Question 1

Parallel circuit fundamentals: which statement correctly describes voltage distribution in an ideal parallel network? Assume one source feeding multiple branches.

Accepted Answer

Introduction / Context: In circuit analysis, recognizing how voltage and current divide among parallel branches is vital. While currents split depending on branch impedances, the source voltage appears unchanged across every branch connected directly in parallel to that source node pair. Given Data / Assumptions: Ideal wires and components (no parasitic drops). Multiple branches share the same two nodes of the source. Steady-state DC or AC behavior is considered. Concept / Approach: Kirchhoff’s Voltage Law (KVL) and the definition of parallel connection ensure that each branch spans the same node-to-node potential difference. Consequently, the voltage across every branch equals the source voltage. Currents differ, but voltages are equal in parallel connection. Step-by-Step Solution: 1) Identify two common nodes: top node and bottom node are shared by all branches.2) Voltage across any branch equals V(top node) − V(bottom node), which is the source voltage.3) Therefore, each branch sees the same voltage as the source.4) Select the option stating equality of branch and total voltage. Verification / Alternative check: Use a simple numerical example: a 12 V source feeding two resistors in parallel (6 Ω and 3 Ω). Both resistors measure 12 V across their terminals despite different currents (2 A and 4 A respectively), confirming the rule. Why Other Options Are Wrong: Average or sum of voltages is incorrect; voltages do not add around parallel branches. “Always less than smallest” is nonsensical for ideal parallel networks. Common Pitfalls: Confusing series voltage division with parallel voltage equality; in series, voltages divide according to resistance, but in parallel they remain equal and currents divide instead. Final Answer: The total voltage of a parallel circuit is the same as the voltages across each branch

Question 2

Atomic composition basics: identify the principal constituents of a neutral atom used in electronics and physics contexts Choose the most complete set.

Accepted Answer

Introduction / Context: Atomic structure underpins semiconductor physics, conductor behavior, and chemical properties. Knowing the three primary subatomic particles—electrons, protons, and neutrons—helps link charge distribution, isotopes, and bonding to device operation in electronics. Given Data / Assumptions: Neutral atoms have equal numbers of electrons and protons. Neutrons contribute mass and nuclear stability but no net electric charge. “Ions” refers to atoms or molecules with net charge, not a fundamental third component. Concept / Approach: An atom consists of a nucleus (protons and neutrons) surrounded by electrons. Electrons are negatively charged, protons positively charged, and neutrons neutral. Ionization changes the electron count, forming ions, but does not add a new fundamental particle type beyond the three listed. Step-by-Step Solution: 1) Identify subatomic building blocks: electrons, protons, neutrons.2) Evaluate options: select the set containing these three.3) Reject “ions” as a fundamental constituent—ions are modified atoms, not basic particles.4) Choose electrons, protons, and neutrons. Verification / Alternative check: Physics references consistently describe these three as the canonical components for basic atomic models (ignoring deeper quark structure at this educational level). Why Other Options Are Wrong: “Protons only” omits electrons and neutrons. Sets including ions misunderstand that ions result from electron gain/loss rather than forming a separate intrinsic part of all atoms. Common Pitfalls: Confusing ion terminology (cation, anion) with subatomic particles; also forgetting that isotopes differ by neutron count but remain the same element due to unchanged proton count. Final Answer: Electrons, protons, and neutrons

Question 3

Inductive reactance units: select the correct SI unit for X_L in AC circuit analysis Assume sinusoidal steady-state and standard definitions.

Accepted Answer

Introduction / Context: Inductive reactance quantifies an inductor’s opposition to AC. While inductance (L) itself is measured in henrys (H), the reactance (X_L) is the frequency-dependent “resistance-like” term appearing in Ohm’s law for AC magnitudes and phasor analysis. Using correct units avoids dimensional mistakes when calculating current and impedance. Given Data / Assumptions: Definition: X_L = 2 * pi * f * L. Frequency f is in hertz (s^-1). Inductance L is in henrys (H). Concept / Approach: Unit analysis: multiply s^-1 by H to obtain ohms. Specifically, 1 H = 1 V·s/A. Therefore, (s^-1) * (V·s/A) = V/A = ohm. Hence, X_L has units of ohms, analogous to resistance but with a phase component (reactive, not dissipative). Step-by-Step Solution: 1) Write formula: X_L = 2 * pi * f * L.2) Insert units: f → s^-1; L → V·s/A.3) Multiply: (s^-1) * (V·s/A) = V/A.4) Recognize V/A as the ohm, confirming the unit of X_L. Verification / Alternative check: Impedance Z is reported in ohms whether resistive (R) or reactive (X). Since X_L is a component of impedance, the same unit applies. Why Other Options Are Wrong: “Volts per second” and “amperes per second” are rate units, not impedance. “Henrys” is the unit of inductance L, not of its reactance X_L. Common Pitfalls: Confusing L (henrys) with X_L (ohms) or forgetting that reactance contributes to magnitude and phase but does not dissipate power in ideal inductors. Final Answer: Ohm

Question 4

Practical sources: what internal resistance characteristic does a real (non-ideal) voltage source exhibit? Select the most accurate description for typical modeling.

