Q: Geometric effect on capacitance: when the plate area of a capacitor increases (all else unchanged), what happens to its capacitance?

Introduction / Context: Capacitance depends on geometry and dielectric properties. Designers frequently adjust plate area and spacing to tune capacitance values for filters, timing circuits, and energy storage. Recognizing the direct proportionality between plate area and capacitance is fundamental. Given Data / Assumptions: Parallel-plate capacitor model applies. Plate separation and dielectric constant remain fixed. Only the effective plate area A is increased. Concept / Approach: The ideal relation is C = ε * A / d, where ε is the permittivity of the dielectric, A is plate area, and d is separation. With ε and d constant, capacitance scales linearly with A: larger plate area yields larger capacitance. Step-by-Step Solution: Begin with C = ε * A / d.Hold ε and d constant; vary only A.As A increases, C increases proportionally. Verification / Alternative check: Practical capacitors: doubling electrode area approximately doubles capacitance for the same dielectric and thickness. This is consistent with data sheets where larger packages (greater effective area) exhibit higher capacitance for identical dielectrics and thicknesses. Why Other Options Are Wrong: The capacitance decreases / unaffected: Oppose the C ∝ A relation. The voltage it can withstand increases: Breakdown voltage depends primarily on dielectric strength and thickness (d), not area. Common Pitfalls: Confusing breakdown voltage with capacitance; they are governed by different parameters. Ignoring fringing and construction details; however, first-order behavior remains C ∝ A. Final Answer: the capacitance increases

Q: Series capacitors charged from 24 V: a 2 µF capacitor has 16 V across it and the other series capacitor has 8 V. What is the value of the unknown capacitor?

Introduction / Context: In series, capacitors share the same charge magnitude, and voltages divide inversely with capacitance. This is routinely used in high-voltage stacks and sensor circuits. Here we find an unknown capacitance from measured voltage division across series caps. Given Data / Assumptions: Two series capacitors across 24 V. C1 = 2 µF has V1 = 16 V. Unknown C2 has V2 = 8 V. Ideal capacitors; same charge magnitude on each. Concept / Approach: For capacitors in series: charge Q is equal on both. Voltage V = Q / C, so the voltage ratio equals the inverse capacitance ratio. Therefore, V1 / V2 = C2 / C1, which can be solved for C2. Step-by-Step Solution: Use V1 / V2 = C2 / C1.Insert numbers: 16 / 8 = C2 / 2.Compute: 2 = C2 / 2 ⇒ C2 = 4 µF.Check sum: V1 + V2 = 16 + 8 = 24 V (consistent). Verification / Alternative check: Compute equivalent C_total: 1/C_total = 1/2 + 1/4 = 0.5 + 0.25 = 0.75 ⇒ C_total = 1.333 µF. Series charge Q = C_total * V_total = 1.333e-6 * 24 = 32 µC. Then V1 = Q/C1 = 32e-6 / 2e-6 = 16 V; V2 = 32e-6 / 4e-6 = 8 V, confirming the solution. Why Other Options Are Wrong: 1 µF or 2 µF: Would not yield the given 16 V/8 V split. 8 µF: Would give V1/V2 = C2/C1 = 8/2 = 4, implying V1 = 4 * V2, not 2:1 as given. Common Pitfalls: Using resistor-style division (direct proportionality) instead of inverse proportionality for capacitors in series. Forgetting that series capacitors carry equal charge. Final Answer: 4 µF

Q: RC charging at t = 0+: an uncharged capacitor and a 1 kΩ resistor are placed in series with a switch and a 6 V battery. Immediately after closing the switch, what is the voltage across the capacitor?

Introduction / Context: Transient analysis of RC circuits starts with initial conditions. Knowing the instantaneous voltage across a capacitor at the moment a switch closes is a foundational concept for timing circuits, debouncing, and analog signal steps. Given Data / Assumptions: An initially uncharged capacitor (Vc(0^-) = 0 V). A series resistor of 1 kΩ and an ideal 6 V DC source. Switch closes at t = 0, no prior pre-charge. Concept / Approach: A capacitor’s voltage cannot change instantaneously; it changes according to the exponential charging law Vc(t) = Vs * (1 - e^(-t/RC)). At the instant just after closing (t = 0+), Vc(0+) equals the initial value Vc(0^-) which is 0 V for an uncharged capacitor. Step-by-Step Solution: State initial condition: Vc(0^-) = 0 V.Apply continuity: Vc(0^+) = Vc(0^-).Therefore, immediately after closing, Vc(0+) = 0 V. Verification / Alternative check: Use the charging equation: Vc(t) = 6 * (1 - e^(-t/(R*C))). At t = 0, e^0 = 1, so Vc(0) = 6 * (1 - 1) = 0 V. Current initially is I(0+) = 6 / R = 6 mA, then decays as the capacitor charges. Why Other Options Are Wrong: 6 V: That is the long-term value as t → ∞, not at t = 0+. 3 V or 2 V: Arbitrary fractions; the initial value is determined by continuity, not averaging. Depends on capacitance value: The initial voltage is zero regardless of C; C changes the time constant, not the initial condition. Common Pitfalls: Assuming immediate charge to source voltage; real capacitors cannot change voltage instantaneously. Confusing initial current (which can be large) with initial capacitor voltage. Final Answer: 0 V

Q: Two equal capacitors in series across a 10 kHz sine source: with two 0.68 µF capacitors in series, what is the total capacitive reactance?

