Capacitors Alıştırma Soruları — Armed Services Vocational Aptitude Battery (ASVAB) — Sayfa 2

Match each capacitor term or specification on the left with its correct electrical description on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Farad (F) matches with the standard unit used to quantify electrical capacitance; Dielectric matches with the insulating material separating the conductive plates; Electrolytic capacitor matches with a polarized capacitor type sensitive to connection polarity; Working voltage rating matches with the maximum continuous voltage that can safely be applied.

Each term accurately pairs with its foundational physical definition: the farad is the SI unit of capacitance, the dielectric is the insulating medium between plates, electrolytic capacitors are polarized devices requiring correct terminal orientation, and the working voltage rating defines the maximum safe operating voltage before dielectric puncture.

Adım Adım Çözüm

1

Identify the standard unit of capacitance

Farad (F) is the metric unit measuring charge storage ability.

Capacitance (

C = Q/V

) is quantified in units of farads.

2

Identify the insulating component inside a capacitor

The dielectric is the insulating layer.

Dielectrics block steady direct current conduction across the plates while concentrating the electric field.

3

Identify the polarized capacitor category

Electrolytic capacitors have designated positive and negative terminals.

Reverse polarization disrupts the chemical oxide layer, potentially destroying the component.

4

Identify the safe maximum voltage parameter

Working voltage rating represents the dielectric breakdown threshold.

Operating above this threshold risks arcing and dielectric breakdown.

Anahtar Kavram

Fundamental capacitor terminology, units, internal construction, and operational ratings

Tahmini Süre:45s

Soru 22Soru →

Once a capacitor placed in a direct current (DC) circuit reaches a fully charged state, how does it affect the continuous flow of current in that branch?

It blocks direct current completely because the accumulated charge creates an opposing voltage equal to the supply voltage.

It allows direct current to conduct freely because the dielectric spacer functions as an electrical conductor.

It transforms the incoming direct current into an oscillating alternating current output.

It sustains ongoing current flow by continuously storing energy within a magnetic field.

Cevabı ve açıklamayı göster

Cevap: It blocks direct current completely because the accumulated charge creates an opposing voltage equal to the supply voltage.

Cevap

A fully charged capacitor completely blocks the continuous flow of direct current by acting as an open circuit.

The correct choice explains that as a capacitor charges, its plate voltage rises until it matches the applied direct current source voltage. At this equilibrium point, no net voltage difference exists to move additional electrons through the circuit branch, completely stopping current flow and causing the capacitor to act as an open circuit to steady DC.

Adım Adım Çözüm

1

Identify the charging behavior of a capacitor connected to a DC source.

Current flows initially as electrons accumulate on one plate and leave the other, creating a potential difference across the dielectric.

Capacitors respond to changes in voltage by charging until their terminal voltage matches the applied source voltage.

2

Determine the state of the circuit when the capacitor is fully charged.

The voltage across the capacitor equals the supply voltage (

V_C = V_S

), so the net voltage driving current becomes zero (

I = 0

).

With no potential difference remaining across the circuit resistance, current ceases entirely, meaning the fully charged capacitor acts as an open circuit to DC.

Anahtar Kavram

DC Blocking Behavior of Capacitors

Tahmini Süre:45s

Soru 23Soru →

An avionics technician constructs a filtering network using three capacitors. Two capacitors, rated at $0.05\text{ }\mu\text{F}$ and $50\text{ nF}$ , are connected in series with each other. This series combination is then connected in parallel with a third capacitor rated at $15\text{ nF}$ . What is the total equivalent capacitance of the network in nanofarads ( $\text{nF}$ )?

Cevabı ve açıklamayı göster

Cevap: 40

Cevap

The total equivalent capacitance of the network is

40\text{ nF}

.

First, convert

0.05\text{ }\mu\text{F}

to nanofarads:

0.05\text{ }\mu\text{F} \times 1000\text{ nF}/\mu\text{F} = 50\text{ nF}

. Next, the two series capacitors (

50\text{ nF}

and

50\text{ nF}

) combine via the reciprocal formula to give

25\text{ nF}

. Finally, adding the parallel capacitor (

15\text{ nF}

) yields

25\text{ nF} + 15\text{ nF} = 40\text{ nF}

.

