \] \(\textbf{Original peak current:}\) \[ I_m=\frac{V_m}{\omega L} \] \[ I_m=\frac{80}{(200)(0.10)} \] \[ I_m=4.0\,\text{A} \] \(\textbf{Changed frequency:}\) The angular frequency doubles: \[ \omega'=400\,\text{rad s}^{-1} \] At fixed peak voltage and inductance, \[ I'_m=\frac{80}{(400)(0.10)} \] \[ I'_m=2.0\,\text{A} \] \(\textbf{New maximum magnetic energy:}\) \[ U'_{\max}=\frac12L(I'_m)^2 \] Substituting, \[ U'_{\max}=\frac12(0.10)(2.0)^2 \] \[ U'_{\max}=0.20\,\text{J} \] \(\textbf{Physical interpretation:}\) Doubling frequency doubles inductive reactance. The peak current therefore falls to half its original magnitude. Because magnetic energy depends on current squared, maximum stored energy falls to one-fourth. The source peak voltage remains unchanged, but the inductor has less time to build current during each part of the faster alternating cycle. This explains the reduced magnetic-energy amplitude at the higher frequency. \(\textbf{Final answer:}\) \[ \boxed{L=0.10\,\text{H},\qquad I_m=4.0\,\text{A},\qquad U'_{\max}=0.20\,\text{J}} \]

Explanation: The stored electric energy of a capacitor depends on the square of instantaneous voltage. Its instantaneous energy-transfer rate depends on both voltage and current. Thus, equal instantaneous voltage magnitudes do not necessarily imply equal directions of energy flow. \(\textbf{Capacitor energy relation:}\) \[ U_C=\frac12Cv^2 \] The applied voltage is \[ v=V_m\sin\theta \] where \[ V_m=100\,\text{V} \] Therefore, \[ U_C=\frac12CV_m^2\sin^2\theta \] \(\textbf{Energy at phase P:}\) At \[ \theta_P=\frac{\pi}{4} \] the voltage is \[ v_P=100\sin\frac{\pi}{4} \] \[ v_P=50\sqrt2\,\text{V} \] The stored energy is \[ U_P=\frac12(100\times10^{-6})(50\sqrt2)^2 \] \[ U_P=\frac12(100\times10^{-6})(5000) \] \[ U_P=0.25\,\text{J} \] \(\textbf{Energy at phase Q:}\) At \[ \theta_Q=\frac{3\pi}{4} \] the voltage is \[ v_Q=100\sin\frac{3\pi}{4} \] \[ v_Q=50\sqrt2\,\text{V} \] Therefore, \[ U_Q=\frac12(100\times10^{-6})(50\sqrt2)^2 \] \[ U_Q=0.25\,\text{J} \] \(\textbf{Determine current directions:}\) For the capacitor, \[ i=C\frac{dv}{dt} \] Differentiating the source voltage, \[ i=\omega CV_m\cos\theta \] At P, \[ \cos\frac{\pi}{4}\gt 0 \] so current is positive while voltage is positive. The instantaneous power absorbed by the capacitor is therefore positive. At Q, \[ \cos\frac{3\pi}{4}\lt 0 \] so current is negative while voltage remains positive. The instantaneous power absorbed is therefore negative. \(\textbf{Physical interpretation:}\) At P, capacitor voltage is increasing toward its positive maximum. The capacitor is accumulating electric-field energy. At Q, capacitor voltage is decreasing from its positive maximum. The capacitor is returning stored energy to the source. The stored energy happens to have the same value at both instants because the voltage magnitudes are equal. The energy-transfer directions differ because the voltage-change rates have opposite signs. The capacitor's instantaneous power is therefore not determined by its instantaneous stored energy alone. The correct conclusion is equal stored energy with opposite directions of energy transfer.