Introduction / Context: Capacitive reactance determines how a capacitor impedes AC. When capacitors are placed in series, their equivalent capacitance drops. This problem connects series-capacitance calculation to reactance at a specified frequency, common in filter and coupling design. Given Data / Assumptions: Two capacitors, each C = 0.68 µF, in series. Frequency f = 10 kHz. Ideal components; no ESR or parasitics. Concept / Approach: For equal capacitors in series, the equivalent capacitance is C_eq = C/2. Capacitive reactance is Xc = 1 / (2 * π * f * C_eq). Compute C_eq first, then apply the reactance formula. Step-by-Step Solution: Compute equivalent: C_eq = 0.68 µF / 2 = 0.34 µF = 0.34 * 10^-6 F.Compute denominator: 2 * π * f * C_eq = 2 * π * 10,000 * 0.34e-6.Numerical step: 10,000 * 0.34e-6 = 0.0034; multiply by 2π ≈ 6.283 → 0.0034 * 6.283 ≈ 0.02136.Reactance: Xc = 1 / 0.02136 ≈ 46.8 Ω. Verification / Alternative check: Check scale: A few tenths of microfarads at 10 kHz gives tens of ohms reactance—reasonable. If you incorrectly used a single 0.68 µF value, you would get half the correct reactance, which flags a series-capacitance mistake. Why Other Options Are Wrong: 23.41 Ω or 11.70 Ω: These correspond to using 0.68 µF (not the series equivalent) or additional division errors. 4.68 Ω: Off by a factor of 10; likely a frequency or unit slip. Common Pitfalls: Forgetting that two equal capacitors in series halve the capacitance. Unit conversion errors (µF to F) leading to decade mistakes. Final Answer: 46.8 Ω

Q: In an AC circuit, a sine-wave voltage is applied across a capacitor. When the frequency of the applied voltage is decreased, what happens to the capacitor current?

Introduction / Context: Capacitors in alternating-current (AC) circuits pass current that depends on both the capacitance and the frequency of the applied sine-wave voltage. This question tests understanding of how frequency affects capacitive reactance and, in turn, the current through a capacitor. Given Data / Assumptions: Sine-wave voltage source across a capacitor. Frequency is decreased from some initial value. Capacitance is constant; voltage amplitude is unchanged. Concept / Approach: Capacitive reactance is Xc = 1 / (2 * pi * f * C). Current magnitude in a pure capacitive branch is I = V / Xc. Using reactance, I can also be written I = 2 * pi * f * C * V. Thus current is directly proportional to frequency when V and C are constant. Step-by-Step Solution: Xc = 1 / (2 * pi * f * C)If f decreases, then Xc increases.I = V / XcAs Xc increases, I must decrease for fixed V.Equivalently, I = 2 * pi * f * C * V shows I decreases when f decreases. Verification / Alternative check: Consider two frequencies f1 and f2 with f2 = f1 / 2. Then Xc doubles, so I halves. This confirms the inverse relationship between current and reactance (and direct proportionality to frequency). Why Other Options Are Wrong: increases: False, because decreasing f reduces I. remains constant: False; I depends on f. ceases: False unless f approaches 0 Hz (true DC), where steady-state current is zero, but not for a mere decrease. Common Pitfalls: Confusing capacitors with resistors (whose current does not depend on frequency) or inductors (which have opposite frequency behavior for reactance). Another mistake is thinking voltage alone sets current without considering reactance variation with frequency. Final Answer: decreases

Q: Basic capacitor relationship: Given electric charge Q = 60 µC on the plates and voltage V = 12 V across a capacitor, compute the capacitance value.

Introduction / Context: This problem checks the fundamental capacitor equation linking charge (Q), voltage (V), and capacitance (C). Such conversions are common in electronics when sizing capacitors for filters or timing networks. Given Data / Assumptions: Charge Q = 60 µC. Voltage V = 12 V. Ideal capacitor; no leakage or series resistance considered. Concept / Approach: The basic relationship is Q = C * V. Solving for C gives C = Q / V. Keep units consistent by converting microcoulombs to coulombs where needed: 1 µC = 10^-6 C. Step-by-Step Solution: Q = 60 µC = 60 * 10^-6 CV = 12 VC = Q / VC = (60 * 10^-6) / 12C = 5 * 10^-6 F = 5 µF Verification / Alternative check: Reverse the calculation: Q' = C * V = 5 µF * 12 V = 60 µC, which matches the given charge. This confirms the computed capacitance is correct. Why Other Options Are Wrong: 720 µF: Exaggerated by a factor of 144; does not satisfy Q = C * V. 50 µF: Ten times too large; would imply Q = 600 µC at 12 V. 12 µF: More than double the correct value; would imply Q = 144 µC. Common Pitfalls: Forgetting to convert microcoulombs to coulombs, or misplacing decimal points during division, commonly leads to errors of 10x or 100x. Always keep track of powers of ten when working with micro-units. Final Answer: 5 µF

Q: Unit conversion in electronics: Convert a capacitance of 0.01 µF into an equivalent value in picofarads (pF).

Introduction / Context: Electronics often requires quick conversion among farad sub-multiples: microfarads (µF), nanofarads (nF), and picofarads (pF). This question checks accurate unit conversion across orders of magnitude. Given Data / Assumptions: Capacitance C = 0.01 µF. Find the equivalent in picofarads. Concept / Approach: Use standard SI relations: 1 µF = 10^-6 F, 1 nF = 10^-9 F, 1 pF = 10^-12 F. Also, 1 µF = 1,000 nF and 1 nF = 1,000 pF, so 1 µF = 1,000,000 pF. Step-by-Step Solution: C = 0.01 µFConvert µF to pF directly: 1 µF = 1,000,000 pFTherefore, 0.01 µF = 0.01 * 1,000,000 pF= 10,000 pF Verification / Alternative check: Convert via nanofarads first: 0.01 µF = 10 nF, and 10 nF = 10 * 1,000 pF = 10,000 pF. Both approaches yield the same result. Why Other Options Are Wrong: 100 pF and 1,000 pF: Undershoot by factors of 100 and 10, respectively. 100,000 pF: Overshoot by a factor of 10. Common Pitfalls: Mixing up nF and pF steps or forgetting that 1 µF equals one million pF. Writing 0.01 µF as 100 pF is a typical decimal-place error. Final Answer: 10,000 pF

Q: Four identical capacitors, each of 0.15 µF, are connected in parallel. Determine the equivalent (total) capacitance of the parallel network.

