Question 1

Series AC drops to peak source: Two resistors in series are connected to an AC source. If the measured voltages are 7.5 V rms across one resistor and 4.2 V rms across the other, what is the peak (maximum) source voltage?

Accepted Answer

Introduction / Context: In basic AC circuit analysis with purely resistive series elements, the root-mean-square (rms) voltages across individual resistors add directly to give the total rms source voltage. Once the rms value is known, the peak value of a sinusoidal source can be found using the standard rms–peak relationship. This problem evaluates your fluency with these two ideas and correct unit handling. Given Data / Assumptions: Two resistors are connected in series across a single sinusoidal AC source. Voltage drops: 7.5 V rms across one resistor and 4.2 V rms across the other. Assume ideal, purely resistive elements (voltages are in phase in series). Waveform is sinusoidal, so Vrms = Vpeak / √2. Concept / Approach: For series resistors in a purely resistive circuit, the same current flows through each and the phase is the same. Therefore, individual rms drops add arithmetically to obtain the total rms source voltage. Then use Vpeak = Vrms * √2 to convert to the peak (maximum) value of the sinusoid. Step-by-Step Solution: Total rms voltage: Vrms_total = 7.5 V + 4.2 V = 11.7 V.Peak source voltage: Vpeak = Vrms_total * √2 = 11.7 * 1.414 ≈ 16.54 V.Therefore, the peak source voltage is approximately 16.54 V. Verification / Alternative check: Back-conversion: 16.54 V / 1.414 ≈ 11.7 V rms, matching the sum of the given drops, so the computation is self-consistent for a sinusoidal source with resistive series elements. Why Other Options Are Wrong: 10.60 V: Close to the rms total, not the peak. 5.93 V and 1.65 V: These are far below the correct peak; they misuse the √2 factor or mix rms and peak values incorrectly. Common Pitfalls: Adding peaks or mixing rms and peak magnitudes inconsistently. Using vector addition unnecessarily—here, resistive drops are in phase, so simple arithmetic addition is correct. Final Answer: 16.54 V

Question 2

Ohm’s law with AC rms values: If 8 mA rms of current flows through a 6.8 kΩ resistor, what is the corresponding rms voltage drop across the resistor?

Accepted Answer

Introduction / Context: In a linear, purely resistive circuit, Ohm’s law applies to instantaneous, rms, or average quantities provided the matching forms are used. With AC sinusoidal signals, rms current multiplied by resistance gives rms voltage. This problem reinforces unit conversion (mA to A, kΩ to Ω) and precise multiplication. Given Data / Assumptions: I_rms = 8 mA = 0.008 A. R = 6.8 kΩ = 6800 Ω. Purely resistive element, so V_rms = I_rms * R. Concept / Approach: Apply Ohm’s law directly with rms quantities. No phase angles or reactive components are involved, so the relationship is linear and scalar. Keep careful track of milli (10^-3) and kilo (10^3) prefixes during the calculation to avoid factor-of-1000 errors. Step-by-Step Solution: Convert units: I_rms = 0.008 A; R = 6800 Ω.Compute V_rms: V_rms = I_rms * R = 0.008 * 6800 = 54.4 V.Therefore, the resistor drop is 54.4 V rms. Verification / Alternative check: Power check: P = I_rms^2 * R = (0.008)^2 * 6800 ≈ 0.435 W. Then V_rms = √(P * R) = √(0.435 * 6800) ≈ √2958 ≈ 54.4 V, confirming the result by a different route. Why Other Options Are Wrong: 5.44 V: Misses a power-of-10 factor (likely using 680 Ω instead of 6800 Ω). 7.07 V and 8 V: Not consistent with the Ohm’s law product for the given values; these resemble unrelated common rms numbers. Common Pitfalls: Mistaking 6.8 kΩ for 680 Ω or 68 kΩ. Leaving current in mA and resistance in kΩ without ensuring the product yields volts (careful with prefixes). Final Answer: 54.4 V

Question 3

Radians to degrees conversion: How many degrees are contained in an angle of π/3 radians (use 180° = π radians)?

