Q.A sinusoidal voltage of peak value 283 V and frequency 50Hz is applied to a series LCR circuit in which R = 3Ω, L = 25.48mH and C = 796 μF. Find

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1. CBSE 2026Set DS1 mark
   Q.The variation between $V$ and $I$ with time ($t$) in two alternative circuits is shown in the following graphs. Which one of the curves of circuit is capacitive and which one is resistive?

   ›Reveal solutionSolution

   In-phase $V$ and $I$ ⇒ resistive circuit; current leading voltage by $90^\circ$ ⇒ capacitive circuit.

   Concept. The phase relation between voltage and current identifies the circuit element:

   • Pure resistor: current is in phase with voltage — both reach their maxima and zeros at the same instant.
   • Pure capacitor: current leads the voltage by a quarter cycle ($90^\circ$, i.e. $\pi/2$) — the current curve peaks before the voltage curve.

   Reading the graphs.

   • In the left graph the current curve $I$ and the voltage curve $V$ rise together and reach their peaks at the same time — they are in phase → this is the resistive circuit. …

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2. CBSE 2026Set ANNUAL1 mark
   Q.Can a battery be charged by using alternating current? Write the reason.

   ›Reveal solutionSolution

   Charging a battery needs current flowing in one fixed direction; AC reverses direction every half-cycle, so used directly it undoes on the second half-cycle whatever it did on the first.

   Charging a cell/battery means forcing conventional current into it against its own e.m.f. so chemical energy is stored, which requires a steady, one-directional (DC) current. Alternating current changes direction periodically (e.g. every 1/100 s for 50 Hz mains), so during one half-cycle it would tend to charge the battery, but during the very next half-cycle the reversed current would discharge it by an equal amount — the net effect over a full cycle is essentially zero charging, and repeatedly reversing cu …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.The time taken by an AC of frequency 50 Hz to reach from zero to the maximum value is

   (a) $50\times10^{-3}$ s

   (b) $5\times10^{-3}$ s

   (c) $1\times10^{-3}$ s

   (d) $2\times10^{-3}$ s

   ›Reveal solutionSolution

   An alternating current varies as $i=I_0\sin\omega t$. It starts at zero and first reaches its maximum ($I_0$) after one quarter of a full cycle, i.e. $t=T/4$.

   Step 1 — Find the time period

   $T=\frac{1}{f}=\frac{1}{50\text{ Hz}}=0.02\text{ s}=20\times10^{-3}\text{ s}$

   Step 2 — Express the current as a function of time

   Since the current is zero at $t=0$ and increases from there, it is described by

   $i(t)=I_0\sin(\omega t), \qquad \omega=\frac{2\pi}{T}$

   $i(t)$ reaches its maximum value $I_0$ when $\sin\omega t=1$, i.e. when

   …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.The figure given below shows the phase relationship between voltage $E$ and current $I$. Which of the following statements is true?

   (a) This phase relationship is for a resistor when AC is passed through it.

   (b) This phase relationship is for a capacitor when AC is passed through it.

   (c) This phase relationship is for an inductor when AC is passed through it.

   (d) This phase relationship is for an L-C-R circuit.

   ›Reveal solutionSolution

   The graph shows current lagging voltage by a quarter cycle ($\pi/2$). Comparing this to the known phase behaviour of R, L, C elements in AC circuits identifies the element as a pure inductor.

   Phase relationships of R, L, C in AC circuits (recap)

   • Pure resistor: $V=IR$ at every instant, so $I$ and $V$ (or $E$) are always in phase (zero phase difference).
   • Pure capacitor: $I=C\,dV/dt$; when $V=V_0\sin\omega t$, $I=\omega C V_0\cos\omega t=\omega C V_0\sin(\omega t+\pi/2)$ — current leads voltage by $\pi/2$.
   • Pure inductor: $V=L\,dI/dt$; when $I=I_0\sin\omega t$, $V=\omega L I_0\cos\omega t = \omega L I_0 \sin(\omega t+\pi/2)$ — equivalently, current lags voltage by $\pi/2$: $I=I_0\sin(\omega t-\pi/2)$ relative to $V=V_0\sin\omega t$.
   • Series L-C-R circuit: the phase difference $\phi$ between $V$ and $I$ is given by $\tan\phi=(X_L-X_C)/R$, which generally lies between $0$ and $\pi/2$ (not exactly $\pi/2$) unless $R=0$.

   Reading the graph

   …

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5. CBSE 2025Set 55/4/11 markMCQ
   Q.Assertion (A): A series LCR circuit behaves as a pure resistive circuit at resonance. Reason (R): At resonance, $X_L = X_C$, which gives $\omega = \frac{1}{\sqrt{LC}}$. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Both Assertion (A) and Reason (R) are false.

   ›Reveal solutionSolution

   At resonance, the inductive and capacitive reactances cancel exactly, making the circuit purely resistive. The reason correctly states the condition $X_L = X_C$ and the resonant frequency formula, so both are true and the reason is the correct explanation.