Introduction / Context: Parallel and series combinations alter the effective capacitance of a network. In parallel, plate areas effectively add, increasing total capacitance. This is a fundamental concept in designing filters and decoupling networks. Given Data / Assumptions: Four capacitors, each 0.15 µF. All connected in parallel. Ideal components assumed. Concept / Approach: For capacitors in parallel: C_total = C1 + C2 + C3 + C4. The voltage across all branches is the same, and charges add, making total capacitance the arithmetic sum. Step-by-Step Solution: C_total = 0.15 µF + 0.15 µF + 0.15 µF + 0.15 µFC_total = 4 * 0.15 µFC_total = 0.60 µF Verification / Alternative check: Group two at a time: two in parallel give 0.30 µF; two such groups in parallel add to 0.30 µF + 0.30 µF = 0.60 µF. Same result confirms correctness. Why Other Options Are Wrong: 0.15 µF: That would be a single capacitor, not four in parallel. 0.30 µF: Equals only two in parallel. 0.8 µF: Exceeds the sum; not achievable here. Common Pitfalls: Confusing series and parallel rules; in series, capacitances combine via reciprocals and result reduces, while in parallel the value increases by addition. Final Answer: 0.6 µF

Q: Time constant calculation: A 220 Ω resistor is in series with a 2.2 µF capacitor. Compute the RC time constant and choose the closest option.

Introduction / Context: The RC time constant quantifies how quickly voltages across a resistor-capacitor network rise or fall during charging or discharging. It is essential for timing circuits and filters. Given Data / Assumptions: R = 220 Ω. C = 2.2 µF. Series RC network; ideal components. Concept / Approach: The time constant is tau = R * C. Use SI units: ohms for resistance and farads for capacitance. Convert microfarads to farads before multiplying. Step-by-Step Solution: C = 2.2 µF = 2.2 * 10^-6 Ftau = R * C = 220 * 2.2 * 10^-6tau = 484 * 10^-6 s = 484 µsClosest listed value: 480 µs Verification / Alternative check: Order-of-magnitude check: 200 Ω * 2 µF ≈ 400 µs, so 220 Ω * 2.2 µF should be slightly higher, about 480 µs. This corroborates the calculation. Why Other Options Are Wrong: 48 µs and 24 µs: Too small by factors of about 10 and 20. 2.42 µs: Off by two orders of magnitude due to unit or decimal error. Common Pitfalls: Forgetting to convert µF to F or misplacing the decimal is the most common source of error. Also, do not confuse milliseconds with microseconds. Final Answer: 480 µs

Question 1

Series combination of capacitors: given 2 µF, 4 µF, and 10 µF in series, the total capacitance is less than which value?

Accepted Answer

Introduction / Context: Capacitors in series behave differently from resistors: the total capacitance decreases and is always less than the smallest individual capacitor. Knowing this qualitative rule helps catch design and exam errors quickly, even before calculating exact values. Given Data / Assumptions: Three capacitors: 2 µF, 4 µF, and 10 µF connected in series. Ideal capacitors; no leakage or parasitics considered. Concept / Approach: The series capacitance relation is 1/C_total = 1/C1 + 1/C2 + 1/C3. Since each term is positive, 1/C_total > 1/C_smallest, which implies C_total < C_smallest. Thus the total must be less than 2 µF, the smallest of the three. Step-by-Step Solution: Identify smallest capacitor: 2 µF.Use inequality insight: series combination is less than the smallest.Therefore, C_total < 2 µF. Verification / Alternative check: If desired, compute: 1/C_total = 1/2 + 1/4 + 1/10 = 0.5 + 0.25 + 0.1 = 0.85 µF^-1, so C_total ≈ 1 / 0.85 ≈ 1.176 µF, which is indeed less than 2 µF. Why Other Options Are Wrong: 4 µF, 10 µF: These are larger than the smallest value; the total cannot exceed the smallest in a series network. 1.5 µF: This is a candidate numeric value but the prompt asks “less than which value?” The correct qualitative bound is the smallest capacitor, 2 µF. Common Pitfalls: Applying resistor intuition (adding values in series) to capacitors; capacitors in series do the opposite. Arithmetic mistakes with reciprocals; the inequality method is a fast sanity check. Final Answer: 2 µF

Question 2

Geometric effect on capacitance: when the plate area of a capacitor increases (all else unchanged), what happens to its capacitance?

Accepted Answer

Introduction / Context: Capacitance depends on geometry and dielectric properties. Designers frequently adjust plate area and spacing to tune capacitance values for filters, timing circuits, and energy storage. Recognizing the direct proportionality between plate area and capacitance is fundamental. Given Data / Assumptions: Parallel-plate capacitor model applies. Plate separation and dielectric constant remain fixed. Only the effective plate area A is increased. Concept / Approach: The ideal relation is C = ε * A / d, where ε is the permittivity of the dielectric, A is plate area, and d is separation. With ε and d constant, capacitance scales linearly with A: larger plate area yields larger capacitance. Step-by-Step Solution: Begin with C = ε * A / d.Hold ε and d constant; vary only A.As A increases, C increases proportionally. Verification / Alternative check: Practical capacitors: doubling electrode area approximately doubles capacitance for the same dielectric and thickness. This is consistent with data sheets where larger packages (greater effective area) exhibit higher capacitance for identical dielectrics and thicknesses. Why Other Options Are Wrong: The capacitance decreases / unaffected: Oppose the C ∝ A relation. The voltage it can withstand increases: Breakdown voltage depends primarily on dielectric strength and thickness (d), not area. Common Pitfalls: Confusing breakdown voltage with capacitance; they are governed by different parameters. Ignoring fringing and construction details; however, first-order behavior remains C ∝ A. Final Answer: the capacitance increases

Question 3

Series capacitors charged from 24 V: a 2 µF capacitor has 16 V across it and the other series capacitor has 8 V. What is the value of the unknown capacitor?