Accepted Answer

Introduction / Context: Engineering and physics problems frequently require switching between radians and degrees. Since many wave, phase, and rotational formulas are naturally expressed in radians, fluency in conversion helps prevent downstream mistakes in AC analysis, phasors, and trigonometric calculations. Given Data / Assumptions: Angle given: π/3 radians. Conversion identity: 180 degrees = π radians. Standard arithmetic; no approximations beyond the exact ratio are necessary. Concept / Approach: Use a proportion based on the fundamental equivalence of π radians to 180 degrees. Multiply the radian value by (180/π) to obtain degrees. This works for any radian measure and preserves exactness for rational multiples of π. Step-by-Step Solution: Start with θ_rad = π/3.Apply conversion: θ_deg = θ_rad * (180/π) = (π/3) * (180/π).Cancel π: θ_deg = 180/3 = 60 degrees. Verification / Alternative check: You can instead note common benchmark angles: π/3 corresponds to 60° (π/6 → 30°, π/4 → 45°, π/2 → 90°). This memorized set confirms the arithmetic conversion. Why Other Options Are Wrong: 6° and 27°: These result from incorrect scaling or dividing by 10 accidentally. 180°: That corresponds to π radians, not π/3. Common Pitfalls: Using 360° = 2π but forgetting to divide both numerator and denominator appropriately. Dropping the π factor prematurely and mixing units. Final Answer: 60°

Question 4

Average value of a pulsed waveform: A signal has a baseline (DC offset) of 3 V, a duty cycle of 20%, and an amplitude of 8 V relative to the baseline during its ON interval. What is the average (mean) output voltage?

Accepted Answer

Introduction / Context: Pulse and digital waveforms often ride on a baseline (DC offset). The average value depends on both the offset and how long the waveform spends at its higher level. Understanding how duty cycle and amplitude interact is essential in power electronics, PWM control, and signal conditioning. Given Data / Assumptions: Baseline (offset) voltage V_base = 3 V. Amplitude during ON relative to baseline A = 8 V (so ON level is V_base + A). Duty cycle D = 20% = 0.2 (fraction of the period at the ON level). Assume ideal rectangular transitions; OFF level equals the baseline V_base. Concept / Approach: The average of a two-level waveform is the time-weighted mean of its levels. If the OFF level is the baseline and the ON level is baseline + amplitude, the average is V_avg = V_base + D * A. This expression naturally separates the steady offset from the pulsed increment contributed by the duty cycle. Step-by-Step Solution: Identify ON level: V_on = V_base + A = 3 + 8 = 11 V.OFF level: V_off = V_base = 3 V.Average: V_avg = D * V_on + (1 − D) * V_off.Compute: V_avg = 0.2 * 11 + 0.8 * 3 = 2.2 + 2.4 = 4.6 V. Verification / Alternative check: Use compact form: V_avg = V_base + D * A = 3 + 0.2 * 8 = 3 + 1.6 = 4.6 V. Both methods agree. This matches intuition: with only 20% ON time, the average lies modestly above the 3 V baseline, not near the 11 V peak. Why Other Options Are Wrong: 4 V: Would correspond to a smaller amplitude or duty cycle; it ignores the specified 8 V amplitude contribution. 1.6 V: This is just D * A, omitting the baseline, so it underestimates the mean. 11 V: That is the instantaneous ON level, not the average across a period with 20% duty. Common Pitfalls: Forgetting to add the baseline to the duty-weighted amplitude. Confusing amplitude with peak-to-peak or with absolute level. Final Answer: 4.6 V

Question 5

Frequency from period for a sawtooth: A sawtooth waveform has a period of 10 ms. What is its frequency in hertz?