Adım Adım Çözüm

1

Convert the microfarad rating to nanofarads to ensure uniform units across all calculations.

0.05\text{ }\mu\text{F} = 0.05 \times 10^3\text{ nF} = 50\text{ nF}

Capacitor combination formulas require all terms to share the same metric prefix.

2

Calculate the equivalent capacitance of the two capacitors connected in series (

C_1 = 50\text{ nF}

and

C_2 = 50\text{ nF}

).

C_{\text{series}} = \frac{1}{\frac{1}{50\text{ nF}} + \frac{1}{50\text{ nF}}} = 25\text{ nF}

Capacitors in series combine reciprocally, meaning two identical series capacitors yield half of an individual capacitor's value.

3

Add the parallel branch capacitance (

C_3 = 15\text{ nF}

) to the series combination equivalent.

C_{\text{total}} = C_{\text{series}} + C_3 = 25\text{ nF} + 15\text{ nF} = 40\text{ nF}

Capacitors in parallel add directly because their plate areas effectively combine to store more charge at a given voltage.

Anahtar Kavram

Series and parallel equivalent capacitance rules with metric prefix conversion

Alternatif Yöntem

Convert all components into microfarads first:

C_1 = 0.05\text{ }\mu\text{F}

,

C_2 = 0.05\text{ }\mu\text{F}

, and

C_3 = 0.015\text{ }\mu\text{F}

. The series pair gives

\frac{0.05}{2} = 0.025\text{ }\mu\text{F}

. Adding the parallel capacitor gives

0.025\text{ }\mu\text{F} + 0.015\text{ }\mu\text{F} = 0.040\text{ }\mu\text{F}

. Finally, converting

0.040\text{ }\mu\text{F}

back to nanofarads gives

0.040 \times 1000 = 40\text{ nF}

.

Tahmini Süre:1m 30s

Soru 24Soru →

During the maintenance of an aerospace navigation system, an avionics technician isolates a filtering network to verify its total capacitance. The network is configured as follows: a $20 \mu\text{F}$ capacitor and a $5 \mu\text{F}$ capacitor are wired in series. This series combination is placed in parallel across a $36 \mu\text{F}$ capacitor. Finally, this entire grouped section is wired in series with a $120 \mu\text{F}$ capacitor. What is the total equivalent capacitance of this filtering network? Enter your answer as a number.

Cevabı ve açıklamayı göster

Cevap: 30

Cevap

30

To find the total equivalent capacitance, you must simplify the circuit step-by-step applying the specific rules for capacitors, which are the exact reverse of resistors. First, the

20 \mu\text{F}

and

5 \mu\text{F}

series capacitors are combined using the product-over-sum method to yield

4 \mu\text{F}

. Next, this

4 \mu\text{F}

equivalent is added directly to the

36 \mu\text{F}

parallel capacitor, resulting in

40 \mu\text{F}

for the entire parallel section. Finally, this

40 \mu\text{F}

equivalent block is in series with the final

120 \mu\text{F}

capacitor. Using the product-over-sum method once more (

\frac{40 \times 120}{160}

), the final total equivalent capacitance equals

30 \mu\text{F}

.

Adım Adım Çözüm

1

Calculate the equivalent capacitance of the

20 \mu\text{F}

and

5 \mu\text{F}

series pair.

4 \mu\text{F}

Capacitors in series are calculated using the reciprocal formula or product-over-sum rule:

\frac{20 \times 5}{20 + 5} = 4

.

2

Calculate the equivalent capacitance of the parallel section.

40 \mu\text{F}

The

4 \mu\text{F}

series equivalent is in parallel with the

36 \mu\text{F}

capacitor. Capacitors in parallel simply add together:

4 + 36 = 40

.

3

Calculate the final total capacitance of the entire network.