Accepted Answer

Introduction / Context: In series, capacitors share the same charge magnitude, and voltages divide inversely with capacitance. This is routinely used in high-voltage stacks and sensor circuits. Here we find an unknown capacitance from measured voltage division across series caps. Given Data / Assumptions: Two series capacitors across 24 V. C1 = 2 µF has V1 = 16 V. Unknown C2 has V2 = 8 V. Ideal capacitors; same charge magnitude on each. Concept / Approach: For capacitors in series: charge Q is equal on both. Voltage V = Q / C, so the voltage ratio equals the inverse capacitance ratio. Therefore, V1 / V2 = C2 / C1, which can be solved for C2. Step-by-Step Solution: Use V1 / V2 = C2 / C1.Insert numbers: 16 / 8 = C2 / 2.Compute: 2 = C2 / 2 ⇒ C2 = 4 µF.Check sum: V1 + V2 = 16 + 8 = 24 V (consistent). Verification / Alternative check: Compute equivalent C_total: 1/C_total = 1/2 + 1/4 = 0.5 + 0.25 = 0.75 ⇒ C_total = 1.333 µF. Series charge Q = C_total * V_total = 1.333e-6 * 24 = 32 µC. Then V1 = Q/C1 = 32e-6 / 2e-6 = 16 V; V2 = 32e-6 / 4e-6 = 8 V, confirming the solution. Why Other Options Are Wrong: 1 µF or 2 µF: Would not yield the given 16 V/8 V split. 8 µF: Would give V1/V2 = C2/C1 = 8/2 = 4, implying V1 = 4 * V2, not 2:1 as given. Common Pitfalls: Using resistor-style division (direct proportionality) instead of inverse proportionality for capacitors in series. Forgetting that series capacitors carry equal charge. Final Answer: 4 µF

Question 4

RC charging at t = 0+: an uncharged capacitor and a 1 kΩ resistor are placed in series with a switch and a 6 V battery. Immediately after closing the switch, what is the voltage across the capacitor?

Accepted Answer

Introduction / Context: Transient analysis of RC circuits starts with initial conditions. Knowing the instantaneous voltage across a capacitor at the moment a switch closes is a foundational concept for timing circuits, debouncing, and analog signal steps. Given Data / Assumptions: An initially uncharged capacitor (Vc(0^-) = 0 V). A series resistor of 1 kΩ and an ideal 6 V DC source. Switch closes at t = 0, no prior pre-charge. Concept / Approach: A capacitor’s voltage cannot change instantaneously; it changes according to the exponential charging law Vc(t) = Vs * (1 - e^(-t/RC)). At the instant just after closing (t = 0+), Vc(0+) equals the initial value Vc(0^-) which is 0 V for an uncharged capacitor. Step-by-Step Solution: State initial condition: Vc(0^-) = 0 V.Apply continuity: Vc(0^+) = Vc(0^-).Therefore, immediately after closing, Vc(0+) = 0 V. Verification / Alternative check: Use the charging equation: Vc(t) = 6 * (1 - e^(-t/(R*C))). At t = 0, e^0 = 1, so Vc(0) = 6 * (1 - 1) = 0 V. Current initially is I(0+) = 6 / R = 6 mA, then decays as the capacitor charges. Why Other Options Are Wrong: 6 V: That is the long-term value as t → ∞, not at t = 0+. 3 V or 2 V: Arbitrary fractions; the initial value is determined by continuity, not averaging. Depends on capacitance value: The initial voltage is zero regardless of C; C changes the time constant, not the initial condition. Common Pitfalls: Assuming immediate charge to source voltage; real capacitors cannot change voltage instantaneously. Confusing initial current (which can be large) with initial capacitor voltage. Final Answer: 0 V

Question 5

Two equal capacitors in series across a 10 kHz sine source: with two 0.68 µF capacitors in series, what is the total capacitive reactance?

Accepted Answer

Introduction / Context: Capacitive reactance determines how a capacitor impedes AC. When capacitors are placed in series, their equivalent capacitance drops. This problem connects series-capacitance calculation to reactance at a specified frequency, common in filter and coupling design. Given Data / Assumptions: Two capacitors, each C = 0.68 µF, in series. Frequency f = 10 kHz. Ideal components; no ESR or parasitics. Concept / Approach: For equal capacitors in series, the equivalent capacitance is C_eq = C/2. Capacitive reactance is Xc = 1 / (2 * π * f * C_eq). Compute C_eq first, then apply the reactance formula. Step-by-Step Solution: Compute equivalent: C_eq = 0.68 µF / 2 = 0.34 µF = 0.34 * 10^-6 F.Compute denominator: 2 * π * f * C_eq = 2 * π * 10,000 * 0.34e-6.Numerical step: 10,000 * 0.34e-6 = 0.0034; multiply by 2π ≈ 6.283 → 0.0034 * 6.283 ≈ 0.02136.Reactance: Xc = 1 / 0.02136 ≈ 46.8 Ω. Verification / Alternative check: Check scale: A few tenths of microfarads at 10 kHz gives tens of ohms reactance—reasonable. If you incorrectly used a single 0.68 µF value, you would get half the correct reactance, which flags a series-capacitance mistake. Why Other Options Are Wrong: 23.41 Ω or 11.70 Ω: These correspond to using 0.68 µF (not the series equivalent) or additional division errors. 4.68 Ω: Off by a factor of 10; likely a frequency or unit slip. Common Pitfalls: Forgetting that two equal capacitors in series halve the capacitance. Unit conversion errors (µF to F) leading to decade mistakes. Final Answer: 46.8 Ω

Question 6

In an AC circuit, a sine-wave voltage is applied across a capacitor. When the frequency of the applied voltage is decreased, what happens to the capacitor current?