Accepted Answer

Introduction / Context: Frequency and period are reciprocals for any repetitive waveform, including sawtooth signals used in timing, ramp generators, and sweep circuits. Converting between milliseconds and hertz correctly is a fundamental time–frequency skill for electronics and instrumentation work. Given Data / Assumptions: Sawtooth period T = 10 ms. Relation: f = 1 / T. Units: 1 ms = 10^-3 s, so 10 ms = 0.01 s. Concept / Approach: Compute the frequency as the reciprocal of the period after converting to seconds. Maintaining unit consistency avoids order-of-magnitude mistakes that are common when switching between milliseconds and seconds. Step-by-Step Solution: Convert period: T = 10 ms = 0.01 s.Compute frequency: f = 1 / T = 1 / 0.01 = 100 Hz.Therefore, the frequency of the sawtooth is 100 Hz. Verification / Alternative check: Check via intuition: A period of 10 ms means 100 cycles occur each second (since 100 * 0.01 s = 1 s). This reaffirms the numerical result without further calculation. Why Other Options Are Wrong: 10 Hz: Would correspond to T = 0.1 s (100 ms), not 10 ms. 50 Hz: Would require T = 20 ms. 1,000 Hz: Would require T = 1 ms. Common Pitfalls: Forgetting to convert milliseconds to seconds before inversion. Inverting and then misplacing the decimal point by a factor of 10. Final Answer: 100 Hz

Question 6

Counting cycles of a sine wave: A sinusoidal signal has a frequency of 100 Hz. Over an interval of 12 s, how many complete cycles does it undergo?

Accepted Answer

Introduction / Context: Frequency represents the number of cycles per second. Converting a time interval into total cycles is a direct multiplication and shows up in sampling, synchronization, and timing design. This question ensures correct interpretation of the hertz unit and appropriate scaling across seconds. Given Data / Assumptions: Frequency f = 100 Hz = 100 cycles per second. Observation interval t = 12 s. We count complete cycles, assuming steady frequency. Concept / Approach: Total cycles N over time t is N = f * t when frequency is constant. This linear relation is fundamental to periodic phenomena and aligns with everyday concepts like RPM and sampling counts. Step-by-Step Solution: Compute N = f * t = 100 * 12 = 1200 cycles.Thus, in 12 s, the sine wave completes 1,200 full cycles. Verification / Alternative check: Cross-check using period: T = 1/f = 1/100 s = 0.01 s. The number of periods in 12 s is 12 / 0.01 = 1200, matching the prior calculation exactly. Why Other Options Are Wrong: 12 cycles or 120 cycles: These confuse seconds with cycles; they are too small by factors of 100 or 10. 1/100 cycle: This is one period’s fraction, not 12 seconds of activity. Common Pitfalls: Treating hertz as radians per second or misplacing the decimal during multiplication. Failing to use consistent units (seconds for time, hertz for cycles per second). Final Answer: 1,200 cycles

Question 7

Peak voltage from rms current and resistance: If a 4.7 kΩ resistor carries 4 mA rms of current, what is the peak (maximum) voltage drop across this resistor?

Accepted Answer

Introduction / Context: Determining peak values from rms measurements is routine when dealing with sinusoidal signals. For a purely resistive element, the rms voltage is I_rms * R, and the peak voltage equals the rms value multiplied by √2. This problem also checks clean handling of milli- and kilo- prefixes. Given Data / Assumptions: I_rms = 4 mA = 0.004 A. R = 4.7 kΩ = 4700 Ω. Sinusoidal signal across a resistor (no reactive effects). Concept / Approach: First compute the rms voltage drop using Ohm’s law. Then convert rms to peak via Vpeak = Vrms * √2. The result represents the maximum instantaneous magnitude of the sinusoidal voltage across the resistor. Step-by-Step Solution: Vrms = I_rms * R = 0.004 A * 4700 Ω = 18.8 V.Vpeak = Vrms * √2 = 18.8 * 1.414 ≈ 26.6 V.Therefore, the peak drop across the 4.7 kΩ resistor is approximately 26.6 V. Verification / Alternative check: Power consistency: P = I_rms^2 * R = (0.004)^2 * 4700 ≈ 0.0752 W. Using Vpeak, the instantaneous maximum power would be (Vpeak^2 / R) at peaks; the average value aligns with the rms-based power, indicating consistent calculations. Why Other Options Are Wrong: 18.8 V: That is the rms drop, not the peak value. 4 V and 2.66 V: Off by factors from neglecting R magnitude or the √2 conversion. Common Pitfalls: Mistaking rms for peak or vice versa. Dropping prefixes: treating 4.7 kΩ as 4.7 Ω produces wildly incorrect results. Final Answer: 26.6 V

Question 8

Duty cycle of a pulse train: A 20 kHz rectangular pulse waveform contains pulses that are 15 µs wide. What is the duty cycle of this waveform?