Explanation: The capacitor's reactance decreases as the applied AC frequency increases. The resistor's resistance remains unchanged. Therefore, the series impedance decreases with increasing frequency, allowing greater RMS current. \(\textbf{Series impedance:}\) For a series resistor-capacitor circuit, \[ Z=\sqrt{R^2+X_C^2} \] where \[ X_C=\frac1{\omega C} \] The RMS current is \[ I_{\text{rms}}=\frac{V_{\text{source}}}{Z} \] The RMS resistor voltage is \[ V_R=I_{\text{rms}}R \] Therefore, \[ V_R=V_{\text{source}}\frac R{\sqrt{R^2+X_C^2}} \] Because \(X_C^2\) is nonnegative, \[ \sqrt{R^2+X_C^2}\geq R \] Consequently, \[ V_R\leq V_{\text{source}} \] For finite positive angular frequency and finite capacitance, \(X_C\) is positive, so the inequality is strict. As frequency increases, \(X_C\) decreases. The denominator of the resistor-voltage expression therefore decreases toward \(R\). Hence, the RMS resistor voltage increases toward the source RMS voltage. \(\textbf{Capacitor-voltage behaviour:}\) The RMS capacitor voltage is \[ V_C=I_{\text{rms}}X_C \] Equivalently, \[ V_C=V_{\text{source}} \frac{X_C}{\sqrt{R^2+X_C^2}} \] As frequency increases and \(X_C\) becomes smaller, this voltage decreases. The resistor and capacitor voltage phasors are perpendicular, so \[ V_{\text{source}}^2=V_R^2+V_C^2 \] An increasing resistor-voltage magnitude therefore accompanies a decreasing capacitor-voltage magnitude. The Reason describes precisely this redistribution of the source voltage. Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Explanation: The impedance of a series resistor-inductor circuit depends on the resistor and the inductive reactance. At zero angular frequency, the inductive reactance tends to zero. The graph intercept therefore reveals the resistance. A second graph point can then be used to determine inductance. \(\textbf{Impedance relation:}\) For a series resistor-inductor circuit, \[ Z=\sqrt{R^2+(\omega L)^2} \] \(\textbf{Determine resistance from the graph:}\) At \[ \omega=0 \] the inductive reactance is zero. Therefore, \[ Z(0)=R \] The graph intercept gives \[ R=20\,\Omega \] \(\textbf{Use the second graph point:}\) At \[ \omega=100\,\text{rad s}^{-1} \] the impedance magnitude is \[ Z=25\,\Omega \] Therefore, \[ 25^2=20^2+(100L)^2 \] \[ 625=400+10000L^2 \] \[ 10000L^2=225 \] \[ L^2=0.0225 \] Since inductance is positive, \[ L=0.15\,\text{H} \] \(\textbf{Changed operating frequency:}\) The angular frequency becomes \[ \omega'=200\,\text{rad s}^{-1} \] The inductive reactance is \[ X'_L=\omega'L \] \[ X'_L=(200)(0.15) \] \[ X'_L=30\,\Omega \] \(\textbf{New impedance magnitude:}\) \[ Z'=\sqrt{20^2+30^2} \] \[ Z'=\sqrt{400+900} \] \[ Z'=10\sqrt{13}\,\Omega \] \(\textbf{Determine RMS current:}\) The source RMS voltage is \[ V_{\text{rms}}=50\,\text{V} \] Therefore, \[ I_{\text{rms}}=\frac{V_{\text{rms}}}{Z'} \] \[ I_{\text{rms}}=\frac{50}{10\sqrt{13}} \] \[ I_{\text{rms}}=\frac5{\sqrt{13}}\,\text{A} \] \(\textbf{Physical interpretation:}\) The graph does not rise linearly because impedance is the square root of the sum of squared resistive and inductive contributions. The zero-frequency intercept identifies the resistance, while the variation above this intercept reveals inductance. At higher frequency, inductive reactance becomes larger and the RMS current decreases for fixed source voltage. The current also lags the source voltage by an angle determined by the ratio of reactance to resistance. The graph therefore encodes both circuit parameters and their frequency-dependent combined effect. \(\textbf{Final answer:}\) \[ \boxed{L=0.15\,\text{H},\qquad I_{\text{rms}}=\frac5{\sqrt{13}}\,\text{A}} \]