Accepted Answer

Introduction / Context: Capacitors in alternating-current (AC) circuits pass current that depends on both the capacitance and the frequency of the applied sine-wave voltage. This question tests understanding of how frequency affects capacitive reactance and, in turn, the current through a capacitor. Given Data / Assumptions: Sine-wave voltage source across a capacitor. Frequency is decreased from some initial value. Capacitance is constant; voltage amplitude is unchanged. Concept / Approach: Capacitive reactance is Xc = 1 / (2 * pi * f * C). Current magnitude in a pure capacitive branch is I = V / Xc. Using reactance, I can also be written I = 2 * pi * f * C * V. Thus current is directly proportional to frequency when V and C are constant. Step-by-Step Solution: Xc = 1 / (2 * pi * f * C)If f decreases, then Xc increases.I = V / XcAs Xc increases, I must decrease for fixed V.Equivalently, I = 2 * pi * f * C * V shows I decreases when f decreases. Verification / Alternative check: Consider two frequencies f1 and f2 with f2 = f1 / 2. Then Xc doubles, so I halves. This confirms the inverse relationship between current and reactance (and direct proportionality to frequency). Why Other Options Are Wrong: increases: False, because decreasing f reduces I. remains constant: False; I depends on f. ceases: False unless f approaches 0 Hz (true DC), where steady-state current is zero, but not for a mere decrease. Common Pitfalls: Confusing capacitors with resistors (whose current does not depend on frequency) or inductors (which have opposite frequency behavior for reactance). Another mistake is thinking voltage alone sets current without considering reactance variation with frequency. Final Answer: decreases

Question 7

Basic capacitor relationship: Given electric charge Q = 60 µC on the plates and voltage V = 12 V across a capacitor, compute the capacitance value.

Accepted Answer

Introduction / Context: This problem checks the fundamental capacitor equation linking charge (Q), voltage (V), and capacitance (C). Such conversions are common in electronics when sizing capacitors for filters or timing networks. Given Data / Assumptions: Charge Q = 60 µC. Voltage V = 12 V. Ideal capacitor; no leakage or series resistance considered. Concept / Approach: The basic relationship is Q = C * V. Solving for C gives C = Q / V. Keep units consistent by converting microcoulombs to coulombs where needed: 1 µC = 10^-6 C. Step-by-Step Solution: Q = 60 µC = 60 * 10^-6 CV = 12 VC = Q / VC = (60 * 10^-6) / 12C = 5 * 10^-6 F = 5 µF Verification / Alternative check: Reverse the calculation: Q' = C * V = 5 µF * 12 V = 60 µC, which matches the given charge. This confirms the computed capacitance is correct. Why Other Options Are Wrong: 720 µF: Exaggerated by a factor of 144; does not satisfy Q = C * V. 50 µF: Ten times too large; would imply Q = 600 µC at 12 V. 12 µF: More than double the correct value; would imply Q = 144 µC. Common Pitfalls: Forgetting to convert microcoulombs to coulombs, or misplacing decimal points during division, commonly leads to errors of 10x or 100x. Always keep track of powers of ten when working with micro-units. Final Answer: 5 µF

Question 8

Unit conversion in electronics: Convert a capacitance of 0.01 µF into an equivalent value in picofarads (pF).

Accepted Answer

Introduction / Context: Electronics often requires quick conversion among farad sub-multiples: microfarads (µF), nanofarads (nF), and picofarads (pF). This question checks accurate unit conversion across orders of magnitude. Given Data / Assumptions: Capacitance C = 0.01 µF. Find the equivalent in picofarads. Concept / Approach: Use standard SI relations: 1 µF = 10^-6 F, 1 nF = 10^-9 F, 1 pF = 10^-12 F. Also, 1 µF = 1,000 nF and 1 nF = 1,000 pF, so 1 µF = 1,000,000 pF. Step-by-Step Solution: C = 0.01 µFConvert µF to pF directly: 1 µF = 1,000,000 pFTherefore, 0.01 µF = 0.01 * 1,000,000 pF= 10,000 pF Verification / Alternative check: Convert via nanofarads first: 0.01 µF = 10 nF, and 10 nF = 10 * 1,000 pF = 10,000 pF. Both approaches yield the same result. Why Other Options Are Wrong: 100 pF and 1,000 pF: Undershoot by factors of 100 and 10, respectively. 100,000 pF: Overshoot by a factor of 10. Common Pitfalls: Mixing up nF and pF steps or forgetting that 1 µF equals one million pF. Writing 0.01 µF as 100 pF is a typical decimal-place error. Final Answer: 10,000 pF

Question 9

Four identical capacitors, each of 0.15 µF, are connected in parallel. Determine the equivalent (total) capacitance of the parallel network.