Accepted Answer

Introduction / Context: Duty cycle indicates the fraction of each period during which a pulse waveform is ON (high). It is crucial in pulse-width modulation (PWM), digital timing, and power control applications, where average power or voltage depends linearly on duty cycle for a fixed amplitude and baseline. Given Data / Assumptions: Frequency f = 20 kHz → period T = 1 / f. Pulse width (ON time) t_on = 15 µs. Rectangular pulse train with a consistent period and width. Concept / Approach: Duty cycle D is defined as D = t_on / T expressed as a percentage. Convert frequency to period in microseconds to match the given width, then compute the ratio and multiply by 100% to get the percentage value. Step-by-Step Solution: Compute period: T = 1 / 20,000 s = 0.00005 s = 50 µs.Compute duty: D = t_on / T = 15 µs / 50 µs = 0.3.Express as percent: D = 0.3 * 100% = 30%. Verification / Alternative check: Reasonableness: A 15 µs pulse within a 50 µs period must correspond to less than half the period. 30% is consistent (since 25 µs would be 50% at 20 kHz, so 15 µs is noticeably smaller). Why Other Options Are Wrong: 1%: Would correspond to t_on = 0.5 µs at 20 kHz, not 15 µs. 100%: Would require a continuous ON signal with zero OFF time (t_on = T). cannot be determined: Not true; both width and period are given explicitly. Common Pitfalls: Failing to convert kHz to µs for an easy ratio. Confusing frequency with angular frequency (20 kHz vs. 2πf). Final Answer: is 30%

Question 9

Peak current from rms voltage and resistance: If the rms voltage across a 15 kΩ resistor is 16 V, what is the peak (maximum) current through the resistor?

Accepted Answer

Introduction / Context: Converting between rms and peak values is routine for sinusoidal steady-state analysis. Given an rms voltage across a resistor, find the rms current via Ohm’s law, then convert that rms current to its peak counterpart by multiplying by √2. This skill is frequently applied in amplitude ratings and current-limiting calculations. Given Data / Assumptions: Vrms = 16 V. R = 15 kΩ = 15,000 Ω. Sinusoidal voltage across a purely resistive element. Concept / Approach: First compute Irms from Ohm’s law: Irms = Vrms / R. For a sinusoid, Ipeak = Irms * √2. Carefully handle kilo-ohm to ohm conversion to obtain milliamperes correctly. Step-by-Step Solution: Irms = Vrms / R = 16 / 15,000 A ≈ 0.0010667 A = 1.0667 mA.Ipeak = Irms * √2 ≈ 1.0667 mA * 1.414 ≈ 1.51 mA ≈ 1.5 mA.Therefore, the peak current is approximately 1.5 mA. Verification / Alternative check: Back-check Vrms via Ipeak: Irms = Ipeak / √2 ≈ 1.5 / 1.414 ≈ 1.06 mA. Then Vrms = Irms * R ≈ 0.00106 * 15,000 ≈ 15.9 V (rounding to 16 V), confirming consistency with the given value. Why Other Options Are Wrong: 15 mA and 10 mA: One or two orders of magnitude too large; they ignore the 15 kΩ resistance. 1 mA: Closer, but this is rms-like rather than peak; it does not match the √2 conversion from 1.0667 mA rms. Common Pitfalls: Mishandling prefixes: using 15 Ω instead of 15 kΩ. Confusing rms with peak values or forgetting the √2 factor for sinusoids. Final Answer: 1.5 mA

Question 10

Signal basics: A periodic signal has a period of 400 µs (microseconds). What is its frequency in hertz (Hz)?

