Alternating Current & RMS Values — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Alternating Current & RMS Values MCQs with step-by-step solutions (21 questions). Part of AC Circuits. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Alternating Current & RMS Values · easy · theory
An alternating current is one whose:
A. Magnitude and direction are both constant
B. Direction changes but magnitude stays fixed
C. Magnitude changes but direction stays fixed
D. Magnitude and direction both change periodically with time ✓ Correct
Solution: A sinusoidal alternating current completes a full reversal in each cycle, which is why its mean value over a cycle is zero.
Q2 — Alternating Current & RMS Values · medium · theory
The RMS value of an alternating current is defined as:
A. The steady current that produces the same heating effect in a given resistance ✓ Correct
B. The average of the current over one full cycle
C. Half the peak value
D. The maximum value reached by the current
Solution: It is also called the effective or virtual value, and it is what AC meters are calibrated to read.
Q3 — Alternating Current & RMS Values · easy · theory
For a sinusoidal alternating current of peak value $I_0$, the RMS value is:
A. $I_0\sqrt{2}$
B. $\dfrac{2I_0}{\pi}$
C. $\dfrac{I_0}{\sqrt{2}}$ ✓ Correct
D. $\dfrac{I_0}{2}$
Solution: Averaging $\sin^2$ over a full cycle gives one-half, so the root mean square is $\dfrac{I_0}{\sqrt{2}} \approx 0.707I_0$.
Q4 — Alternating Current & RMS Values · easy · theory
The average value of a sinusoidal alternating current over one complete cycle is:
A. $\dfrac{2I_0}{\pi}$
B. Zero ✓ Correct
C. $I_0$
D. $\dfrac{I_0}{\sqrt{2}}$
Solution: The positive and negative half cycles are identical in shape, so they cancel exactly.
Q5 — Alternating Current & RMS Values · medium · theory
The average value of a sinusoidal alternating current over half a cycle is:
A. $\dfrac{I_0}{\sqrt{2}}$
B. Zero
C. $\dfrac{I_0}{2}$
D. $\dfrac{2I_0}{\pi}$ ✓ Correct
Solution: Integrating $\sin\theta$ over half a period gives $\dfrac{2I_0}{\pi} \approx 0.637I_0$.
Q6 — Alternating Current & RMS Values · medium · theory
An AC ammeter connected in a circuit reads the:
A. Instantaneous value
B. Peak value of the current
C. RMS value of the current ✓ Correct
D. Average value over a full cycle
Solution: AC meters respond to the heating effect, which depends on the mean square, so they are calibrated in RMS.
Q7 — Alternating Current & RMS Values · hard · theory
The form factor of a sinusoidal alternating current is:
A. $1.414$
B. $0.637$
C. $0.707$
D. $1.11$ ✓ Correct
Solution: Form factor is $\dfrac{I_{rms}}{I_{avg}} = \dfrac{0.707I_0}{0.637I_0} \approx 1.11$.
Q8 — Alternating Current & RMS Values · medium · theory
The peak factor of a sinusoidal alternating current is:
A. $\dfrac{2}{\pi}$
B. $1.11$
C. $\dfrac{1}{\sqrt{2}}$
D. $\sqrt{2}$ ✓ Correct
Solution: Peak factor is $\dfrac{I_0}{I_{rms}} = \sqrt{2} \approx 1.414$.
Q9 — Alternating Current & RMS Values · medium · numerical
An alternating current is given by $i = 10\sqrt{2}\sin(100\pi t)\text{ A}$. Its RMS value and frequency are:
A. $10\sqrt{2}\text{ A}$ and $50\text{ Hz}$
B. $14.14\text{ A}$ and $100\text{ Hz}$
C. $10\text{ A}$ and $100\text{ Hz}$
D. $10\text{ A}$ and $50\text{ Hz}$ ✓ Correct
Solution: Peak $= 10\sqrt{2}$, so $I_{rms} = 10\text{ A}$. From $\omega = 100\pi = 2\pi f$, $f = 50\text{ Hz}$.
Q10 — Alternating Current & RMS Values · easy · numerical
An alternating voltage is $V = 200\sqrt{2}\sin(100t)\text{ V}$. Its RMS value is:
A. $141.4\text{ V}$
B. $200\text{ V}$ ✓ Correct
C. $100\text{ V}$
D. $282.8\text{ V}$
Solution: $V_{rms} = \dfrac{V_0}{\sqrt{2}} = \dfrac{200\sqrt{2}}{\sqrt{2}} = 200\text{ V}$.