   Concept and Intuition

   In a series LCR circuit, the total impedance is $Z = R + j(X_L - X_C)$, where $X_L = \omega L$ and $X_C = 1/(\omega C)$. The phase angle $\phi$ between voltage and current is given by $\tan \phi = (X_L - X_C)/R$. For the circuit to behave as a pure resistor, the imaginary part must vanish — that is, $X_L = X_C$. This condition defines resonance. When it holds, the impedance is minimum ($Z = R$), current is maximum, and voltage and current are in phase ($\phi = 0$). The reason correctly identifies this condition and the resulting angular frequency $\omega = 1/\sqrt{LC}$.

   Watch out

   A common mistake is to think that at resonance the circuit has zero impedance. It does not — the impedance equals $R$, which is purely resistive but not zero (unless $R=0$). The key is that the reactive parts cancel, not that reactance disappears.

   Step-by-step reasoning

   1. Assertion (A) states that a series LCR circuit behaves as a pure resistive circuit at resonance. This is true because at resonance, $X_L = X_C$, so the net reactance is zero. The impedance becomes $Z = R + j0 = R$, a purely real quantity. Hence voltage and current are in phase, exactly as in a resistor.

   2. Reason (R) states that at resonance, $X_L = X_C$, which gives $\omega = 1/\sqrt{LC}$. This is also true. Setting $X_L = X_C$:

   $\omega L = \frac{1}{\omega C} \quad\Rightarrow\quad \omega^2 = \frac{1}{LC} \quad\Rightarrow\quad \omega = \frac{1}{\sqrt{LC}}$

   (taking the positive root, since frequency is positive). …

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6. CBSE 2025Set X11 markMCQ
   Q.The variation of voltage and current through an a.c. circuit with time is as shown in the figure. Along with the a.c. source, the circuit :

   (a) has a series combination of resistance and capacitance

   (b) has only inductance

   (c) has only capacitance

   (d) may have only resistance or may have a suitable series combination of inductance $(L)$, capacitance $(C)$ and resistance $(R)$

   ›Reveal solutionSolution

   The printed graph shows voltage $v$ and current $i$ crossing zero together and peaking at the same $\omega t$ — they are in phase (phase difference $\phi = 0$). A zero phase angle means the net reactance is zero, which is true for a pure resistor or for a series $LCR$ circuit at resonance. Hence the correct choice is (d).

   What the graph tells us

   Both sinusoids start at the origin, rise together, reach their peaks ($v_0$ and $i_0$) at the same instant, and cross the $\omega t$ axis at the same points. So:

   $v = v_0 \sin(\omega t), \qquad i = i_0 \sin(\omega t)$

   The phase difference between voltage and current is $\phi = 0$. (Only the amplitudes differ, $v_0 > i_0$, which is irrelevant to the phase.)

   Linking phase to the circuit

   For a general a.c. circuit, $\tan\phi = \dfrac{X_L - X_C}{R}$. A phase difference of zero requires $X_L - X_C = 0$, i.e. the net reactance vanishes. This can happen in two ways:

   • Only resistance ($R$): there is no reactance at all, so $v$ and $i$ are always in phase.
   • A series $LCR$ circuit at resonance: here $X_L = X_C$, the reactances cancel, the impedance becomes purely resistive ($Z = R$), and $v$ and $i$ come back into phase.

   Why the other options are wrong …

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7. CBSE 2025Set JS1 markMCQ
   Q.The resistance of 8 ohm and inductive reactance of 6 ohm are connected in series in an alternating current circuit. The impedance of circuit will be: (A) 2 ohm (B) 14 ohm (C) $14\sqrt{2}$ ohm (D) 10 ohm

   ›Reveal solutionSolution

   Impedance of a series R–L circuit is $Z=\sqrt{R^2+X_L^2}=\sqrt{64+36}=10\ \Omega$ — option (D).

   Concept. In a series AC circuit the resistance $R$ and reactance $X_L$ add like the two perpendicular sides of a right triangle (because the voltage across $R$ is in phase with the current while the voltage across $L$ leads it by $90^\circ$). The magnitude of the total opposition, the impedance, is the hypotenuse: …

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8. CBSE 2025Set JS1 mark
   Q.Voltage ($V$) equation in an alternating current circuit is represented by $V = 40 \sin (100 \pi t)$ volt. Here $t$ is in second. Draw the time-voltage graph for one cycle with proper scale.

   ›Reveal solutionSolution

   

   Peak $40$ V, period $T=0.02$ s ($20$ ms); plot a sine curve rising to $+40$ V at $t=5$ ms, back to $0$ at $10$ ms, down to $-40$ V at $15$ ms, and returning to $0$ at $20$ ms.