Explanation: The capacitance of a parallel-plate capacitor depends on the permittivity of the material between its plates. For identical plate geometry, a dielectric with relative permittivity \(3\) increases capacitance by a factor of three. The applied voltage amplitude and frequency remain unchanged in the two experiments. \(\textbf{Statement I:}\) The RMS current through an ideal capacitor under sinusoidal voltage is \[ I_{\text{rms}}=\omega CV_{\text{rms}} \] Thus, at fixed angular frequency and RMS voltage, current is directly proportional to capacitance. The dielectric-filled capacitor has three times the capacitance. It therefore carries three times the RMS alternating current. Statement I is correct. \(\textbf{Statement II:}\) The maximum electric energy stored in a capacitor is \[ U_{\max}=\frac12CV_m^2 \] Both capacitors experience the same peak voltage. Therefore, their maximum stored energies are proportional to capacitance. The dielectric-filled capacitor stores three times the maximum electric energy. Statement II is correct. \(\textbf{Statement III:}\) An ideal insulating dielectric does not permit free charge to conduct directly from one capacitor plate to the other. The alternating current in the external circuit arises because the plate charges change with time. Charge accumulates on and leaves the plates as the applied voltage varies. The changing electric field through the dielectric is associated with this alternating behaviour. A larger capacitance produces a larger plate-charge variation for the same voltage variation. Direct conduction through the ideal insulating gap is not required. Statement III is incorrect. \(\textbf{Physical significance:}\) Capacitor current must not be confused with a continuous transport of free charge through the insulating material. The dielectric changes how much charge can be stored at a given voltage. This changes both charging current and electric-field energy. The correct combination is Statements I and II only.

Explanation: The applied voltage is positive and increasing. This means that its sinusoidal phase is between zero and the positive voltage maximum. The current phases of ideal inductors and capacitors differ by a quarter-cycle from the applied voltage, but in opposite directions. \(\textbf{Inductor current:}\) For sinusoidal voltage, \[ v=V_m\sin\theta \] The steady current through an ideal inductor lags voltage by \(90^\circ\). Therefore, \[ i_L=I_m\sin\left(\theta-\frac{\pi}{2}\right) \] Using the trigonometric identity, \[ i_L=-I_m\cos\theta \] Since voltage is positive and increasing, \[ 0\lt \theta\lt \frac{\pi}{2} \] and \[ \cos\theta\gt 0 \] Hence, \[ i_L\lt 0 \] \(\textbf{Capacitor current:}\) For an ideal capacitor, \[ i_C=C\frac{dv}{dt} \] Differentiating, \[ i_C=\omega CV_m\cos\theta \] Since the voltage is increasing, \[ i_C\gt 0 \] \(\textbf{Energy-transfer directions:}\) Instantaneous power absorbed by an element is \[ p=vi \] under the passive sign convention. The applied voltage is positive. For the inductor, current is negative, so \[ p_L\lt 0 \] The inductor returns magnetic energy to its source. For the capacitor, current is positive, so \[ p_C\gt 0 \] The capacitor absorbs energy and increases its stored electric energy. \(\textbf{Physical interpretation:}\) The inductor's negative current is moving toward zero as the positive voltage drives its current upward. Its magnetic-energy magnitude therefore decreases. The capacitor's voltage magnitude is increasing, so its electric-field energy increases. Equal reactance magnitudes make the two current amplitudes equal, but do not make their instantaneous directions or energy-transfer roles identical. The two ideal reactive elements can exchange energy with their sources in opposite directions at the same applied-voltage phase. The correct conclusion is negative inductor current with energy release and positive capacitor current with energy absorption.