Accepted Answer

Introduction / Context: Frequency–period relationships are foundational in electrical engineering and signal processing. Knowing either the period (time for one cycle) or the frequency (cycles per second) allows you to compute the other. This question tests conversion from a given period in microseconds to frequency in hertz. Given Data / Assumptions: Period T = 400 µs. 1 µs = 10^-6 s. Ideal, perfectly periodic waveform (no jitter). Concept / Approach: The period T and frequency f are reciprocals: f = 1 / T. Convert microseconds to seconds first to avoid scale errors, then take the reciprocal to obtain frequency in hertz. Step-by-Step Solution: Convert period: T = 400 µs = 400 * 10^-6 s = 4.00 * 10^-4 s.Apply reciprocal relation: f = 1 / T.Compute: f = 1 / (4.00 * 10^-4) = 2.5 * 10^3 Hz = 2,500 Hz. Verification / Alternative check: A period of 1 ms corresponds to 1,000 Hz. Since 400 µs is less than 1 ms, the frequency should be greater than 1,000 Hz. 2,500 Hz fits this expectation. Why Other Options Are Wrong: 250 Hz corresponds to T = 4 ms. 25,000 Hz corresponds to T = 40 µs. 400 Hz corresponds to T = 2.5 ms. None match 400 µs. Common Pitfalls: Forgetting to convert µs to s; placing the decimal incorrectly when taking reciprocals; confusing kHz and Hz. Final Answer: 2,500 Hz

Question 11

Sine wave fundamentals: For a pure sine wave with a peak value of 12 V, what is the average (mean) value over one complete cycle?

Accepted Answer

Introduction / Context: The average (mean) value of an alternating sinusoidal voltage over a full cycle is a frequent point of confusion. While RMS and average-rectified values are nonzero, the algebraic average across one complete cycle of a symmetric sine is zero. This question checks recognition of that property. Given Data / Assumptions: Sine wave peak Vp = 12 V. Waveform is symmetrical about zero (no DC offset). Average is taken over one full period (0 to 2π in angle). Concept / Approach: For v(t) = Vp * sin(ωt), the positive half-cycle area equals the negative half-cycle area. Over 0 to 2π, the integral of sin is zero, hence the average value is zero volts. Step-by-Step Solution: Average over one cycle: V_avg = (1/T) * ∫_0^T v(t) dt.With v(t) = Vp sin(ωt), the integral over 0→T (or 0→2π) is zero.Therefore, V_avg = 0 V regardless of Vp (if there is no DC offset). Verification / Alternative check: The RMS of this wave is Vp/√2 (≈ 8.49 V here) and the average of the absolute (half-wave average doubled) is 2Vp/π (≈ 7.64 V), but the algebraic average over a full cycle remains 0 V. Why Other Options Are Wrong: 6.37 V is the RMS of a 9 V peak, not relevant. 7.64 V is the average-rectified value for 12 V peak, not the algebraic average. 1.27 V is unrelated. Common Pitfalls: Confusing average-rectified or RMS with algebraic average; overlooking DC offsets. Final Answer: 0 V

Question 12

Peak-to-peak computation: If the peak of a sine wave is 13 V, what is the peak-to-peak voltage (Vpp)?

Accepted Answer

Introduction / Context: Engineers often switch between peak (Vp), peak-to-peak (Vpp), and RMS values. Peak-to-peak measures the total excursion from the most negative to the most positive instant of a sinusoid. This item checks quick conversion from Vp to Vpp. Given Data / Assumptions: Peak voltage Vp = 13 V. Ideal sinusoid centered at 0 V (no DC offset). Concept / Approach: For a symmetric sine wave centered at zero, negative peak is −Vp and positive peak is +Vp. Therefore, Vpp = (+Vp) − (−Vp) = 2 * Vp. Step-by-Step Solution: Use identity: Vpp = 2 * Vp.Compute: Vpp = 2 * 13 V = 26 V.Report result exactly: 26 V. Verification / Alternative check: If RMS were needed, Vrms = Vp/√2 ≈ 9.19 V, and Vpp would still be 2 * 13 V = 26 V. Different amplitude measures serve different purposes, but Vpp depends only on Vp for a centered sine. Why Other Options Are Wrong: 13 V is the peak, not peak-to-peak. 6.5 V would be Vp/2. ‘‘none of the above’’ is incorrect because 26 V is valid. Common Pitfalls: Confusing RMS or average with peak-to-peak; forgetting the factor of 2 when converting. Final Answer: 26 V