Q11 — Alternating Current & RMS Values · easy · numerical
An alternating current has a peak value of $5\text{ A}$. Its RMS value is approximately:
A. $3.54\text{ A}$ ✓ Correct
B. $7.07\text{ A}$
C. $3.18\text{ A}$
D. $2.5\text{ A}$
Solution: $I_{rms} = \dfrac{5}{\sqrt{2}} \approx 3.54\text{ A}$.
Q12 — Alternating Current & RMS Values · medium · numerical
The RMS value of the mains supply is $220\text{ V}$. Its peak value is approximately:
A. $155\text{ V}$
B. $440\text{ V}$
C. $220\text{ V}$
D. $311\text{ V}$ ✓ Correct
Solution: $V_0 = V_{rms}\sqrt{2} = 220 \times 1.414 \approx 311\text{ V}$.
Q13 — Alternating Current & RMS Values · easy · numerical
An alternating voltage has angular frequency $100\pi\text{ rad/s}$. Its frequency is:
A. $314\text{ Hz}$
B. $25\text{ Hz}$
C. $100\text{ Hz}$
D. $50\text{ Hz}$ ✓ Correct
Solution: $f = \dfrac{\omega}{2\pi} = \dfrac{100\pi}{2\pi} = 50\text{ Hz}$.
Q14 — Alternating Current & RMS Values · easy · numerical
An alternating current has angular frequency $314\text{ rad/s}$. Its frequency is approximately:
A. $50\text{ Hz}$ ✓ Correct
B. $100\text{ Hz}$
C. $314\text{ Hz}$
D. $157\text{ Hz}$
Solution: $f = \dfrac{314}{2\pi} \approx 50\text{ Hz}$.
Q15 — Alternating Current & RMS Values · medium · numerical
An alternating current is $i = 2\sin(314t)\text{ A}$. Its RMS value is approximately:
A. $2.83\text{ A}$
B. $1.27\text{ A}$
C. $2\text{ A}$
D. $1.41\text{ A}$ ✓ Correct
Solution: $I_{rms} = \dfrac{2}{\sqrt{2}} = \sqrt{2} \approx 1.41\text{ A}$.
Q16 — Alternating Current & RMS Values · hard · numerical
An alternating current has a peak value of $10\text{ A}$. Its average value over half a cycle is approximately:
A. $7.07\text{ A}$
B. Zero
C. $6.37\text{ A}$ ✓ Correct
D. $5\text{ A}$
Solution: $I_{avg} = \dfrac{2I_0}{\pi} = \dfrac{20}{3.1416} \approx 6.37\text{ A}$.
Q17 — Alternating Current & RMS Values · medium · numerical
An alternating current has RMS value $7.07\text{ A}$ and average value over half a cycle $6.37\text{ A}$. Its form factor is approximately:
A. $0.90$
B. $1.11$ ✓ Correct
C. $0.64$
D. $1.41$
Solution: Form factor $= \dfrac{7.07}{6.37} \approx 1.11$.
Q18 — Alternating Current & RMS Values · easy · numerical
For an alternating voltage, the ratio of the peak value to the RMS value is:
A. $0.707$
B. $1.11$
C. $2$
D. $1.414$ ✓ Correct
Solution: This ratio, the peak factor, equals $\sqrt{2}$ for a sinusoidal waveform.
Q19 — Alternating Current & RMS Values · easy · numerical
An alternating supply has a frequency of $50\text{ Hz}$. Its time period is:
A. $50\text{ s}$
B. $0.05\text{ s}$
C. $0.1\text{ s}$
D. $0.02\text{ s}$ ✓ Correct
Solution: $T = \dfrac{1}{f} = \dfrac{1}{50} = 0.02\text{ s}$.
Q20 — Alternating Current & RMS Values · medium · numerical
An alternating voltage is $V = 100\sin(200\pi t)\text{ V}$. Its frequency is:
A. $200\text{ Hz}$
B. $628\text{ Hz}$
C. $100\text{ Hz}$ ✓ Correct
D. $50\text{ Hz}$
Solution: $\omega = 200\pi = 2\pi f$, so $f = 100\text{ Hz}$.
Q21 — Alternating Current & RMS Values · medium · numerical
An alternating current of RMS value $2\text{ A}$ flows through a $10\,\Omega$ resistor. The power dissipated is:
A. $5\text{ W}$
B. $20\text{ W}$
C. $80\text{ W}$
D. $40\text{ W}$ ✓ Correct
Solution: $P = I_{rms}^2R = 4 \times 10 = 40\text{ W}$.