Q.a) Prove that the peak value (I_m) of an alternating current is √2 times of its root mean square (rms) value. b) If alternating current I = 4 sin ωt and voltage V = 200 sin(ωt + π/3), then calculate the average power dissipated in the circuit. OR

Showing the 12 most recent of 40 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.The voltage of domestic ac is 220 V. What does this represent? (A) Peak value voltage (B) Mean value voltage (C) Root mean voltage (D) Root mean square voltage

   ›Reveal solutionSolution

   The 220 V of domestic mains is the RMS (root-mean-square) voltage.

   An AC voltage varies sinusoidally, so it is specified by an effective value that produces the same heating as an equivalent DC — the root-mean-square (RMS) value. Household ratings such as "220 V" are RMS values. The peak val …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.The ratio of root mean square (rms) value and peak value of an alternating current is

   (a) 1 : 1

   (b) 1 : 2

   (c) √2 : 1

   (d) 1 : √2

   ›Reveal solutionSolution

   For a sinusoidal alternating current i = i0 sin(omega t), the rms value is i0/root2, so the ratio rms:peak is 1:root2.

   RMS (root mean square) value is defined so that it produces the same heating effect as an equivalent DC. For i = i0 sin(omega t), averaging i^2 over a …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.The mean value of an alternating current in a half cycle is

   (a) I0/sqrt(2)

   (b) I0/2

   (c) 2*I0/pi

   (d) none of these

   ›Reveal solutionSolution

   Averaging I = I0sin(omegat) over one half cycle (0 to pi/omega) gives 2*I0/pi - this is the standard 'mean/average value of AC'.

   For a sinusoidal current I = I0sin(omegat), the average over a FULL cycle is zero (positive and negative halves cancel exactly). So the 'mean value' of AC is conventionally defined over just a HALF cycle, where the current keeps one sign throughout. Averaging:

   I_mean = (1/T') * integral of I0sin(omegat) dt, over one half period T' = pi/omega

   …

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4. CBSE 2026Set SEM31 markMCQ
   Q.Statement I : Direct current (DC) is less dangerous than alternating current (AC). Statement II : The rms value of the alternating current (AC) is 70·7% of the peak value.

   (a) Only Statement I is true.

   (b) Only Statement II is true.

   (c) Both Statements I and II are true.

   (d) Both Statements I and II are false.

   ›Reveal solutionSolution

   Statement I is true (AC of equal rated voltage is generally more dangerous than DC), and Statement II is true (rms value = peak/√2 = 70·7% of peak). Hence option (c).

   Statement I: For the same magnitude, alternating current is generally considered more dangerous than direct current, largely because AC can cause sustained muscular contraction and its effective (rms) value acts continuously. So DC being 'less dangerous' is accepted as true.

   …

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5. CBSE 2026Set SEM31 markMCQ
   Q.The equation of an alternating electromotive force is E = 220 sin(100πt − π/15) V, here t is in second. Its rms value and frequency are respectively

   (a) (220/√2) V, 50 Hz

   (b) 220 V, 50 Hz

   (c) (220/√2) V, 100 Hz

   (d) 220√2 V, 50 Hz

   ›Reveal solutionSolution

   Compare E = 220 sin(100πt − π/15) with E = E₀ sin(ωt + φ): E₀ = 220 V and ω = 100π. Then E_rms = E₀/√2 = 220/√2 and f = ω/2π = 50 Hz. Option (a).

   Step 1 — peak value: E₀ = 220 V, so the rms value E_rms = E₀/√2 = 220/√2 V.

   Step 2 — angular frequency: ω = 100π rad/s.

   Step 3 — frequency: f = ω/(2π) = 100π/(2π) = 50 Hz.

   …

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6. CBSE 2026Set SEM31 markMCQ
   Q.The ratio of rms value and average value of current for a half cycle of an AC circuit is

   (a) √2 : π

   (b) √2 : 1

   (c) 2√2 : π

   (d) π : 2√2

   ›Reveal solutionSolution

   For a sinusoid, I_rms = I₀/√2 and the half-cycle average I_avg = 2I₀/π. Their ratio is (1/√2)/(2/π) = π/(2√2), i.e. π : 2√2. Option (d).

   Step 1 — rms value of a sinusoidal current: I_rms = I₀/√2.

   Step 2 — average value over a half cycle: I_avg = 2I₀/π (NCERT/CBSE Class 12 Physics, Alternating Current).