   Concept. For $V=V_0\sin(\omega t)$, the amplitude is $V_0$ and the angular frequency is $\omega$, giving period $T=\dfrac{2\pi}{\omega}$.

   Read off the values.

   $V_0=40\ \text{V},\qquad \omega=100\pi\ \text{rad s}^{-1},$

   $T=\frac{2\pi}{\omega}=\frac{2\pi}{100\pi}=0.02\ \text{s}=20\ \text{ms},\qquad f=\frac1T=50\ \text{Hz}.$

   Key points of one cycle (scale: $x$-axis $1$ div $=5$ ms, $y$-axis $1$ div $=10$ V):

   | $t$ (ms) | 0 | 5 | 10 | 15 | 20 | …

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9. CBSE 2025Set ANNUAL1 markMCQ
   Q.The value of alternating voltage in the following circuit is

   (a) 2 V

   (b) 4 V

   (c) 8 V

   (d) 10 V

   ›Reveal solutionSolution

   In a series R-L-C a.c. circuit, $V_L$ and $V_C$ are 180 degrees out of phase with each other and both are 90 degrees out of phase with $V_R$, so the source voltage is the phasor sum $V=\sqrt{V_R^2+(V_L-V_C)^2}$, not the plain arithmetic sum.

   In a series a.c. circuit, the voltage across R is in phase with the current, the voltage across L leads the current by $90^\circ$, and the voltage across C lags the current by $90^\circ$ ($V_L$ and $V_C$ are therefore antiphase to each other). The applied (source) voltage is t …

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10. CBSE 2025Set ANNUAL1 markMCQ
    Q.The unit of reactance is

    (a) ohm

    (b) mho

    (c) farad

    (d) ampere

    ›Reveal solutionSolution

    Reactance X (whether inductive, XL = wL, or capacitive, XC = 1/wC) has the same dimensions as resistance, since it appears in V = IX exactly like Ohm's law.

    …

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11. CBSE 2025Set ANNUAL1 mark
    Q.The equation of an alternating current is $I = 50\sin100t$. Find the frequency. OR A 100 hertz alternating current is flowing in a 14 mH coil. Find its reactance.

    ›Reveal solutionSolution

    Reading off $\omega$ from the given equation and using $f=\omega/2\pi$ gives about 15.9 Hz; for the alternative, $X_L=2\pi fL$ gives about 8.8 $\Omega$.

    Finding the frequency

    The standard form of an alternating current is

    $I = I_0 \sin(\omega t)$

    Comparing with the given equation $I = 50\sin(100t)$:

    $\omega = 100\ \text{rad/s}$

    Since $\omega = 2\pi f$,

    $f = \frac{\omega}{2\pi} = \frac{100}{2\pi} \approx 15.9\ \text{Hz}$

    …

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12. CBSE 2024Set 55/2/11 markMCQ
    Q.A resistor and an ideal inductor are connected in series to a $100\sqrt{2}$ V, 50 Hz ac source. When a voltmeter is connected across the resistor or the inductor, it shows the same reading. The reading of the voltmeter is : (A) $100\sqrt{2}$ V (B) $100$ V (C) $50\sqrt{2}$ V (D) $50$ V

    ›Reveal solutionSolution

    In a series RL circuit, equal voltmeter readings across R and L imply equal voltage drops, which forces the impedance triangle to be isosceles — the common reading is $100$ V.

    The key idea here is that the voltmeter reads the RMS voltage across each component. For a series RL circuit driven by an AC source, the source voltage $V_s$ is the phasor sum of $V_R$ and $V_L$, not the arithmetic sum — because the resistor voltage and inductor voltage are $90^\circ$ out of phase.

    When the voltmeter shows the same reading across R and L, we have $V_R = V_L = V$ (say). The source RMS voltage is given as $100\sqrt{2}$ V. The phasor relationship is:

    $V_s^2 = V_R^2 + V_L^2 = V^2 + V^2 = 2V^2$

    So $V = \frac{V_s}{\sqrt{2}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100$ V.

    That is the reading on the voltmeter — $100$ V.

    ---

    1. Understand the circuit and the source.

       The source is $100\sqrt{2}$ V, 50 Hz AC. The voltmeter reads RMS values (standard for AC voltmeters). The resistor and ideal inductor are in series.

    2. Phase relationship.

       In a series RL circuit, current $I$ is common. Voltage across resistor $V_R = IR$ is in phase with current. Voltage across inductor $V_L = I X_L$ leads current by $90^\circ$. So $V_R$ and $V_L$ are perpendicular phasors.

    3. Given condition: equal readings.

       The voltmeter reads the same value across R and L. So $V_R = V_L = V$ (say).

    4. Apply phasor addition for the source voltage.

       The source RMS voltage $V_s$ is the magnitude of the phasor sum:

    $V_s = \sqrt{V_R^2 + V_L^2} = \sqrt{V^2 + V^2} = V\sqrt{2}$

    5. Solve for $V$. …

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