Question 13

Phase comparison: Sine wave A has a positive-going zero crossing at 45°, while sine wave B crosses zero positively at 60°. Which statement about their phase relationship is correct?

Accepted Answer

Introduction / Context: Phase lead/lag is central to AC analysis. The positive-going zero crossing is a clear reference marker: the waveform that crosses zero earlier in angle is said to lead the other. This question compares two given crossing angles. Given Data / Assumptions: Wave A positive-going zero crossing at 45°. Wave B positive-going zero crossing at 60°. Both are sinusoidal and at the same frequency. Concept / Approach: If wave A reaches a reference point earlier (smaller angle), A leads B by the difference in their angles. If A reached later (larger angle), it would lag. Here, the angles are measured consistently from the same reference. Step-by-Step Solution: Identify earlier crossing: 45° (A) vs 60° (B).Compute phase difference: Δφ = 60° − 45° = 15°.Since A is earlier, A leads B by 15°. Verification / Alternative check: Visualize rotating phasors: A's phasor angle is 15° ahead of B at any instant for equal frequencies; equivalently, B lags A by 15°. Why Other Options Are Wrong: ‘‘Lags by 15°’’ reverses the relationship. 105° differences are incorrect and exceed the small offset implied by the given crossings. Common Pitfalls: Swapping lead and lag; subtracting in the wrong order; confusing negative-going with positive-going crossings. Final Answer: Wave A leads wave B by 15°.

Question 14

Phasor concept: In AC analysis, a phasor is used to represent which of the following properties of a sinusoidal quantity?

Accepted Answer

Introduction / Context: Phasors are rotating vectors used to represent steady-state sinusoidal signals in the frequency domain. They compress time-domain sinusoids into a complex-number form that captures both magnitude and phase at a single frequency, simplifying circuit analysis with impedances. Given Data / Assumptions: Sinusoidal steady state is assumed (single frequency). We are representing voltage or current as a phasor. No transient or non-sinusoidal content is considered. Concept / Approach: A phasor can be written as M∠φ or as a complex number M(cosφ + j sinφ). It encodes both magnitude M and phase angle φ relative to a reference sinusoid, while the time dependence is suppressed. Step-by-Step Solution: Identify what a phasor contains: amplitude (magnitude) and phase angle.Note that 'width' is not a standard signal descriptor for phasors.Recognize that magnitude-only or phase-only descriptions are incomplete. Verification / Alternative check: Converting back to time domain: v(t) = Re{V∠φ * e^{jωt}} = Vm cos(ωt + φ). The phasor V∠φ determines both Vm and φ, confirming the two key pieces of information. Why Other Options Are Wrong: Phase angle only or magnitude only omit essential information. 'Width' is irrelevant to phasor representation and pertains to pulses, not sinusoids. Common Pitfalls: Confusing phasor magnitude with RMS vs peak (either can be used consistently); treating time-varying transients as phasors (phasors require steady state). Final Answer: the magnitude and the phase angle of a quantity

Question 15

Duty cycle calculation: A pulse waveform has a high time (Ton) of 8 ms and a period (pulse width from start of one pulse to the start of the next) of 32 ms. What is the duty cycle?