   Step 3 — form the ratio: …

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7. CBSE 2025Set 55/4/11 markMCQ
   Q.An ammeter connected in series in an ac circuit reads 10 A. The maximum value of current at any instant in the circuit is: (A) $10\sqrt{2}$ A (B) $\frac{10}{\sqrt{2}}$ A (C) $\frac{10}{\pi}$ A (D) $\frac{10}{\sqrt{2}\,\pi}$ A

   ›Reveal solutionSolution

   An AC ammeter reads the RMS (root-mean-square) value of current. For a sinusoidal AC, the peak (maximum) current is $\sqrt{2}$ times the RMS value. Given RMS = 10 A, the maximum current is $10\sqrt{2}$ A.

   The key here is understanding what an AC ammeter actually measures. Unlike a DC ammeter, which reads the average current, an AC ammeter is calibrated to read the RMS value of the current. For a sinusoidal alternating current, the RMS value is the "effective" value — it tells you the equivalent DC current that would produce the same heating effect in a resistor.

   The relationship between the RMS value ($I_{\text{rms}}$) and the peak or maximum value ($I_0$) for a pure sine wave is:

   $I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \quad \text{or} \quad I_0 = I_{\text{rms}} \times \sqrt{2}$

   This comes from averaging the square of the sine function over one cycle. The factor $\sqrt{2}$ (approximately 1.414) is a fixed mathematical result for sinusoidal waveforms.

   Now let's apply this directly to the problem.

   1. The ammeter reading is given as 10 A. Since it's an AC ammeter, this is the RMS current: $I_{\text{rms}} = 10$ A.

   2. We want the maximum instantaneous current, which is the peak value $I_0$. Using the formula above:

   $I_0 = I_{\text{rms}} \times \sqrt{2} = 10 \times \sqrt{2} \text{ A}$

   3. That's it. No further calculation needed. The maximum value is simply $10\sqrt{2}$ amperes. …

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8. CBSE 2025Set D1 markMCQ
   Q.The peak value of an alternating current is 10 A. Its root mean square value will be (A) 5 A (B) 7.07 A (C) 10 A (D) 14.14 A

   ›Reveal solutionSolution

   The rms value of a sinusoidal current is the peak value divided by √2: 10/√2 ≈ 7.07 A.

   For a sinusoidal alternating current, the root-mean-square value relates to the peak (amplitude) value by

   $I_{rms} = \frac{I_0}{\sqrt{2}}$

   …

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9. CBSE 2025Set A1 mark
   Q.Write True or False: Average value of alternating current is zero for full cycle.

   ›Reveal solutionSolution

   The statement is True: over a full cycle the positive and negative halves of an AC waveform exactly cancel, giving zero average.

   For a sinusoidal alternating current $i = I_0\sin\omega t$, the average value over one complete cycle is found by integrating over a full period T:

   $I_{avg} = \dfrac{1}{T}\int_0^T I_0\sin\omega t\, dt = 0$

   …

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10. CBSE 2025Set A1 mark
    Q.Write answer in one sentence: Write the root mean square value of alternating current.

    ›Reveal solutionSolution

    The rms value of a sinusoidal AC is I₀/√2, about 70.7% of its peak value.

    For an alternating current $i = I_0\sin\omega t$, the root-mean-square (rms) value is obtained by taking the square root of the mean of $i^2$ over one complete cycle:

    $I_{rms} = \sqrt{\dfrac{1}{T}\int_0^T I_0^2\sin^2\omega t\, dt} = \dfrac{I_0}{\sqrt{2}} \approx 0.707\,I_0$

    …

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11. CBSE 2025Set ANNUAL1 markMCQ
    Q.If the peak value of alternating voltage in a circuit is E0, then the root mean square value will be

    (a) E0/2

    (b) E0

    (c) E0/sqrt(2)

    (d) E0^2/2

    ›Reveal solutionSolution

    For a sinusoidal quantity, the root-mean-square value is obtained by averaging the square of the waveform over a cycle and taking the square root, which gives peak/sqrt(2).

    For e(t) = E0 sin(wt), the mean of sin^2(wt) over a full cycle is 1/2. So:

    E_rms = sqrt( mean of e^2 ) = sqrt( E0^2 x 1/2 ) = E0 / sqrt(2)

    …

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12. CBSE 2025Set ANNUAL1 mark
    Q.The equation of an alternating current is I = 15 sin 100t. Find its root mean square value.

    ›Reveal solutionSolution

    Comparing I = 15 sin(100t) with the standard form I = I0 sin(wt) gives peak current I0 = 15 A, and the rms value of any sinusoidal current is I0/sqrt(2).

    The given alternating current is I = 15 sin(100t) A, matching the standard form I = I0 sin(wt) with:

    I0 = 15 A (peak/amplitude), w = 100 rad/s

    …

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