30 \mu\text{F}

The

40 \mu\text{F}

equivalent section is in series with the

120 \mu\text{F}

capacitor. Using the product-over-sum rule again:

\frac{40 \times 120}{40 + 120} = 30

.

Anahtar Kavram

Calculating equivalent capacitance in compound series-parallel networks

Tahmini Süre:3m 0s

Soru 25Soru →

Inside a standard capacitor, what is the primary role of the dielectric material placed between the two conductive plates?

To conduct direct current continuously between the plates to complete the circuit

To produce an electromagnetic field that resists sudden changes in current

To prevent physical contact between the plates while supporting an electrostatic field

To convert incoming direct current directly into alternating current

Cevabı ve açıklamayı göster

Cevap: To prevent physical contact between the plates while supporting an electrostatic field

Cevap

To prevent physical contact between the plates while supporting an electrostatic field

The correct answer accurately states that the dielectric serves as an insulating barrier that prevents the conductive plates from touching (which would cause a short circuit) while enabling an electrostatic field to establish between the accumulated charges.

Adım Adım Çözüm

1

Identify the basic physical structure of a capacitor

A capacitor consists of two conducting plates separated by an insulating layer called a dielectric.

Understanding the physical layout helps identify the specific function of each layer.

2

Determine the electrical properties and purpose of the dielectric

The dielectric material is a non-conductive insulator that blocks direct electron flow between the plates, allowing positive and negative charges to accumulate and establish an electrostatic field.

Without an insulating dielectric, the plates would touch or conduct, creating a short circuit instead of storing energy.

Anahtar Kavram

Function and Properties of Capacitor Dielectric Materials

Tahmini Süre:45s

Soru 26Soru →

Electronics technicians must select the proper capacitor type for specific circuit requirements. Match each capacitor type on the left with its characteristic construction or typical circuit application on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Ceramic Disc Capacitor pairs with non-polarized low-capacitance RF bypassing; Aluminum Electrolytic Capacitor pairs with polarized high-capacitance power supply filtering; Tantalum Capacitor pairs with polarized low-leakage precision low-voltage applications; Trimmer (Variable) Capacitor pairs with adjustable capacitance for tuning resonant circuits.

Each capacitor type is tailored to specific circuit roles based on dielectric material, capacitance range, polarity, and frequency response: ceramic disc for high-frequency bypassing, aluminum electrolytic for bulk power smoothing, tantalum for stable low-leakage applications, and trimmer capacitors for variable circuit calibration.

Adım Adım Çözüm

1

Identify the non-polarized capacitor designed for high frequencies.

Ceramic disc capacitors have low capacitance and excellent high-frequency response, matching the description for RF bypassing.

Ceramic dielectrics maintain stability at high frequencies without polarity constraints.

2

Identify the high-capacity polarized capacitor used in bulk power filtering.

Aluminum electrolytic capacitors provide the large microfarad values necessary to smooth DC ripple in rectifiers.

An electrochemical oxide layer allows very high capacitance per unit volume, but requires strict polarity observation.

3

Identify the polarized capacitor known for stability, low leakage, and high density.

Tantalum capacitors provide high capacitance in small packages with minimal leakage current for precision circuits.

Tantalum pentoxide dielectric layers are extremely thin and uniform compared to standard liquid electrolytes.

4

Identify the adjustable capacitor used for calibration.

Trimmer capacitors are variable capacitors used to tune resonant and RF circuits.

Movable plates or screw-adjusted dielectric spacing allow fine calibration of the total capacitance.

Anahtar Kavram

Classification, dielectric characteristics, polarity, and circuit roles of different capacitor types.

Tahmini Süre:1m 0s

Soru 27Soru →

An automotive audio technician installs two filter capacitors connected in parallel across a power line to suppress voltage ripple. If the individual capacitor values are $47\text{ }\mu\text{F}$ and $33\text{ }\mu\text{F}$ , what is the total equivalent capacitance of this parallel combination in microfarads ( $\mu\text{F}$ )?