Accepted Answer

Introduction / Context: Parallel and series combinations alter the effective capacitance of a network. In parallel, plate areas effectively add, increasing total capacitance. This is a fundamental concept in designing filters and decoupling networks. Given Data / Assumptions: Four capacitors, each 0.15 µF. All connected in parallel. Ideal components assumed. Concept / Approach: For capacitors in parallel: C_total = C1 + C2 + C3 + C4. The voltage across all branches is the same, and charges add, making total capacitance the arithmetic sum. Step-by-Step Solution: C_total = 0.15 µF + 0.15 µF + 0.15 µF + 0.15 µFC_total = 4 * 0.15 µFC_total = 0.60 µF Verification / Alternative check: Group two at a time: two in parallel give 0.30 µF; two such groups in parallel add to 0.30 µF + 0.30 µF = 0.60 µF. Same result confirms correctness. Why Other Options Are Wrong: 0.15 µF: That would be a single capacitor, not four in parallel. 0.30 µF: Equals only two in parallel. 0.8 µF: Exceeds the sum; not achievable here. Common Pitfalls: Confusing series and parallel rules; in series, capacitances combine via reciprocals and result reduces, while in parallel the value increases by addition. Final Answer: 0.6 µF

Question 10

Time constant calculation: A 220 Ω resistor is in series with a 2.2 µF capacitor. Compute the RC time constant and choose the closest option.

Accepted Answer

Introduction / Context: The RC time constant quantifies how quickly voltages across a resistor-capacitor network rise or fall during charging or discharging. It is essential for timing circuits and filters. Given Data / Assumptions: R = 220 Ω. C = 2.2 µF. Series RC network; ideal components. Concept / Approach: The time constant is tau = R * C. Use SI units: ohms for resistance and farads for capacitance. Convert microfarads to farads before multiplying. Step-by-Step Solution: C = 2.2 µF = 2.2 * 10^-6 Ftau = R * C = 220 * 2.2 * 10^-6tau = 484 * 10^-6 s = 484 µsClosest listed value: 480 µs Verification / Alternative check: Order-of-magnitude check: 200 Ω * 2 µF ≈ 400 µs, so 220 Ω * 2.2 µF should be slightly higher, about 480 µs. This corroborates the calculation. Why Other Options Are Wrong: 48 µs and 24 µs: Too small by factors of about 10 and 20. 2.42 µs: Off by two orders of magnitude due to unit or decimal error. Common Pitfalls: Forgetting to convert µF to F or misplacing the decimal is the most common source of error. Also, do not confuse milliseconds with microseconds. Final Answer: 480 µs

Question 11

Capacitance unit conversion: Convert a capacitance of 0.00022 F into microfarads (µF).

Accepted Answer

Introduction / Context: Converting between farads and sub-multiples (µF, nF, pF) is routine in circuit design. This question assesses precision with scientific notation and unit scaling. Given Data / Assumptions: C = 0.00022 F. Find the equivalent in microfarads. Concept / Approach: 1 µF = 10^-6 F, hence to convert farads to microfarads, divide by 10^-6 (or multiply by 10^6). Step-by-Step Solution: C(µF) = C(F) * 10^6C(µF) = 0.00022 * 10^6C(µF) = 220 µF Verification / Alternative check: Back-convert: 220 µF = 220 * 10^-6 F = 0.00022 F, matching the original value. Why Other Options Are Wrong: 22 µF: 10x too small. 2,200 µF: 10x too large. 22,200 µF: 100x too large. Common Pitfalls: Misreading decimal places or confusing milli (10^-3) with micro (10^-6) leads to power-of-ten mistakes. Always track exponents explicitly. Final Answer: 220 µF

Question 12

Dielectric comparison: Among air, mica, glass, and paper, identify the capacitor dielectric with the highest typical dielectric constant.

Accepted Answer

Introduction / Context: Dielectric materials determine a capacitor’s capacitance for a given geometry. A higher dielectric constant (relative permittivity) allows more charge storage at the same voltage, enabling physically smaller capacitors for the same value. Given Data / Assumptions: Dielectrics listed: air, mica, glass, paper. Compare typical relative permittivity ranges used in general-purpose capacitors. Concept / Approach: Relative permittivity (epsilon_r) indicates how much the dielectric increases capacitance compared to vacuum. Typical values: air ~ 1, paper ~ 2–3, mica ~ 5–7, many glasses ~ 5–10 (and specialized glasses even higher). Step-by-Step Solution: Identify approximate ranges: air ≈ 1paper ≈ 2–3mica ≈ 5–7glass ≈ 5–10 (or higher depending on composition)Highest among given standard choices is glass. Verification / Alternative check: Reference component data sheets: general-purpose glass dielectrics often outpace mica and paper in epsilon_r, explaining compact high-value glass capacitors and feedthrough designs. Why Other Options Are Wrong: air: Lowest epsilon_r (~1), used mainly where minimal dielectric loading is desired. mica: Stable and low-loss, but typical epsilon_r below common glasses. paper: Lower epsilon_r than mica and glass. Common Pitfalls: Confusing dielectric constant with breakdown voltage or loss tangent. A higher dielectric constant does not automatically mean better for all applications; losses and stability also matter. Final Answer: glass

Question 13

RC charging time: A 4.7 MΩ resistor is in series with a 0.015 µF capacitor across a 12 V source. Estimate how long it takes for the capacitor to be effectively fully charged (use the standard 5τ rule).

Accepted Answer

Introduction / Context: Capacitors in RC networks charge exponentially. Engineers often use the 5τ guideline (five time constants) to approximate the time for a capacitor to be essentially fully charged (over 99%). Given Data / Assumptions: R = 4.7 MΩ. C = 0.015 µF. Series RC driven by a 12 V DC source. Use t_full ≈ 5 * tau. Concept / Approach: Time constant tau = R * C. Convert units carefully before multiplying: megaohms to ohms, microfarads to farads. Then multiply tau by 5 to approximate full charge time. Step-by-Step Solution: R = 4.7 MΩ = 4.7 * 10^6 ΩC = 0.015 µF = 15 * 10^-9 Ftau = R * C = 4.7 * 10^6 * 15 * 10^-9tau = 70.5 * 10^-3 s = 70.5 mst_full ≈ 5 * tau = 5 * 70.5 ms = 352.5 msClosest option is 352 ms. Verification / Alternative check: At t = 5τ, the capacitor reaches about 99.3% of the final voltage, which is the standard engineering rule-of-thumb for “fully” charged in practice. Why Other Options Are Wrong: 70.5 ms: This is only 1τ, not near full charge. 35 ms: Less than 1τ; underestimates by a factor of about 10. 3.5 s: Overestimates by an order of magnitude. Common Pitfalls: Using 3τ instead of 5τ when the question explicitly asks for fully charged, or mis-converting µF and MΩ. Always convert to base units before multiplying. Final Answer: 352 ms

Question 14

Capacitive reactance and AC magnitude: A 12 kHz sinusoidal voltage is applied to a 0.33 µF capacitor, and the measured rms current is 200 mA. Determine the rms voltage magnitude across the capacitor.