Accepted Answer

Introduction / Context: Ideal voltage sources maintain fixed voltage regardless of load current. Real sources—batteries, regulators, power supplies—deviate from ideal behavior due to internal resistance and limitations. Modeling this with a small series resistance helps predict voltage sag and short-circuit current. Given Data / Assumptions: Considering a practical voltage source under load. Internal resistance causes voltage drop proportional to current: V_load = V_th − I * R_s. “Small” here means nonzero but low enough to approximate a stiff source for moderate loads. Concept / Approach: Thevenin modeling represents any linear source as an ideal voltage source in series with an internal resistance. For a good voltage source, R_s is small so that load current changes do not strongly affect output voltage. Norton modeling gives an equivalent current source in parallel with a large resistance; both are interconvertible. Step-by-Step Solution: 1) Recognize that zero internal resistance is an idealization, not physically attainable.2) Infinite internal resistance corresponds to an ideal current source, not a voltage source.3) Practical voltage sources exhibit a finite, typically small series resistance.4) Choose “a small internal resistance.” Verification / Alternative check: Measure source under two loads; compute R_s from ΔV / ΔI. High-quality bench supplies show milliohm-level effective R_s, confirming “small.” Why Other Options Are Wrong: Zero is ideal and unattainable. Infinite resistance contradicts the voltage-source model (that would block current). “Large” resistance would produce severe voltage droop, acting more like a poor source. Common Pitfalls: Confusing internal resistance of a source with load resistance or with source output impedance at AC, which may be frequency-dependent. Final Answer: A small internal resistance

Question 5

In semiconductor devices (e.g., a silicon PN junction or each PN junction in a silicon BJT) at room temperature, what is the typical built-in barrier potential across the depletion region?

Accepted Answer

Introduction / Context: The barrier potential (also called built-in potential) is the internal electric potential that forms across the depletion region of a PN junction when P-type and N-type materials are brought together. Knowing its approximate value in common materials such as silicon is essential for understanding diode conduction thresholds, BJT base-emitter voltage, and rectifier behavior in power and signal circuits. Given Data / Assumptions: Material considered: silicon. Condition: room temperature (around 25 °C). Device examples: silicon diode, each PN junction in a silicon BJT (emitter–base and collector–base). Concept / Approach: The built-in potential arises from diffusion of majority carriers and the resulting space charge region that establishes an opposing electric field. Its value depends on temperature, doping levels, and material properties. For practical electronics education and design calculations, standard reference values are used: roughly 0.7 V for silicon and about 0.3 V for germanium at room temperature. This is consistent with the typical forward conduction “knee” seen in silicon diodes and the base-emitter voltage V_BE in BJTs biased in the forward-active region. Step-by-Step Solution: Identify the semiconductor material: silicon.Recall typical barrier potential at room temperature for silicon: approximately 0.7 V.Match this standard value to the closest listed option.Select 0.7 V as the correct answer. Verification / Alternative check: Design handbooks and device datasheets consistently use V_BE ≈ 0.7 V as a rule of thumb for silicon BJTs at moderate currents and room temperature. Similarly, silicon rectifier diodes exhibit about 0.7 V forward drop near their rated current, reflecting the same underlying physics of the PN junction barrier and series effects. Why Other Options Are Wrong: 0 is impossible because a depletion region always has a built-in field in equilibrium. 0.8 V can occur under some doping/temperature conditions but the widely accepted nominal at room temperature is 0.7 V. 20 V is several orders of magnitude too large and pertains to breakdown or external bias, not the built-in potential. “None of the above” is invalid because 0.7 V is correct. Common Pitfalls: Confusing barrier potential with forward drop under load (which includes series resistance and current dependence); forgetting that temperature and current slightly shift V_BE; mixing germanium and silicon values. Final Answer: 0.7 V

Question 6

Load-line analysis in a BJT amplifier: at cutoff (no collector current), where is the Q-point located on the DC load line?