Cevabı ve açıklamayı göster

Cevap: 80

Cevap

80 microfarads (

80\text{ }\mu\text{F}

)

When capacitors are wired in parallel, the total plate area increases, allowing more charge to be stored at a given voltage. Therefore, the equivalent capacitance is the simple arithmetic sum of the individual capacitances:

47\text{ }\mu\text{F} + 33\text{ }\mu\text{F} = 80\text{ }\mu\text{F}

.

Adım Adım Çözüm

1

Identify the network connection type and select the appropriate rule.

Capacitors are connected in parallel.

Connecting capacitors in parallel increases the total effective plate area, so their capacitances add directly.

2

Calculate the total capacitance using the sum formula

C_{\text{total}} = C_1 + C_2

.

C_{\text{total}} = 47\text{ }\mu\text{F} + 33\text{ }\mu\text{F} = 80\text{ }\mu\text{F}

Direct addition of the individual capacitance ratings yields the total circuit capacitance.

Anahtar Kavram

Parallel Capacitance Calculation

Tahmini Süre:45s

Soru 28Soru →

An audio technician connects a capacitor in series with a high-frequency speaker (tweeter) inside a multi-speaker enclosure. What is the primary electrical reason for using a capacitor in this configuration?

It stores energy in a magnetic field to boost low-frequency bass signals delivered to the speaker.

It easily passes high-frequency signals due to lower capacitive reactance while blocking low frequencies.

It blocks high-frequency signals and allows direct current and low audio frequencies to pass freely.

It acts as a permanent insulator that converts excess electrical voltage into heat at all audio frequencies.

Cevabı ve açıklamayı göster

Cevap: It easily passes high-frequency signals due to lower capacitive reactance while blocking low frequencies.

Cevap

The capacitor is used because it easily passes high-frequency signals due to lower capacitive reactance while blocking low frequencies.

Capacitive reactance (

X_C = \frac{1}{2\pi f C}

) is inversely proportional to the frequency of the electrical signal. For high-frequency signals directed to a tweeter, the capacitor offers very low opposition (low reactance), allowing the high-frequency audio to pass smoothly. At low frequencies or DC, the reactance becomes extremely high, blocking low-frequency bass currents that could overheat or damage the delicate tweeter voice coil.

Adım Adım Çözüm

1

Analyze the relationship between frequency and capacitive reactance (

X_C

).

Capacitive reactance is inversely proportional to frequency, governed by the formula

X_C = \frac{1}{2\pi f C}

.

Understanding how reactance changes with frequency is essential for determining how AC audio signals behave across the component.

2

Determine the effect on high versus low audio frequencies.

At high frequencies (treble),

X_C

is very low, presenting minimal opposition to current; at low frequencies (bass) and DC (

f = 0

),

X_C

becomes extremely large, effectively blocking the signal.

Tweeters are designed to handle only high-frequency signals, so placing a capacitor in series forms a high-pass filter that protects the tweeter from damaging low frequencies.

Anahtar Kavram

Capacitors exhibit capacitive reactance inversely proportional to frequency, acting as high-pass filters in series audio crossover circuits.

Tahmini Süre:1m 0s

Soru 29Soru →

Which of the following statements correctly identifies the primary function and operating principle of a capacitor in an electronic circuit?

It stores energy in a magnetic field to oppose sudden changes in electrical current.

It converts alternating current into direct current by permitting electron flow in only one direction.

It stores electrical energy in an electrostatic field and opposes sudden changes in voltage.

It allows continuous direct current flow across its internal dielectric layer to dissipate energy as heat.

Cevabı ve açıklamayı göster

Cevap: It stores electrical energy in an electrostatic field and opposes sudden changes in voltage.

Cevap

It stores electrical energy in an electrostatic field and opposes sudden changes in voltage.

A capacitor is designed with two conductive plates separated by an insulating dielectric material. When a voltage is applied, opposite charges accumulate on the plates, establishing an electrostatic field that stores electrical energy. Because charge accumulation requires finite time, a capacitor naturally opposes any sudden change in voltage across its terminals.