Accepted Answer

Introduction / Context: Duty cycle quantifies the fraction of time a digital or pulse waveform stays high during each cycle. It is widely used in PWM control, digital timing, and communications. The duty cycle is expressed as a percentage of the period. Given Data / Assumptions: High (on) time Ton = 8 ms. Period T = 32 ms (time between successive pulse starts). Single, stable repetition rate; ignore rise/fall times. Concept / Approach: Duty cycle D is defined as D = (Ton / T) * 100%. Insert the given values directly to compute the percentage. Step-by-Step Solution: Write formula: D = (Ton / T) * 100%.Substitute: D = (8 ms / 32 ms) * 100%.Compute: D = 0.25 * 100% = 25%. Verification / Alternative check: Sanity check: Since the high time is one quarter of the period, the duty cycle must be 25%. The waveform is high for 8 ms and low for 24 ms each 32 ms cycle. Why Other Options Are Wrong: 50% would require Ton = T/2 = 16 ms. 1% implies Ton ≈ 0.32 ms. 100% means always high (no low time), which is not the case here. Common Pitfalls: Confusing 'pulse width' as Ton; here, the problem clarifies the period is 32 ms. Always confirm definitions in context. Final Answer: 25%

Question 16

Amplitude relationships: A sinusoidal current has an RMS value of 14 mA. What is its peak-to-peak current (Ipp)?

Accepted Answer

Introduction / Context: Converting between RMS, peak, and peak-to-peak values is routine in AC analysis. RMS reflects equivalent DC heating, while peak and peak-to-peak describe waveform extrema. This problem converts a given RMS current to peak-to-peak. Given Data / Assumptions: RMS current Irms = 14 mA. Pure sinusoid with zero DC offset. Concept / Approach: For a sine wave: Ip = Irms * √2, and Ipp = 2 * Ip. Combine to get Ipp = 2 * √2 * Irms. Step-by-Step Solution: Compute peak: Ip = Irms * √2 = 14 mA * 1.4142 ≈ 19.80 mA.Compute peak-to-peak: Ipp = 2 * Ip ≈ 2 * 19.80 mA = 39.60 mA.Round sensibly to two decimals: 39.6 mA. Verification / Alternative check: Reverse check: Irms = Ipp / (2√2) ≈ 39.6 / 2.828 ≈ 14.0 mA, confirming consistency. Why Other Options Are Wrong: 16 mA confuses peak with RMS; 22.6 mA corresponds to √2 * 16 mA, not applicable. 45.12 mA would arise from using √(3) or other incorrect factor. Common Pitfalls: Using √3 or 2 instead of √2; mixing up RMS and average-rectified values. Final Answer: 39.6 mA

Question 17

Average over a half-cycle: For a sine wave with peak value 40 V, what is the average (mean) value over one half-cycle (i.e., the average-rectified value)?

Accepted Answer

Introduction / Context: Two common averages for sinusoids are the full-cycle algebraic average (zero for a centered sine) and the average-rectified value (over a half-cycle, or equivalently the average of the absolute value over a full cycle). Rectifier and power supply calculations often use the half-cycle average. Given Data / Assumptions: Sine wave peak Vp = 40 V. No DC offset; pure sinusoid. We want the average value over a single half-cycle. Concept / Approach: The half-cycle average of a sine is V_avg(half) = (2/π) * Vp ≈ 0.63662 * Vp. This comes from integrating sin(θ) from 0 to π and dividing by π. Step-by-Step Solution: Write formula: V_half_avg = (2/π) * Vp.Substitute: V_half_avg ≈ 0.63662 * 40 V.Compute: ≈ 25.4648 V ≈ 25.48 V. Verification / Alternative check: Compare with RMS: Vrms = Vp/√2 ≈ 28.28 V. It is logical that the RMS exceeds the half-cycle average for a sine. Values are consistent. Why Other Options Are Wrong: 6.37 V corresponds to 10 V peak's half-cycle average. 14.14 V is the RMS of a 20 V peak, not applicable. 50.96 V is 1.274 * Vp, not a standard sine factor. Common Pitfalls: Confusing RMS with average-rectified; forgetting to multiply by 2/π instead of 1/π. Final Answer: 25.48 V

Question 18

AC machine frequency: To generate an 800 Hz sine wave with a four-pole (P = 4) generator, at what rotational speed in revolutions per second (rps) must the rotor run?