Explanation: The inductor and capacitor carry equal RMS currents, but they do not have equal reactance magnitudes. Therefore, the peak voltages required to produce those currents are different. Their maximum stored energies depend on the element values and the corresponding current or voltage amplitudes. \(\textbf{Determine the inductance:}\) For the ideal inductor, \[ X_L=\omega L \] Therefore, \[ L=\frac{X_L}{\omega} \] Substituting, \[ L=\frac{80}{200} \] \[ L=0.40\,\text{H} \] \(\textbf{Peak inductor current:}\) The common RMS current is \[ I_{\text{rms}}=1.5\,\text{A} \] For sinusoidal current, \[ I_m=\sqrt2I_{\text{rms}} \] \[ I_m=1.5\sqrt2\,\text{A} \] \(\textbf{Maximum magnetic energy:}\) \[ U_{L,\max}=\frac12LI_m^2 \] Substituting, \[ U_{L,\max} =\frac12(0.40)(1.5\sqrt2)^2 \] \[ U_{L,\max}=0.20(4.5) \] \[ U_{L,\max}=0.90\,\text{J} \] \(\textbf{Determine the capacitance:}\) For an ideal capacitor, \[ X_C=\frac1{\omega C} \] Therefore, \[ C=\frac1{\omega X_C} \] Substituting, \[ C=\frac1{(200)(40)} \] \[ C=1.25\times10^{-4}\,\text{F} \] \[ C=125\,\mu\text{F} \] \(\textbf{Determine the capacitor's peak voltage:}\) The capacitor RMS voltage is \[ V_{C,\text{rms}}=I_{\text{rms}}X_C \] \[ V_{C,\text{rms}}=(1.5)(40) \] \[ V_{C,\text{rms}}=60\,\text{V} \] Therefore, \[ V_{C,m}=\sqrt2V_{C,\text{rms}} \] \[ V_{C,m}=60\sqrt2\,\text{V} \] \(\textbf{Maximum electric energy:}\) \[ U_{C,\max}=\frac12CV_{C,m}^2 \] Substituting, \[ U_{C,\max} =\frac12(125\times10^{-6})(60\sqrt2)^2 \] \[ U_{C,\max} =\frac12(125\times10^{-6})(7200) \] \[ U_{C,\max}=0.45\,\text{J} \] \(\textbf{Compare the stored energies:}\) \[ \frac{U_{L,\max}}{U_{C,\max}} =\frac{0.90}{0.45} \] \[ \frac{U_{L,\max}}{U_{C,\max}}=2 \] \(\textbf{Physical interpretation:}\) Equal current amplitudes do not imply equal stored energies in different reactive elements. The inductor stores energy according to inductance and current squared. The capacitor stores energy according to capacitance and voltage squared. Its voltage amplitude depends on capacitive reactance at the specified current amplitude. These distinct relations lead to a maximum-energy ratio of two in the stated experiments. The result also agrees with the fact that, at equal angular frequency and RMS current, maximum stored energy is proportional to reactance magnitude for either ideal element. \(\textbf{Final answer:}\) \[ \boxed{U_{L,\max}=0.90\,\text{J},\qquad U_{C,\max}=0.45\,\text{J},\qquad \frac{U_{L,\max}}{U_{C,\max}}=2} \]

Explanation: The current in a series resistor-capacitor circuit leads the applied source voltage. Therefore, at an upward zero crossing of the source voltage, the circuit current can already be positive. The capacitor and resistor voltages must still add algebraically to the instantaneous source voltage. \(\textbf{Determine circuit impedance:}\) \[ Z=\sqrt{R^2+X_C^2} \] Substituting, \[ Z=\sqrt{12^2+16^2} \] \[ Z=20\,\Omega \] \(\textbf{Peak circuit current:}\) The source peak voltage is \[ V_m=100\,\text{V} \] Therefore, \[ I_m=\frac{V_m}{Z} \] \[ I_m=\frac{100}{20} \] \[ I_m=5.0\,\text{A} \] \(\textbf{Current-leading angle:}\) \[ \tan\phi=\frac{X_C}{R} \] \[ \tan\phi=\frac{16}{12} \] \[ \tan\phi=\frac43 \] From the impedance triangle, \[ \sin\phi=\frac{X_C}{Z} \] \[ \sin\phi=\frac{16}{20} \] \[ \sin\phi=\frac45 \] \(\textbf{Instantaneous current:}\) Since current leads voltage, \[ i(t)=I_m\sin(200t+\phi) \] At \(t=0\), \[ i(0)=I_m\sin\phi \] \[ i(0)=(5.0)\left(\frac45\right) \] \[ i(0)=+4.0\,\text{A} \] \(\textbf{Instantaneous resistor voltage:}\) \[ v_R=iR \] Substituting, \[ v_R=(4.0)(12) \] \[ v_R=+48\,\text{V} \] \(\textbf{Instantaneous capacitor voltage:}\) Kirchhoff's voltage law applies at every instant: \[ v_{\text{source}}=v_R+v_C \] At \(t=0\), \[ v_{\text{source}}=0 \] Therefore, \[ 0=48+v_C \] \[ v_C=-48\,\text{V} \] \(\textbf{Electric-energy change rate:}\) For an ideal capacitor, \[ U_C=\frac12Cv_C^2 \] Differentiating and using \(i=C\,dv_C/dt\), \[ \frac{dU_C}{dt}=v_Ci \] Substituting, \[ \frac{dU_C}{dt}=(-48)(+4.0) \] \[ \frac{dU_C}{dt}=-192\,\text{W} \] \(\textbf{Energy balance at this instant:}\) The resistor heating power is \[ P_R=i^2R \] \[ P_R=(4.0)^2(12) \] \[ P_R=192\,\text{W} \] The instantaneous source voltage is zero, so the source supplies zero instantaneous power. The capacitor releases electric energy at \(192\,\text{W}\), exactly matching resistor heating. \(\textbf{Physical interpretation:}\) A capacitor can supply energy to a resistor during part of an AC cycle. This does not mean that the capacitor permanently generates energy. It is returning energy previously stored in its electric field. The negative capacitor voltage and positive current show that the magnitude of its stored charge is decreasing. The source-voltage zero crossing is therefore compatible with nonzero series current, nonzero resistor heating and a decreasing capacitor energy. \(\textbf{Final answer:}\) \[ \boxed{i(0)=+4.0\,\text{A},\qquad v_C(0)=-48\,\text{V},\qquad \frac{dU_C}{dt}=-192\,\text{W}} \]

Explanation: The heating measurement determines the RMS current, whereas the charge measurement involves the area under one positive half-cycle of the current-time graph. Combining these independent measurements permits both the current amplitude and frequency to be determined. \(\textbf{Recover average heating power:}\) The given heat and duration are \[ H=1920\,\text{J},\qquad t=120\,\text{s} \] Therefore, \[ P_{\text{avg}}=\frac{H}{t} \] \[ P_{\text{avg}}=\frac{1920}{120} \] \[ P_{\text{avg}}=16\,\text{W} \] \(\textbf{Determine RMS current:}\) For a fixed resistor, \[ P_{\text{avg}}=I_{\text{rms}}^2R \] Substituting, \[ 16=I_{\text{rms}}^2(8.0) \] \[ I_{\text{rms}}^2=2.0\,\text{A}^2 \] \[ I_{\text{rms}}=\sqrt2\,\text{A} \] \(\textbf{Recover peak current:}\) For a sinusoidal current, \[ I_m=\sqrt2I_{\text{rms}} \] \[ I_m=\sqrt2(\sqrt2) \] \[ I_m=2.0\,\text{A} \] \(\textbf{Use the half-cycle charge measurement:}\) Represent the current as \[ i=I_m\sin\omega t \] The charge transferred during a positive half-cycle is \[ q_+=\int_0^{\pi/\omega}I_m\sin\omega t\,dt \] Integrating, \[ q_+=\left[-\frac{I_m}{\omega}\cos\omega t\right]_0^{\pi/\omega} \] \[ q_+=\frac{2I_m}{\omega} \] The measured value is \[ q_+=\frac{0.020}{\pi}\,\text{C} \] Therefore, \[ \frac{0.020}{\pi}=\frac{2(2.0)}{\omega} \] \[ \omega=\frac{4.0\pi}{0.020} \] \[ \omega=200\pi\,\text{rad s}^{-1} \] \(\textbf{Determine frequency:}\) \[ f=\frac{\omega}{2\pi} \] \[ f=\frac{200\pi}{2\pi} \] \[ f=100\,\text{Hz} \] \(\textbf{Physical interpretation:}\) Resistive heating depends on the mean square of current and does not directly reveal its frequency when resistance and current amplitude are fixed. Charge transferred during a single half-cycle depends on both amplitude and half-cycle duration. The two measurements therefore provide complementary information. Once heating determines the amplitude, the half-cycle charge reveals the frequency. The algebraic charge over a complete cycle would be zero and would not provide the same information. \(\textbf{Final answer:}\) \[ \boxed{I_m=2.0\,\text{A},\qquad f=100\,\text{Hz}} \]