Adım Adım Çözüm

1

Identify the fundamental physical construction and energy storage mechanism of a capacitor.

A capacitor consists of two conducting plates separated by an insulating dielectric, which accumulates opposite charges to establish an electrostatic field.

Capacitors store potential electrical energy in an electrostatic field rather than a magnetic field.

2

Determine how a capacitor interacts with circuit voltage and current over time.

Because electrical charge requires finite time to build up or dissipate across the plates (

q = C \cdot V

), the component naturally resists instantaneous changes in voltage.

This property makes capacitors essential for filtering voltage ripples in power supplies and setting time constants in electronic circuits.

Anahtar Kavram

Capacitor Energy Storage and Voltage Regulation

Tahmini Süre:45s

Soru 30Soru →

An industrial electronics technician is servicing a timing circuit on an automated assembly line. The circuit network contains two $12\text{ }\mu\text{F}$ capacitors connected in series, and this combination is connected in parallel with a third capacitor rated at $14\text{ }\mu\text{F}$ . What is the total equivalent capacitance of this circuit network in microfarads ( $\mu\text{F}$ )?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

20\text{ }\mu\text{F}

For the series branch, two identical

12\text{ }\mu\text{F}

capacitors yield an equivalent capacitance of

6\text{ }\mu\text{F}

. When placed in parallel with the

14\text{ }\mu\text{F}

capacitor, the capacitances add directly (

6\text{ }\mu\text{F} + 14\text{ }\mu\text{F}

), giving a total equivalent capacitance of

20\text{ }\mu\text{F}

.

Adım Adım Çözüm

1

Calculate the equivalent capacitance of the two

12\text{ }\mu\text{F}

capacitors connected in series.

C_s = 6\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}

, or

C_s = \frac{C}{n} = \frac{12}{2} = 6\text{ }\mu\text{F}

.

2

Add the capacitance of the parallel branch to the series equivalent.

C_{\text{total}} = 20\text{ }\mu\text{F}

Capacitors in parallel add directly together:

C_{\text{total}} = C_s + C_3 = 6\text{ }\mu\text{F} + 14\text{ }\mu\text{F} = 20\text{ }\mu\text{F}

.

Anahtar Kavram

Series and parallel capacitor combinations and equivalent capacitance calculations

Tahmini Süre:1m 0s

Soru 31Soru →

In a multi-stage radio receiver, an electronics technician places a coupling capacitor in series along the signal line between two amplifier stages. What is the primary function of this capacitor regarding the combined Direct Current (DC) bias and Alternating Current (AC) signal?

Allows the continuous DC bias to pass freely while completely stopping all fluctuating AC signal frequencies.

Permits the fluctuating AC signal to pass through while completely blocking the steady DC bias voltage.

Transforms the AC signal into a higher DC voltage by generating a strong magnetic field across its plates.

Conducts the steady DC voltage directly across its dielectric material while dissipating the AC signal as heat.

Cevabı ve açıklamayı göster

Cevap: Permits the fluctuating AC signal to pass through while completely blocking the steady DC bias voltage.

Cevap

Permits the fluctuating AC signal to pass through while completely blocking the steady DC bias voltage.

A capacitor acts as an open circuit to steady Direct Current (DC) because the insulating dielectric stops continuous charge conduction once the plates reach the bias voltage level. Simultaneously, the alternating fluctuations of an AC signal continuously charge and discharge the plates, allowing the alternating signal to transfer cleanly to the next amplifier stage without shifting its DC bias point.

Adım Adım Çözüm

1

Analyze capacitor behavior in a steady Direct Current (DC) state.

Because the dielectric layer is an electrical insulator, steady DC charges the plates until the potential difference equals the supply voltage, reducing steady-state current to zero (

X_C = \infty

at

0\text{ Hz}

). The capacitor acts as an open circuit to DC.

Determines how the DC operating bias of the first amplifier stage is isolated from the second stage.

2

Analyze capacitor behavior when exposed to an Alternating Current (AC) signal.