Accepted Answer

Introduction / Context: Synchronous generator frequency depends on the number of poles and mechanical speed. This relation is used to design alternators and set turbine speeds. Knowing how to back-calculate speed from a desired frequency is a practical skill. Given Data / Assumptions: Desired electrical frequency f = 800 Hz. Number of poles P = 4. We assume synchronous operation and rps (revolutions per second). Concept / Approach: The standard relation is f = (P/2) * n, where n is speed in rps. Solve for n = f / (P/2). Step-by-Step Solution: Compute pole pairs: P/2 = 4/2 = 2.Solve speed: n = f / (P/2) = 800 / 2 = 400 rps.Thus, the four-pole machine must run at 400 revolutions per second. Verification / Alternative check: In rpm, multiply by 60: 400 rps * 60 = 24,000 rpm, consistent with high-frequency generation requirements. Why Other Options Are Wrong: 200 rps would give 400 Hz with four poles. 800 rps would produce 1,600 Hz. 1,600 rps would produce 3,200 Hz. None match 800 Hz. Common Pitfalls: Using rpm in the formula without converting; forgetting to divide poles by 2 (pole pairs). Final Answer: 400 rps

Question 19

Period from cycles over time: If a sine wave completes 10 cycles in 20 µs (microseconds), what is the period of one cycle?

Accepted Answer

Introduction / Context: Period and frequency calculations often start from counts of cycles over a measured time window. Converting counts to per-cycle period is straightforward division, but unit care (microseconds, milliseconds) matters to avoid scale errors. Given Data / Assumptions: Total time window Δt = 20 µs. Number of cycles N = 10. Uniform sinusoid over the interval (no frequency drift). Concept / Approach: The period T is the time per cycle: T = Δt / N. Insert the given total time and the number of cycles completed. Step-by-Step Solution: Compute: T = 20 µs / 10 = 2 µs per cycle.Report in microseconds: T = 2 µs.Optionally, frequency f = 1 / T = 1 / (2 µs) = 500 kHz (for context). Verification / Alternative check: If each cycle is 2 µs, then 10 cycles take 20 µs, consistent. The reciprocal frequency calculation corroborates the period. Why Other Options Are Wrong: 4 µs would yield only 5 cycles in 20 µs. 20 µs implies one cycle only. 100 µs is unrelated and would correspond to much lower frequency. Common Pitfalls: Dividing the wrong way (N/Δt); mixing µs with ms or s; failing to reduce fraction correctly. Final Answer: 2 µs

Question 20

AC waveform with DC offset: A sinusoid with an rms value of 12 V is superimposed on a steady 18 V DC level. What is the maximum (peak positive) value of the resulting voltage waveform?

Accepted Answer

Introduction / Context: Many practical signals are an AC sine wave riding on a DC offset. To find the instantaneous maximum, you must combine the DC level with the positive peak of the sinusoid. This question checks comfort with rms–peak conversion and how offsets shift waveform extrema in electronics and instrumentation. Given Data / Assumptions: Sine wave rms value: 12 V. DC offset (baseline): 18 V. Ideal sinusoid; standard rms–peak relation for sine applies. Concept / Approach: For a sine wave: V_peak = V_rms * sqrt(2). The waveform with offset has maximum value V_max = V_DC + V_peak and minimum value V_min = V_DC − V_peak. Only the maximum is asked here. Step-by-Step Solution: Compute peak of AC: V_peak = 12 * 1.414 ≈ 16.97 V.Add DC offset to get maximum: V_max = 18 + 16.97 ≈ 34.97 V.Round sensibly to listed option: ≈ 35 V. Verification / Alternative check: Check the minimum for sanity: V_min = 18 − 16.97 ≈ 1.03 V > 0, consistent with a positive-offset waveform never going negative. Why Other Options Are Wrong: 30 V: This ignores part of the AC peak contribution. 6 V or 0 V: Numerically inconsistent; these confuse rms/peak or ignore the DC offset entirely. Common Pitfalls: Using V_rms directly instead of converting to peak. Adding peak-to-peak to DC instead of the single-sided peak. Final Answer: 35 V

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