Accepted Answer

Introduction / Context: In AC analysis, the magnitude of a capacitor’s impedance is given by its reactance Xc, which depends on frequency and capacitance. With known current and reactance, the voltage magnitude follows Ohm’s law for AC magnitudes: V = I * Xc. Given Data / Assumptions: f = 12 kHz. C = 0.33 µF. Irms = 200 mA = 0.200 A. Pure capacitive branch; ignore series resistance. Concept / Approach: Capacitive reactance: Xc = 1 / (2 * pi * f * C). Then Vrms = Irms * Xc. Keep units consistent (Hz, farads, amperes). Step-by-Step Solution: C = 0.33 µF = 0.33 * 10^-6 FXc = 1 / (2 * pi * 12,000 * 0.33 * 10^-6)Xc ≈ 40.19 Ω (approximate)Vrms = Irms * Xc = 0.200 * 40.19 ≈ 8.04 VRounded to the nearest option: 8 V Verification / Alternative check: Proportional reasoning: A larger capacitance or frequency would reduce Xc and therefore reduce Vrms for a fixed current. The computed value is consistent with typical magnitudes for these parameters. Why Other Options Are Wrong: 80 V: Ten times too large; would require Xc ≈ 400 Ω at 0.2 A. 800 mV and 80 mV: Too small by factors of 10 and 100 relative to the computed 8 V. Common Pitfalls: Forgetting to convert microfarads to farads or miscalculating 2 * pi * f * C. Also, do not confuse peak and rms values; the problem is clearly rms-based. Final Answer: 8 V

Question 15

Parallel addition of capacitance: Four capacitors of 10 µF, 20 µF, 22 µF, and 100 µF are connected in parallel. Find the total capacitance of the combination.

Accepted Answer

Introduction / Context: Capacitors in parallel add directly, a rule exploited to realize non-standard values or to increase total capacitance for energy storage and filtering applications. Given Data / Assumptions: Four capacitors: 10 µF, 20 µF, 22 µF, 100 µF. All are connected in parallel. Ideal components assumed. Concept / Approach: For parallel capacitors, the equivalent capacitance is the sum: C_total = sum(Ci). The voltage is common, and the individual stored charges add up, yielding a larger net capacitance. Step-by-Step Solution: C_total = 10 + 20 + 22 + 100 µFC_total = 152 µF Verification / Alternative check: Group addition: (10 + 20) = 30 µF; (22 + 100) = 122 µF; then 30 + 122 = 152 µF. Same result confirms correctness. Why Other Options Are Wrong: 2.43 µF and 4.86 µF: Far too small; perhaps confuse with series rules. 100 µF: Ignores the additional capacitors in parallel. Common Pitfalls: Applying series formulas to parallel networks or overlooking unit consistency. In series, capacitance decreases; in parallel, it increases by direct addition. Final Answer: 152 µF

Question 16

Capacitor types in electronics: which of the following capacitor constructions is polarized (has a required +/− orientation)?

Accepted Answer

Introduction / Context: Polarized capacitors must be connected with the correct polarity because their dielectric structure relies on a formed oxide layer that fails if reverse-biased. Knowing which capacitor families are polarized is essential for safe PCB design and power-supply filtering. Given Data / Assumptions: Options include common capacitor types used in electronics. We assume ideal, general-purpose parts (no specialized non-polar electrolytics). Question asks which construction is inherently polarized. Concept / Approach: Electrolytic capacitors (aluminum or tantalum) are normally polarized because the dielectric is a thin oxide grown on the anode foil; reverse voltage can damage or destroy them. By contrast, mica, ceramic, and plastic-film capacitors use solid dielectrics that are normally non-polarized and can be connected either way in DC circuits. Step-by-Step Solution: Identify capacitor families that require polarity: electrolytic.Confirm others: mica, ceramic, plastic-film are non-polarized in standard forms.Conclude that only electrolytic is polarized among the listed options. Verification / Alternative check: Datasheets show polarity stripe and maximum reverse voltage ~0 V for standard electrolytics; non-polar electrolytics exist but are explicitly labeled as such and are not implied here. Why Other Options Are Wrong: Mica: stable, non-polar dielectric. Ceramic: Class 1 and Class 2 types are non-polarized. Plastic-film: polypropylene, polyester, etc., are non-polarized. None of the above: incorrect because electrolytic is polarized. Common Pitfalls: Assuming all capacitors can be reversed; electrolytics usually cannot. Confusing specialized non-polar electrolytics with standard polarized parts. Final Answer: electrolytic

Question 17

Capacitor charge–voltage relation: when the voltage across a capacitor is tripled, how does the stored charge change?