The alternating polarity causes continuous charging and discharging cycles, allowing the AC signal to transfer across the component with low capacitive reactance (

X_C = \frac{1}{2\pi f C}

).

Determines how dynamic information and audio/RF waveforms propagate across the circuit.

3

Evaluate the combined circuit function of a series coupling capacitor.

The component blocks the steady DC bias voltage while coupling the alternating signal waveform directly to the next amplifier stage.

Confirms the core operational role of coupling capacitors in electronic signal chains.

Anahtar Kavram

Capacitor DC Blocking and AC Coupling

Tahmini Süre:1m 0s

Soru 32Soru →

Match each capacitor circuit application with its primary functional role in electronic systems.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Coupling Capacitor matches transferring AC signal voltages between amplifier stages while blocking DC bias levels; Filter Capacitor matches smoothing pulsating direct current output from a rectifier; Bypass Capacitor matches diverting AC noise or signals directly to ground around a component; Tuning Capacitor matches providing adjustable capacitance to set resonant frequency in RF circuits.

Each application corresponds directly to fundamental capacitive behaviors: coupling capacitors leverage DC-blocking and AC-passing in series with the signal path; filter capacitors utilize charge storage to smooth rectified DC voltage ripple; bypass capacitors create a low-reactance parallel path to ground for AC signals; and tuning capacitors provide variable capacitance for frequency selection in resonant LC circuits.

Adım Adım Çözüm

1

Determine the function of a coupling capacitor in multi-stage signal amplification.

Coupling capacitors block DC bias voltages while passing AC information signals between amplifier stages.

Capacitors provide infinite opposition (reactance) to zero-frequency DC while exhibiting low capacitive reactance to high-frequency AC signals.

2

Determine the function of a filter capacitor in a DC power supply.

Filter capacitors reduce voltage ripple by charging at peak rectified voltages and discharging during valleys.

This energy storage smooths pulsating DC output into a steady DC voltage for electronic circuits.

3

Determine the function of a bypass capacitor across a circuit element.

Bypass capacitors shunt alternating current signals to ground around a component without disturbing the DC operating point.

The capacitor provides an alternate low-reactance path for AC currents.

4

Determine the function of a tuning capacitor in radio frequency systems.

Tuning capacitors vary circuit capacitance to alter the resonant frequency of an LC circuit.

Resonant frequency varies inversely with the square root of capacitance, allowing station or frequency selection.

Anahtar Kavram

Operational roles and circuit applications of capacitors across DC power filtering, signal coupling, AC bypassing, and RF tuning

Tahmini Süre:1m 0s

Soru 33Soru →

In a linear DC power supply circuit, a technician places a smoothing filter capacitor in parallel with the load resistor following a bridge rectifier. What primary function does this capacitor perform in this circuit?

It converts the oscillating alternating current input directly into pulsating direct current.

It stores energy in a magnetic field to maintain a continuous, uninterrupted current flow.

It charges during peak rectified voltage and releases stored electrical energy when voltage falls, reducing ripple voltage.

It acts as a low-resistance conductor to continuously dissipate excess DC voltage into ground.

Cevabı ve açıklamayı göster

Cevap: It charges during peak rectified voltage and releases stored electrical energy when voltage falls, reducing ripple voltage.

Cevap

It charges during peak rectified voltage and releases stored electrical energy when voltage falls, reducing ripple voltage.

A parallel filter capacitor functions as an electrostatic energy reservoir across the load. During peak outputs from the rectifier, the capacitor absorbs electrical charge. As the pulsating rectified input falls toward zero, the capacitor supplies stored energy to the load, filling the troughs between pulses and significantly attenuating the unwanted AC ripple voltage.

Adım Adım Çözüm

1

Identify the component and its circuit role after the rectifier stage.

The component is a capacitor placed in parallel with the DC load resistor as a power supply filter.

The output of a rectifier consists of pulsating DC with substantial AC ripple that must be smoothed into a steady DC voltage level.

2

Analyze the electrostatic charge and discharge mechanism of the capacitor across the voltage cycle.