Accepted Answer

Introduction / Context: In transistor biasing, the quiescent operating point (Q-point) determines linearity and headroom. The DC load line graphically relates the collector current and collector-emitter voltage for a given collector resistor and supply. Understanding where the Q-point sits in cutoff and saturation helps with quick reasoning about amplifier states and troubleshooting bias problems. Given Data / Assumptions: Common-emitter configuration with supply voltage V_CC and collector resistor R_C. Cutoff means base current is essentially zero and collector current I_C ≈ 0. The DC load line extends from (I_C = 0, V_CE = V_CC) to (I_C = V_CC / R_C, V_CE = 0). Concept / Approach: Cutoff corresponds to no conduction through the transistor. If I_C ≈ 0, then the voltage drop across R_C is I_C * R_C ≈ 0 and the collector-emitter voltage rises to nearly V_CC. On the I_C–V_CE axes, this is the bottom-right or “lower-end” point at I_C = 0 and V_CE = V_CC on the load line. Step-by-Step Solution: Recognize cutoff: base drive absent → I_C ≈ 0.Compute V_CE at I_C = 0: V_CE ≈ V_CC (maximum).Locate on load line: point at the I_C = 0 intercept.Identify this as the lower end of the plotted line if current is the vertical axis. Verification / Alternative check: Oscilloscope or curve tracer plots show that as bias is reduced, the operating point slides toward I_C = 0 along the same load line; measured V_CE approaches V_CC, confirming the cutoff location. Why Other Options Are Wrong: Upper end corresponds to saturation (I_C maximum, V_CE near 0). Middle is quiescent for symmetrical swing but not cutoff. “Infinity” has no meaning on the finite load line. “None” is wrong since the lower end is correct. Common Pitfalls: Reversing axis interpretations; forgetting that terms “upper”/“lower” depend on convention—here “lower end” refers to the end with zero I_C. Final Answer: lower end of the load line

Question 7

Magnetic materials comparison: which class of materials exhibits the highest magnetic permeability under normal conditions?

Accepted Answer

Introduction / Context: Permeability measures how readily a material supports the formation of a magnetic field within itself. Selecting core materials for transformers, inductors, motors, and magnetic shields requires a clear grasp of which material types offer the greatest permeability and therefore strongest magnetic response to an applied magnetizing force. Given Data / Assumptions: Comparing typical classes: diamagnetic, paramagnetic, and ferromagnetic materials. Normal operating conditions; no extreme temperature or fields forcing phase transitions. Relative permeability μ_r indicates response: greater than 1 shows attraction, less than 1 shows weak repulsion. Concept / Approach: Diamagnetic materials have μ_r slightly less than 1 and are weakly repelled by magnetic fields (e.g., bismuth). Paramagnetic materials have μ_r slightly greater than 1 (e.g., aluminum), showing weak attraction. Ferromagnetic materials (e.g., iron, nickel, cobalt, and their alloys) possess domains that align cooperatively, creating very large relative permeabilities—often in the hundreds to thousands or more—until saturation is reached. Thus, ferromagnets produce the strongest flux for a given magnetizing force H. Step-by-Step Solution: Identify what “highest permeability” means: largest μ_r.Recall typical μ_r: diamagnetic < 1, paramagnetic slightly > 1, ferromagnetic ≫ 1.Conclude that ferromagnetic materials exhibit the highest permeability.Select “A ferromagnetic material.” Verification / Alternative check: Core material datasheets (e.g., silicon steel, ferrites) list μ_r values from hundreds to tens of thousands, far exceeding μ_r of paramagnetic materials (~1.x) and diamagnetic materials (~0.999…); practical transformer and inductor designs rely on this contrast. Why Other Options Are Wrong: Diamagnetic materials have the lowest response (μ_r slightly less than 1). Paramagnetic are only mildly above 1. “All of the above” cannot be right because the magnitudes differ greatly; only ferromagnetic truly exhibits the highest permeability. Common Pitfalls: Confusing permeability with saturation magnetization or coercivity; assuming all metals are equally magnetic; ignoring that ferromagnetic μ_r depends on frequency, flux level, and material processing. Final Answer: A ferromagnetic material

Question 8

Finding Norton current (I_N): to determine the Norton equivalent current at a pair of terminals, what should you do with the external load connected to those terminals?

Accepted Answer

Introduction / Context: Norton’s theorem represents a linear network by a current source I_N in parallel with a resistance R_N as seen from two terminals. Determining I_N correctly is fundamental for simplifying networks and predicting how any load will behave when attached to those terminals. Given Data / Assumptions: We consider a linear circuit with independent/dependent sources and resistors. The load initially connects to the terminals of interest. The goal is to measure or calculate Norton current I_N. Concept / Approach: By definition, I_N equals the short-circuit current at the output terminals (the current that would flow if those terminals were shorted together). Therefore, to find I_N, you remove the actual external load and replace it with a short across the output terminals, then compute the resulting current through that short. Afterward, R_N can be found by deactivating sources or via R_N = V_th / I_sc if you also know V_th (Thevenin voltage). Step-by-Step Solution: Disconnect the load so it does not influence the measurement.Short the output terminals together.Compute the current through the short; this is I_sc.Set I_N = I_sc as the Norton current. Verification / Alternative check: Use duality with Thevenin: if V_th is known, then R_N = V_th / I_sc. Simulated or measured results using a circuit tool confirm that the short-circuit current equals the current source value in the Norton model. Why Other Options Are Wrong: Opening the load gives open-circuit voltage V_th, not I_N. Shorting or opening sources (C, D) applies to finding R_N by deactivation, not to finding I_N directly. “None” is wrong because shorting the terminals is exactly correct. Common Pitfalls: Forgetting dependent sources remain active during R_N determination via test sources; confusing open-circuit voltage with short-circuit current. Final Answer: Short the load resistor