Accepted Answer

Introduction / Context: The fundamental relationship for capacitors links charge, capacitance, and voltage. This is central to energy storage, ADC sample-and-hold design, and timing circuits. Understanding proportional scaling avoids sizing and stress errors in practice. Given Data / Assumptions: Capacitance C is constant (same physical device). Voltage V across the capacitor is increased by a factor of 3. Ideal capacitor (no leakage or dielectric absorption considered). Concept / Approach: The governing relation is Q = C * V, where Q is stored charge. If C is fixed and V is multiplied by 3, the stored charge scales by the same factor. Energy (E = 0.5 * C * V^2) would scale by the square of the factor, but the question asks specifically about charge, not energy. Step-by-Step Solution: Start with Q = C * V.Let V′ = 3 * V with C unchanged.Then Q′ = C * V′ = C * (3 * V) = 3 * (C * V) = 3Q. Verification / Alternative check: Pick numbers: let C = 1 µF, V = 2 V → Q = 2 µC. Tripling V to 6 V → Q = 6 µC, exactly 3× larger. Why Other Options Are Wrong: Is cut to one-third / doubles / quadruples: incorrect scaling for Q = C * V. Stays the same: only true if V or C do not change; V is tripled here. Common Pitfalls: Confusing charge scaling (linear with V) with energy scaling (quadratic with V). Assuming non-ideal effects change this first-order proportionality. Final Answer: triples

Question 18

Series RC driven by a sine source: initially XC = R so VR = VC. If the frequency is increased, how do the resistor and capacitor voltages compare?

Accepted Answer

Introduction / Context: Voltage division in AC circuits depends on impedances, not just resistances. In a series RC, the capacitor’s reactance varies with frequency, changing the voltage distribution. This concept underlies filters, phase shifters, and sensor conditioning. Given Data / Assumptions: A series RC circuit driven by a sine wave. At the starting frequency, XC = R, so VR = VC. Frequency is then increased; R and C values are unchanged. Concept / Approach: Capacitive reactance is XC = 1 / (2 * π * f * C). As frequency f increases, XC decreases. In a series network, the magnitude of voltage across a component is proportional to its impedance magnitude. Therefore, with higher f, the capacitor’s impedance shrinks and takes a smaller share of the source voltage, while the resistor takes a larger share. Step-by-Step Solution: Start: XC = R ⇒ |VR| = |VC|.Increase frequency: XC decreases (inverse to f).Voltage division shifts so |VR| > |VC|. Verification / Alternative check: Numerical example with Vs = 1 V, R = 1 kΩ, C chosen so XC = 1 kΩ at f0. At 2f0, XC halves to 500 Ω. Voltage division yields VR ≈ 0.894 V, VC ≈ 0.447 V, confirming VR > VC. Why Other Options Are Wrong: VC > VR: Opposite of frequency effect on XC. VR = VC: Only true when XC = R. VR and VC = 0: Not in a driven sinusoidal circuit. Common Pitfalls: Treating XC as constant with frequency. Ignoring that phase angles change but here only magnitudes are compared. Final Answer: VR > VC

Question 19

RC charging time constant: in a standard first-order RC charging circuit, the time to reach practical full charge is approximately which multiple of RC?

Accepted Answer

Introduction / Context: The time constant τ = R * C sets the speed of exponential charging in RC networks. Engineers often use a practical benchmark for “full” charge to estimate delays, settling times, and sampling windows in analog systems. Given Data / Assumptions: Standard RC charging from a DC source through a resistor. No significant leakage or series resistance beyond R. “Full charge” means about 99% of final value (common engineering convention). Concept / Approach: Capacitor voltage during charging is Vc(t) = Vs * (1 - e^(-t/RC)). At t = 5RC, e^(-5) ≈ 0.0067, so Vc ≈ 99.33% of Vs. This is widely accepted as the time to essentially reach the final value for practical purposes. Step-by-Step Solution: Write charging law: Vc(t) = Vs * (1 - e^(-t/RC)).Choose t = 5RC → Vc/Vs = 1 - e^(-5).Compute e^(-5) ≈ 0.0067 → Vc ≈ 0.9933 * Vs (≈ 99.3%). Verification / Alternative check: At 4RC, Vc ≈ 98.2% Vs; at 6RC, ≈ 99.75% Vs. Many references standardize on 5RC as a convenient “practical full charge.” Why Other Options Are Wrong: RC: Only ~63.2% charged. 6 RC: Also near full, but the conventional benchmark used in design notes is 5RC. None of the above: Incorrect because 5RC is the conventional answer. Common Pitfalls: Confusing time constant (63.2%) with practical full charge (~99%). Assuming exact “100%” is achievable in finite time; exponential never truly reaches it. Final Answer: 5 RC

Question 20

Compute capacitive reactance: a 0.47 µF capacitor across a 2 kHz sine-wave source—what is Xc?

Accepted Answer

Introduction / Context: Capacitive reactance indicates how a capacitor impedes AC at a given frequency. It is critical for sizing coupling capacitors, setting filter cutoffs, and calculating impedance in signal paths. Given Data / Assumptions: Capacitance C = 0.47 µF = 0.47 × 10^-6 F. Frequency f = 2 kHz. Ideal capacitor (neglect ESR). Concept / Approach: The formula for capacitive reactance is Xc = 1 / (2 * π * f * C). Plug in values carefully, keeping units consistent, to avoid power-of-ten mistakes. Step-by-Step Solution: Compute the product: 2 * π * f * C = 2 * π * 2000 * 0.47e-6.2000 * 0.47e-6 = 0.00094; multiply by 2π ≈ 6.283 → ≈ 0.005907.Xc = 1 / 0.005907 ≈ 169.3 Ω ≈ 170 Ω. Verification / Alternative check: Sanity check: At audio kHz and sub-microfarad capacitance, Xc should be on the order of hundreds of ohms; result is reasonable. Why Other Options Are Wrong: 17 Ω or 1.7 Ω: Off by decade(s), likely unit conversion errors. 0.000169 Ω: Physically unrealistic; would imply a near-short. 340 Ω: Would correspond to half the frequency or half the capacitance; not our case. Common Pitfalls: Losing micro (10^-6) in calculations. Rounding early instead of at the end. Final Answer: 170 Ω

Capacitors Questions