The capacitor quickly charges to peak output voltage during the rising edge of each rectified pulse, then discharges its stored charge into the load when the rectifier output drops below the capacitor voltage.

This continuous charge-discharge cycle fills in the valleys between rectified pulses, stabilizing the output voltage and minimizing ripple.

Anahtar Kavram

Capacitive filtering and ripple voltage reduction in DC power supplies

Tahmini Süre:1m 0s

Soru 34Soru →

An avionics technician is assembling a filtering network using three capacitors connected in series. The capacitors have ratings of $30\text{ }\mu\text{F}$ , $20\text{ }\mu\text{F}$ , and $60\text{ }\mu\text{F}$ . What is the total equivalent capacitance of this series network in microfarads ( $\mu\text{F}$ )?

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

The total equivalent capacitance of the series circuit is

10\text{ }\mu\text{F}

.

For capacitors connected in series, the total capacitance is found using the formula

\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

. Substituting

30\text{ }\mu\text{F}

,

20\text{ }\mu\text{F}

, and

60\text{ }\mu\text{F}

gives

\frac{2}{60} + \frac{3}{60} + \frac{1}{60} = \frac{6}{60} = \frac{1}{10\text{ }\mu\text{F}}

. Inverting this fraction yields the correct total capacitance of

10\text{ }\mu\text{F}

, which is less than the smallest individual capacitor (

20\text{ }\mu\text{F}

).

Adım Adım Çözüm

1

State the series capacitance relationship

\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

Capacitors in series combine inversely because the total dielectric separation increases, reducing total capacitance.

2

Calculate the sum of the reciprocals using a common denominator

\frac{1}{30} + \frac{1}{20} + \frac{1}{60} = \frac{2 + 3 + 1}{60} = \frac{6}{60} = \frac{1}{10}

Finding a least common multiple of

60

allows direct fractional addition.

3

Invert the result to find

C_{\text{total}}

C_{\text{total}} = 10\text{ }\mu\text{F}

The final equivalent capacitance is the inverse of the sum of the reciprocals.

Anahtar Kavram

Series capacitance follows the reciprocal sum rule, resulting in a total capacitance smaller than any individual capacitor in the string.

Tahmini Süre:1m 0s

Soru 35Soru →

An electronics technician is testing an optical pulse circuit and measures that a capacitor stores an electrostatic charge of $450\text{ }\mu\text{C}$ when connected across a potential difference of $15\text{ V}$ . What is the capacitance of this component?

6{}750\text{ }\mu\text{F}

30\text{ }\mu\text{H}

30\text{ }\mu\text{F}

0.03\text{ }\mu\text{F}

Cevabı ve açıklamayı göster

Cevap: $30\text{ }\mu\text{F}$

Cevap

30\text{ }\mu\text{F}

Capacitance is defined by the fundamental relationship

C = \frac{Q}{V}

. Dividing the stored charge of

450\text{ }\mu\text{C}

by the applied potential difference of

15\text{ V}

yields

30\text{ }\mu\text{F}

.

Adım Adım Çözüm

1

Identify the relationship between charge, capacitance, and voltage

The governing equation is

Q = C \times V

, which rearranges to

C = \frac{Q}{V}

.

Capacitance is defined as the amount of electric charge stored per unit of electrical potential difference.

2

Substitute the given values into the formula and solve for capacitance

C = \frac{450\text{ }\mu\text{C}}{15\text{ V}} = 30\text{ }\mu\text{F}

Dividing the charge in microcoulombs by the potential difference in volts directly yields the capacitance in microfarads.

Anahtar Kavram

The capacitance (

C

) of a component is the ratio of electric charge (

Q

) stored on its plates to the applied voltage (

V

), measured in Farads (F).

Tahmini Süre:1m 0s

Soru 36Soru →

An electronics technician is replacing components on a circuit board and selects an electrolytic capacitor for the DC power supply filtering stage. Why must the technician observe strict polarity when installing an electrolytic capacitor into a DC circuit?

They store electrical energy in a magnetic field that collapses and damages surrounding components if polarity is reversed.