Question 9

Frequency behavior of a capacitor: how does capacitive reactance (Xc) vary as the operating frequency increases, assuming capacitance remains constant?

Accepted Answer

Introduction / Context: Understanding how a capacitor opposes AC is fundamental for filter design, coupling/decoupling, and impedance matching. The parameter that captures this opposition is capacitive reactance, denoted Xc, which depends on both capacitance and frequency. Designers rely on the frequency dependence of Xc to shape signal spectra and stabilize power rails. Given Data / Assumptions: Ideal capacitor with fixed capacitance C. Sinusoidal steady-state operation at frequency f. We examine magnitude relationships only. Concept / Approach: The magnitude of capacitive reactance is Xc = 1 / (2 * pi * f * C). As f increases, the denominator increases; therefore Xc decreases. At DC (f = 0), Xc tends toward infinity, meaning an ideal capacitor blocks steady current. At very high frequencies, Xc can become very small, allowing significant AC current for a given voltage amplitude (subject to real-world ESL/ESR limits). Step-by-Step Solution: Write the formula: Xc = 1 / (2 * pi * f * C).Observe inverse proportionality to frequency.Conclude: increasing f causes decreasing Xc.Select option stating “decreases as frequency increases.” Verification / Alternative check: Compute a simple example: with C = 1 µF, at 50 Hz Xc ≈ 3183 Ω; at 1 kHz Xc ≈ 159 Ω; at 10 kHz Xc ≈ 15.9 Ω, confirming the inverse relationship. Why Other Options Are Wrong: A misstates the DC behavior (Xc is maximum at DC). C denies the clear mathematical dependence. D is the opposite trend. “None” is invalid because the relationship is well known. Common Pitfalls: Ignoring nonideal ESR/ESL that dominate at very high frequencies; confusing reactance with resistance; forgetting phase shift of −90 degrees in ideal capacitors. Final Answer: Capacitive reactance decreases as the operating frequency increases

Question 10

Transformer current relation: a step-down transformer has a turns ratio of 5:1 (primary:secondary). If the secondary current is 1 A rms, what is the primary current (assume ideal behavior)?

Accepted Answer

Introduction / Context: Transformers exchange voltage and current according to their turns ratio. In an ideal transformer, power is conserved (ignoring losses), so an increase in current on one side corresponds to a decrease on the other when voltage is stepped down or up. This question reinforces the current relationship given a known turns ratio and measured secondary current. Given Data / Assumptions: Turns ratio N_p : N_s = 5 : 1 (step-down). Secondary current I_s = 1 A rms. Ideal transformer (no losses): V_p * I_p ≈ V_s * I_s. Concept / Approach: For an ideal transformer, voltage ratio equals turns ratio: V_p / V_s = N_p / N_s. Current ratio is the inverse: I_p / I_s = N_s / N_p. Therefore I_p = I_s * (N_s / N_p). With N_p/N_s = 5, N_s/N_p = 1/5, so the primary current is one-fifth of the secondary current in this step-down case. Step-by-Step Solution: Identify ratios: N_p/N_s = 5/1.Use current relation: I_p = I_s * (N_s / N_p).Compute: I_p = 1 A * (1 / 5) = 0.2 A.Select 0.2 A. Verification / Alternative check: Voltage is stepped down by 5, current stepped up by 5 on the secondary (relative to primary). Since I_s is larger, the primary must carry less current: 0.2 A is consistent with power balance (V down by 5, I up by 5, P roughly constant). Why Other Options Are Wrong: 0.3 A, 0.4 A, and 0.6 A do not follow the exact inverse ratio of 5:1 and would violate ideal power conservation given the stated secondary current. “None” is incorrect because 0.2 A is correct. Common Pitfalls: Confusing which side is primary vs secondary; using the same ratio for both voltage and current; ignoring that real transformers have losses but the ideal relation is still the design baseline. Final Answer: 0.2 A

Question 11

Basic power calculation: how much current flows through a 60 W incandescent lamp operating from a 120 V supply (assume purely resistive behavior)?

Accepted Answer

Introduction / Context: In household electrical calculations, determining current from known power and voltage is a foundational skill for sizing switches, fuses, and conductors. Incandescent lamps behave approximately as resistive loads at operating temperature, so the simple power law applies directly. Given Data / Assumptions: Rated power P = 60 W. Supply voltage V = 120 V. Assume resistive load so P = V * I. Concept / Approach: For a resistive load, average power equals the product of RMS voltage and RMS current: P = V * I. Rearranging gives I = P / V. This relation is widely used for quick estimates in residential and commercial electrical design and safety checks. Step-by-Step Solution: Start from P = V * I.Solve for current: I = P / V.Substitute values: I = 60 / 120 = 0.5 A.Select 0.5 A. Verification / Alternative check: Compute the approximate resistance: R = V^2 / P = 120^2 / 60 = 240 Ω. Using I = V / R = 120 / 240 = 0.5 A confirms the same result, reinforcing consistency between formulas. Why Other Options Are Wrong: 180 A and 7200 A are wildly unrealistic for a household lamp and would imply catastrophic faults. 2 A would correspond to a 240 W load at 120 V, not a 60 W bulb. “None” is wrong because 0.5 A is correct. Common Pitfalls: Confusing peak with RMS quantities; ignoring that real incandescent resistance varies with filament temperature (but nameplate power is rated at operating conditions, so the calculation remains valid). Final Answer: 0.5 A

Question 12

Magnetic circuit analogy: which property of a magnetic circuit is directly analogous to voltage in an electrical circuit?

Accepted Answer

Introduction / Context: Magnetic circuits are often taught using an analogy to electrical circuits to make design intuition easier. In this analogy, one must map each magnetic quantity to its electrical counterpart to reason about cores, air gaps, and excitation requirements. Knowing which parameter corresponds to “voltage” is key to understanding how coils drive flux. Given Data / Assumptions: Magnetic circuits are treated using lumped-parameter models. The analogy compares voltage, current, and resistance to their magnetic counterparts. We seek the magnetic quantity that “pushes” flux through a magnetic path. Concept / Approach: In the analogy: magnetomotive force (MMF, usually NI where N = turns and I = current) corresponds to voltage; magnetic flux Φ corresponds to current; reluctance ℜ corresponds to resistance. Flux density B relates to flux per unit area and is not directly the “driving force.” MMF is the cause, flux is the effect, and reluctance is the opposition—precisely mirroring Ohm’s law as Φ = MMF / ℜ in magnetic form. Step-by-Step Solution: Recall definitions: MMF drives magnetic flux through a magnetic path.Apply the analogy: Voltage ↔ MMF, Current ↔ Flux, Resistance ↔ Reluctance.Identify the required counterpart of voltage: MMF.Select “Magnetomotive force.” Verification / Alternative check: Using Φ = MMF / ℜ, increasing NI (more turns or current) raises flux for a given reluctance, just as raising voltage increases current for a given resistance per Ohm’s law I = V / R. Why Other Options Are Wrong: Flux density is analogous to current density, not voltage. “Number of lines of force” loosely relates to total flux, not the driving force. Reluctance is the magnetic analog of resistance, not voltage. “None” is wrong because MMF is the correct counterpart. Common Pitfalls: Confusing B (flux density) with H (magnetizing field); forgetting that the analogy is useful but has limits when materials saturate or at high frequencies where losses become dominant. Final Answer: Magnetomotive force

Question 13

Primary vs secondary cells: what is the primary difference between primary-cell batteries and secondary-cell batteries in practical use?

Accepted Answer

Introduction / Context: Battery selection impacts device cost, maintenance, and environmental footprint. A key decision is whether a device needs disposable cells for occasional, low-drain use or rechargeable cells for frequent cycles. Understanding the fundamental distinction between primary and secondary cells helps match the chemistry to the application. Given Data / Assumptions: Primary cells (e.g., alkaline, zinc-carbon) are generally intended for single-use discharge. Secondary cells (e.g., Li-ion, NiMH, lead-acid) are designed for repeated charge/discharge cycles. We compare practical recharge capability and typical use patterns. Concept / Approach: The electrochemical reactions in primary cells are not readily reversible under practical, safe charging conditions; attempting to recharge can be unsafe or damaging. Secondary cells are engineered for reversibility: electrode materials, electrolytes, and separators are chosen to support cycling with manageable degradation. Consequently, secondary batteries are suited to high-cycle applications, whereas primary batteries excel in shelf life and low self-discharge for intermittent use. Step-by-Step Solution: Define the distinguishing property: rechargeability.Recognize that secondary cells are designed to be recharged multiple times.Note that primary cells generally cannot be properly or safely recharged.Select the option that states this difference explicitly. Verification / Alternative check: Manufacturer datasheets list cycle life for secondary cells and explicitly warn against recharging primary cells. Real-world examples: smartphones use Li-ion (secondary), TV remotes often use alkaline (primary) due to long shelf life and convenience. Why Other Options Are Wrong: A reverses the truth. C mischaracterizes secondary cells as non-chemical. D is not a defining difference; charge rates vary within chemistries. “None” is incorrect because a clear statement exists in option B. Common Pitfalls: Confusing low self-discharge with rechargeability; assuming “rechargeable alkalines” are typical (they are specialized and uncommon compared to mainstream secondary chemistries). Final Answer: Secondary cells can be recharged and used many times, whereas primary cells cannot be properly recharged

Question 14

Ideal transformer behavior: if a power transformer’s secondary voltage is twice the applied primary voltage, how does the secondary current compare with the primary current (assume ideal conditions)?

Accepted Answer

Introduction / Context: Transformers trade voltage for current while approximately conserving power in the ideal case. When the secondary voltage is raised relative to the primary, the available current on the secondary drops proportionally. This principle underlies distribution networks and power supplies where different voltage levels are required without changing power significantly (aside from losses). Given Data / Assumptions: V_s = 2 * V_p (secondary voltage is twice primary voltage). Ideal transformer: P_in ≈ P_out (ignoring losses). Current relation follows the inverse of the voltage ratio. Concept / Approach: For an ideal transformer, V_s / V_p = N_s / N_p and I_s / I_p = N_p / N_s. If V_s is twice V_p, then N_s is twice N_p, so I_s must be half I_p to keep apparent power roughly the same (S ≈ V * I on each side). Thus stepping voltage up by 2 steps current down by 2. Step-by-Step Solution: Given: V_s / V_p = 2 → N_s / N_p = 2.Current ratio: I_s / I_p = N_p / N_s = 1 / 2.Therefore I_s = 0.5 * I_p.Select “One-half the primary current.” Verification / Alternative check: Check power balance: If primary has 100 V and 2 A (200 VA), secondary would have ~200 V and 1 A (200 VA). This validates the inverse relationship between voltage and current in the ideal model. Why Other Options Are Wrong: “Same” or “twice” contradicts the inverse proportionality. “No more than 10% less” is arbitrary and incorrect in the ideal case. “None” is invalid because a correct statement is provided. Common Pitfalls: Forgetting that real transformers have copper and core losses leading to small deviations; mixing up which side is stepped up; ignoring that phase and power factor also matter for AC power calculations. Final Answer: One-half the primary current

Question 15

In basic circuit theory, which statement about the equivalent resistance of a parallel network is correct? Select the one that best describes how total resistance compares with the individual branch resistances in a parallel circuit.

Accepted Answer

Introduction / Context: Understanding how resistors combine is a core skill in electrical engineering and electronics. In parallel circuits, multiple resistive paths connect across the same two nodes. The question asks you to identify the correct relationship between the total (equivalent) resistance and the individual branch resistances in such a network. Given Data / Assumptions: A resistive network with two or more branches in parallel. All components are ideal resistors. Direct current or the resistive part of an AC circuit is considered (no reactive elements). Concept / Approach: For resistors in parallel, conductances (the reciprocals of resistances) add. If R_eq is the equivalent resistance, then 1 / R_eq = 1 / R1 + 1 / R2 + ... + 1 / Rn. Because each term on the right is positive, the sum of reciprocals is greater than the reciprocal of any single branch. Consequently, R_eq is less than the smallest branch resistance. This is the intuitive result of adding extra current paths that reduce overall opposition to current. Step-by-Step Solution: Write the parallel formula: 1 / R_eq = Σ (1 / Ri).Observe that Σ (1 / Ri) > 1 / R_min, where R_min is the smallest Ri.Take reciprocals: R_eq < R_min.Therefore, the equivalent resistance is always less than the smallest branch resistance. Verification / Alternative check: Consider two equal resistors R in parallel: 1 / R_eq = 1 / R + 1 / R = 2 / R ⇒ R_eq = R / 2, clearly less than R. Adding more parallel branches continues to decrease R_eq. Why Other Options Are Wrong: Equal to the sum: that rule applies to series, not parallel.Equal to the total voltage: resistance is not equal to voltage; they are different quantities.Equal to the average: there is no averaging rule for parallel resistance; the value is always below the smallest branch.None of the above: incorrect because the correct statement is provided. Common Pitfalls: Confusing series and parallel rules, mixing up units (ohms vs volts), or forgetting that adding parallel branches decreases total resistance. Final Answer: The total resistance of a parallel circuit is always less than the value of the smallest resistance

Question 16

Inductive reactance: choose the statement that most accurately describes its nature in practical circuits (assume sinusoidal steady-state conditions unless stated otherwise).

Accepted Answer

Introduction / Context: Reactance is a frequency-dependent opposition to current flow caused by energy storage in inductors and capacitors. For inductors, this opposition is called inductive reactance and it grows with frequency. Distinguishing inductive reactance from related concepts like back emf helps clarify AC circuit behavior. Given Data / Assumptions: Sinusoidal steady-state AC unless noted otherwise. Ideal inductor with reactance X_L = 2 * π * f * L. Direct current (dc) corresponds to f = 0 Hz. Concept / Approach: Inductive reactance X_L quantifies how an inductor resists changes in current. Since X_L = 2 * π * f * L, it is zero at dc (f = 0) and increases linearly with frequency. This means inductors impede AC but act as short circuits in ideal dc conditions (ignoring winding resistance). Step-by-Step Solution: Recall definition: X_L = 2 * π * f * L (ohms).For dc: f = 0 ⇒ X_L = 0, so no reactive opposition occurs.For ac: f > 0 ⇒ X_L > 0, providing frequency-dependent opposition.Therefore, the correct description is that inductive reactance opposes current flow in AC circuits. Verification / Alternative check: Impedance of an inductor is jX_L. The magnitude increases with f, consistent with reduced current for a given AC voltage (I = V / X_L). Time-domain view: the induced emf opposes changes in current, manifesting as reactance in the frequency domain. Why Other Options Are Wrong: Opposition in dc circuits: incorrect; ideal inductive reactance is zero at dc.Another term for counter emf: back emf is a phenomenon; reactance is a scalar magnitude in ohms.A force overcoming counter emf: not a meaningful electrical definition.None of the above: incorrect since (b) is accurate. Common Pitfalls: Equating physical back emf with the numerical reactance; forgetting the frequency dependence; ignoring winding resistance that exists in real inductors. Final Answer: Inductive reactance is an opposition to current flow in ac circuits

Question 17

Series inductor networks: identify the correct rule for finding the total inductance when multiple inductors are connected in series (assume weak coupling and ideal components).

Accepted Answer

Introduction / Context: Combining inductors is a frequent task in filter design, power electronics, and RF circuits. For series connections, engineers need a reliable rule that does not depend on frequency. The question probes your understanding of inductance addition in series networks. Given Data / Assumptions: Inductors are connected in series. Mutual coupling is negligible (spaced apart or oriented to minimize coupling). Inductors are linear and ideal over the operating current range. Concept / Approach: Inductance is a property independent of frequency for an ideal inductor. In series, stored magnetic energies add, and the equivalent inductance is the algebraic sum: L_eq = L1 + L2 + ... + Ln. This contrasts with reactance, which is frequency-dependent: X_L = 2 * π * f * L. You do not add reactances to find total inductance; you add the inductance values directly. Step-by-Step Solution: Write the series rule: L_eq = Σ Li.Note that this holds for any frequency because L is a component parameter.Conclude that the correct statement is that total inductance equals the sum of individual inductances. Verification / Alternative check: Energy viewpoint: W = 1/2 * L * I^2. For series currents equal in each inductor, energies add, implying L_eq is the sum of Li. Why Other Options Are Wrong: Sum of inductive reactances: mixes frequency-dependent reactance with inductance; not the requested quantity.Less than the smallest: characteristic of parallel, not series.Source voltage divided by total current: that yields impedance at a given frequency, not inductance directly. Common Pitfalls: Confusing inductance with reactance; overlooking coupling which can add or subtract via mutual inductance terms. Final Answer: equal to the sum of the individual inductance values

Question 18

Dielectric constant (relative permittivity): what property of a material does it quantify in the context of capacitors and electrostatics?

Accepted Answer

Introduction / Context: Capacitors store energy in an electric field. The dielectric constant (relative permittivity, ε_r) of the material between plates directly determines how much electric field is supported for a given voltage and how much charge can be stored per volt. Selecting materials with appropriate ε_r is central to capacitor design and high-frequency electronics. Given Data / Assumptions: Linear, isotropic dielectric in standard capacitor applications. We interpret “dielectric constant” as relative permittivity, ε_r = ε / ε_0. Breakdown strength is a different specification from ε_r. Concept / Approach: The dielectric constant measures how a material polarizes in response to an electric field, effectively concentrating electric flux lines and increasing capacitance. For a parallel-plate capacitor, C = ε * A / d, with ε = ε_r * ε_0. Larger ε_r means more capacitance for the same geometry. It does not directly quantify breakdown voltage or magnetic behavior and is not a resistance-like opposition to current (that is resistivity). Step-by-Step Solution: Relate capacitance to permittivity: C ∝ ε.Recognize ε_r captures a material’s ability to support electric field via polarization.Thus, ε_r quantifies how well the dielectric concentrates electrostatic lines of force. Verification / Alternative check: Compare vacuum (ε_r = 1) to ceramics (ε_r > 1000 in some types): ceramics produce much larger capacitance for the same dimensions, confirming the role of ε_r. Why Other Options Are Wrong: Breakdown strength: described by dielectric strength (V/m), not ε_r.Opposition to current flow: that is resistance or impedance, not dielectric constant.Magnetic lines: property of magnetic permeability, not permittivity.None of the above: incorrect because option (b) matches the definition. Common Pitfalls: Confusing permittivity with dielectric strength; mixing electric and magnetic material properties; assuming high ε_r guarantees low loss (loss tangent is a separate parameter). Final Answer: how well the material concentrates electrostatic lines of force

Question 19

Parallel RL network phase angle: A 100 Ω resistor is connected in parallel with an inductor whose reactance magnitude is 200 Ω at the operating frequency. What is the phase shift between the total current and the applied voltage?

Accepted Answer

Introduction / Context: In a parallel RL circuit, the resistive branch current is in phase with the voltage, while the inductive branch current lags the voltage by 90 degrees. The total current is the phasor sum of these two components. The question asks for the overall phase angle between the total current and the applied voltage. Given Data / Assumptions: Resistor R = 100 Ω. Inductive reactance magnitude X_L = 200 Ω. Ideal components; sinusoidal steady state. Voltage is the reference phasor. Concept / Approach: Use admittance to combine parallel branches: Y_total = G + jB. For a resistor, G = 1 / R. For an inductor, susceptance B_L = −1 / X_L (negative because current lags). The total current I_total = V * Y_total; the angle of I_total relative to V is φ = arctan(B / G). A negative φ means the current lags the voltage. Step-by-Step Solution: Compute conductance: G = 1 / R = 1 / 100 = 0.01 S.Compute susceptance: B = −1 / X_L = −1 / 200 = −0.005 S.Find angle: φ = arctan(B / G) = arctan(−0.005 / 0.01) = arctan(−0.5).Evaluate: arctan(−0.5) ≈ −26.6 degrees ⇒ current lags voltage by 26.6 degrees. Verification / Alternative check: Phasor diagram: the in-phase component equals I_R = V / R; the quadrature lagging component equals I_L = V / X_L. The tangent of the lag angle is I_L / I_R = (V / X_L) / (V / R) = R / X_L = 100 / 200 = 0.5, matching arctan(0.5) = 26.6 degrees. Why Other Options Are Wrong: 0 degree: only for purely resistive or perfectly balanced susceptance to zero.90 degree or 180 degree: would require pure reactance or phase inversion; not applicable here.None of the above: incorrect because 26.6 degree is correct. Common Pitfalls: Using impedance instead of admittance for parallel combination; forgetting the negative sign for inductive susceptance; mixing degrees and radians. Final Answer: 26.6 degree

Question 20

Sine-wave characteristics: which option correctly defines the peak value of a sinusoidal waveform in electrical engineering?

Accepted Answer

Introduction / Context: Signal specifications such as peak, peak-to-peak, and RMS values are ubiquitous in power systems and electronics. Misinterpreting these can lead to incorrect component ratings and analysis errors. The question focuses on the correct definition of the peak value of a sinusoid. Given Data / Assumptions: The waveform is a pure sine wave. Standard engineering definitions apply (peak, peak-to-peak, RMS). Amplitude refers to instantaneous magnitude from the zero reference. Concept / Approach: The peak value (sometimes called the amplitude for symmetric sinusoids) is the largest instantaneous magnitude measured from zero. For a sine v(t) = V_peak * sin(ωt), the maximum is +V_peak and the minimum is −V_peak. RMS is V_peak / √2 ≈ 0.707 * V_peak, not 70.7% of “maximum” in the sense used by option (a). Step-by-Step Solution: Define peak value: the maximum magnitude reached by the waveform.Relate to RMS: V_RMS = V_peak / √2.Relate to peak-to-peak: V_pp = 2 * V_peak.Therefore, peak is simply the maximum positive or negative value reached. Verification / Alternative check: Consider v(t) = 10 sin(ωt): peak = 10, RMS ≈ 7.07, peak-to-peak = 20. These relationships highlight the distinctions among common measures. Why Other Options Are Wrong: 70.7% of maximum: that statement describes RMS relative to peak, not the peak definition.90% of maximum: arbitrary and incorrect.1.41 times the maximum: 1.414 relates peak to RMS, but direction is reversed (peak = 1.414 * RMS).None of the above: incorrect because option (d) is correct. Common Pitfalls: Confusing RMS, average, and peak values; mixing peak and peak-to-peak ratings on datasheets. Final Answer: the maximum positive or negative value

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