They possess an electrochemical insulating oxide layer that breaks down and can cause the component to overheat or rupture under reverse voltage.

They are constructed to allow direct current to flow continuously while blocking all alternating current signals.

They operate through direct electrical conduction between the two internal plates without an insulating dielectric barrier.

Cevabı ve açıklamayı göster

Cevap: They possess an electrochemical insulating oxide layer that breaks down and can cause the component to overheat or rupture under reverse voltage.

Cevap

Electrolytic capacitors have an electrochemical oxide dielectric layer that breaks down and causes the component to overheat or rupture if connected with reversed polarity.

Electrolytic capacitors are polarized components because their extremely thin dielectric layer (aluminum or tantalum oxide) is formed and maintained electrochemically in one specific voltage direction. If connected in reverse polarity, this insulating layer breaks down rapidly, allowing heavy leakage current to generate boiling electrolyte gases, resulting in device failure, venting, or explosion.

Adım Adım Çözüm

1

Identify the internal construction of electrolytic capacitors.

Electrolytic capacitors utilize an extremely thin dielectric film of aluminum oxide or tantalum oxide maintained by a chemical electrolyte.

Understanding the unique dielectric mechanism explains why electrolytic capacitors provide high capacitance in a compact size but possess polarity restrictions.

2

Analyze the electrical behavior of the dielectric layer under reverse polarity.

When reversed voltage is applied, the electrochemical reaction breaks down and de-forms the insulating oxide layer.

The formation and insulation of the oxide layer are unidirectional.

3

Evaluate the practical consequence of reverse installation in a circuit.

With the dielectric compromised, large leakage currents flow through the electrolyte, producing excessive heat, gas buildup, and catastrophic rupture or short circuit.

This necessitates strict observance of the positive and negative terminal markings during assembly and maintenance.

Anahtar Kavram

Electrolytic Capacitor Polarity and Dielectric Integrity

Tahmini Süre:1m 0s

Soru 37Soru →

An electronics technician is assembling an energy-storage circuit for an industrial sensor module. The circuit consists of two $100\text{ }\mu\text{F}$ capacitors connected in series, and this entire combination is wired in parallel with a single $150\text{ }\mu\text{F}$ capacitor. What is the total equivalent capacitance of this circuit in microfarads ( $\mu\text{F}$ )?

Cevabı ve açıklamayı göster

Cevap: 200

Cevap

The total equivalent capacitance of the circuit is

200\text{ }\mu\text{F}

.

For capacitors in series, total capacitance decreases and is calculated as

\frac{1}{C_{\text{series}}} = \frac{1}{100\text{ }\mu\text{F}} + \frac{1}{100\text{ }\mu\text{F}} = \frac{2}{100\text{ }\mu\text{F}}

, giving

C_{\text{series}} = 50\text{ }\mu\text{F}

. When this combination is connected in parallel with the

150\text{ }\mu\text{F}

capacitor, parallel capacitances add directly:

C_{\text{total}} = 50\text{ }\mu\text{F} + 150\text{ }\mu\text{F} = 200\text{ }\mu\text{F}

.

Adım Adım Çözüm

1

Calculate the equivalent capacitance of the two

100\text{ }\mu\text{F}

capacitors connected in series using the reciprocal formula

\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2}

.

C_{\text{series}} = 50\text{ }\mu\text{F}

Capacitors in series combine inversely, meaning two identical capacitors in series have half the capacitance of one capacitor.

2

Add the capacitance of the parallel branch to the series equivalent:

C_{\text{total}} = C_{\text{series}} + C_3

.

C_{\text{total}} = 50\text{ }\mu\text{F} + 150\text{ }\mu\text{F} = 200\text{ }\mu\text{F}

Capacitors connected in parallel add directly because their effective plate areas combine, increasing overall charge storage capacity.

Anahtar Kavram

Series and parallel equivalent capacitance rules (series capacitors combine inversely; parallel capacitors add directly).

Tahmini Süre:1m 0s

Capacitors

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Alternatif Yöntem

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar