3.9: Capacitance Flashcards
- The ability of an electronic component to store electric charge is called its:
(a) Addmittance
(b) Reluctance
(c) Capacitance
(c) Capacitance
- The amount of electricity a capacitor can store is directly proportional to :
(a) distance between the plates and inversely proportional to the plate area.
(b) plate area and is not affected by the distance between the plates.
(c) plate area and inversely proportional to the distance between the plates.
(c) plate area and inversely proportional to the distance between the plates.
- The subsatance that exists between the plates of a capacitor is more commonly
known as the:
(a) Electrode
(b) Dialectric
(c) Core
(b) Dialectric
- Placing two charges further from each other, will cause their resultant field force
to:
(a) Increase
(b) Remain the same.
(c) Decrease
(c) Decrease
- A 1 μF capacitor is equivalent to:
(a) 1,000,000 Farads
(b) 1/1,000,000 Farads
(c) 1/1,000 Farads
(b) 1/1,000,000 Farads
- A capacitor that stores 6 coulombs of electrons when a potential of 2 volts is
applied across its terminals has what total value of capacitance?
(a) 12 farads
(b) 3 farads
(c) 6 farads
(b) 3 farads
**Note: **
Q = C * V
where Q = Charge in Coulombs
C = Capacitance in Farads
V = Voltage in Volts
- What is the charge on a capacitor of value 1μF when a voltage across it is 500
volts?
(a) 5 Coulombs
(b) 500 μ Coulombs
(c) 5,000 μ Coulombs
(b) 500 μ Coulombs
Note:
Q = C * V
where Q = Charge in Coulombs
C = Capacitance in Farads
V = Voltage in Volts
- A 28VDC supply is placed across a 100 microfarad capacitor. What will the charge
be placed on it when fully charged?
(a) 100 microcoulombs
(b) 2.8 millicoulombs
(c) 28 Coulombs
(b) 2.8 millicoulombs
Note:
Q = C * V
where Q = Charge in Coulombs
C = Capacitance in Farads
V = Voltage in Volts
Prefixes - Micro (u) (Micro-Pen** 6 inches) ( 10^-6)
- Nano (n) (Nanu steak house -90e for a steak) (10^-9)
- Pico (p) (10^-12)
- If the distance between the plates of a parallel plate capacitor is increased and the
area of overlap of the plates is decreased the capacitance will be:
(a) unaffected
(b) increased
(c) decreased
(c) decreased
- When a 2μF and 100 nF capacitor are connected in parallel, the total capacitance
is equal to:
(a) 3 μF
(b) 102 μF
(c) 2100 nF
(c) 2100 nF
Remember: Capacitors in parallel behave like resistors in series
2μF + 100 nF
(2x10-9)F + 100nF
(2000x10^-9)F + 100F = 2100nF
- The capacitance of two capacitors of 12 μF and 4 μF are connected in series. The
total capacitance is?
(a) 16 μF
(b) 3 μF
(c) 8μF
(b) 3 μF
Remember: Capacitors in series behave like resistors in parallel
1/Ctotal = 1/C1 + 1/C2 ….
Ctotal = (C1)(C2) / (C1 + C2)
Ctotal = 48 / 16 = 3
- Three capacitors are connected in series. They are 470 μF, 750μF and 0.00033
μF respectively. The total capacitance is?
(a) more than 330 pF
(b) less than 330 pF
(c) Approximately 1550 pF
(b) less than 330 pF
Note: Capacitors in series behave like resistors in parallel
Question too complex to work out mathematically, remember if resistors are in parallel, the maximum resistance is less than the smallest resistor value.
- Which statement is correct about polarity markings (+/-) on capacitors?
(a) All capacitors have polarity markings.
(b) Some capacitors have polarity markings.
(c) Capacitors never have polarity markings
(b) Some capacitors have polarity markings.
- When the plates of a Capacitive Type Sensor are placed closer together the
capacitance:
(a) is increased
(b) is decreased
(c) will not change until the dialectric changes
(a) is increased
The capacitance of a capacitor depends on three factors:
1. The area of the plates
2. The distance between the plates
3. The material (dielectric) between the plates
- Which statement is correct about colour coding on capacitors?
(a) All capacitors are colour coded..
(b) Capacitors are never colour coded..
(c) Some capacitors may be colour coded..
(c) Some capacitors may be colour coded..
- Which of the following will decrease the value of a capacitor.
(a) Increase the area of the plates
(b) Increase the thickness of the dielectric.
(c) Decrease the spacing between the two plates.
(b) Increase the thickness of the dielectric.
The capacitance of a capacitor depends on three factors:
1. The area of the plates
2. The distance between the plates
3. The material (dielectric) between the plates
- Which of the following characteristics of a capacitor can be varied WITHOUT
altering its capacitance?
(a) Thickness of the plates
(b) Area of the plates
(c) Material of the dielectric
(a) Thickness of the plates
The capacitance of a capacitor depends on three factors:
1. The area of the plates
2. The distance between the plates
3. The material (dielectric) between the plates
- If a 3μF capacitor is placed in series with a 6μF capacitor, the total capacitance
is:
(a) 1.5 μF
(b) 2.0 μF
(c) 4.5 μF
(b) 2.0 μF
Remember: Capacitors in series behave like resistors in parallel
1/Ctotal = 1/C1 + 1/C2 + …..
Ctotal = (C1*C2) / (C1 + C2)
Ctotal = (3 * 6) / (3 + 6)
= 18/9 = 2
- A capacitor which stores a charge of one millionth of a coulomb when the emf
of one volt is applied has a value of:
(a) 1 farad.
(b) 1 microfarad.
(c) 1 picofarad.
(b) 1 microfarad.
Capacitance is measured in units called farads. A one-farad capacitor stores one coulomb (6.28 x10^18 electrons) of charge when a potential of 1 volt is applied across the terminals of the capacitor.
- A circuit has a 1μF and 2 μF capacitor in series what is the total capacitance?
(a) .66 μF
(b) 0.99 μF
(c) 1.0 μF
(a) .66 μF
Remember: Capacitors in series behave like resistors in parallel
1/Ctotal = 1/C1 + 1/C2 + ….
Ctotal = (C1 * C2) / (C1 + C2)
Ctotal = (1 * 2) / (1 + 2)
=2/3 = 0.6666
- A circuit has a 1 μF and 2 μF connected in parallel . What is the total capacitance
of the circuit?
(a) 1 μF
(b) 2 μF
(c) 3 μF
(c) 3 μF
Remember: Capacitors in paralle behave like resistors in series
Ctotal = C1 + C2 + ….
Ctotal = 1 + 2
= 3
- A 500Ω resistor is connected in series with a 2 μF What is the time constant of the
circuit.
(a) .5 second
(b) 1 msecond
(c) 2 seconds
(b) 1 msecond
1 TC = C * R
Where TC is Time Constant in seconds
C = Capacitance in Farads
R = Resistance in Ohms
1 TC = time required to charge capacitor to 63% of full charge OR to discharge it to 37% of it’s initial voltage
5 TC = time required to be fully charged / discharged
- The capacitance of a capacitor is the ratio of :
(a) voltage between plates to charge
(b) voltage between plates to plate spacing
(c) charge to voltage between plates
(c) charge to voltage between plates
Remember: Q = C * V
Where Q = Charge in Coulombs
C = Capacitance in Farads
V = Voltage in Volts
- Four 8 microfarad capacitors are connected in parallel. The equivalent
capacitance is :
(a) 2 microfarads
(b) 32 microfarads
(c) 8 microfarads
(b) 32 microfarads
Remember: Capacitors in parallel behave like resistors in series
Ctotal = C1 + C2 + ….
Ctotal = 8 + 8 + 8 + 8
Ctotal = 32 microfarads
- A 1 M ohm resistor is connected in series with a 1 μF capacitor across a 10V
battery. What is the voltage across the capacitor 1 second after the capacitor starts
charging.
(a) 5.2 V
(b) 6.32 V
(c) 8.65 V
(b) 6.32 V
1 TC = C * R
Where TC is Time Constant, C is capacitance in Farads and R is resistance in ohms
1 TC = (1x10^-6) * (1x10^6)
Remember (10^a) *(10^b) = 10^(a+b)
1 TC = 1 second
1 TC = 63.2% of charge or 37% of discharge, therefore after 1 second,
1 TC = 63.2% of 10V = 6.32V
- The switch on a DC circuit containing a fully charged capacitor is opened. The
voltage across the capacitor:
(a) Starts to fall exponentially to zero.
(b) Remains equal to the original charging voltage supplied
(c) Drops immediately to zero.
(b) Remains equal to the original charging voltage supplied
Option B: 7.36 volts
1 TC = 63.2% of full charge OR 36.8% discharge of its initial voltage
A capacitor is considered to be fully charged or discharged after 5 Time Constants
1 TC = C (Capacitance in Farads) * R (resistance in Ohms)
Option A: the max voltage that can be constantly applied
Note: Exceed the voltage at which the capacitor is rated, and you destroy the capacitor
Option B: the relative permittivity of the dielectric in relation to a vacuum
Note: ‘Relative’ means relative to ‘free space’ (a ‘vacuum’ in other words)
Option A: leaking dielectric capacitor
Option A: 10.01001 millifarads
Remember: Capacitors in parallel behave like resistors in series
Ctotal = C1 + C2 +….
Option B: 12.5 milliJoules
Energy Stored in a Capacitor
E = 1/2 * C * (V^2)
Where E is energy in Joules, C is capacitance in Farads and V is Voltage in Volts
A capacitor can store electric energy when disconnected from its charging circuit, so it can be used like a temporary battery. Capacitors are commonly used in electronic devices to maintain power supply while batteries are being changed. This prevents loss of information in volatile memory.
Option B: the same capacitance
The capacitance of a capacitor depends on three factors:
1. The area of the plates
2. The distance between the plates
3. The material (dielectric) between the plates
Option B: there is a polarized input
Wiring an electrolytic capacitor wrong polarity will destroy it. AC will therefore destroy it. Current must be DC (i.e polarised) and the right way.
Option B: Put a 6 F capacitor in series
Remember: Capacitors in series behave like resistors in parallel.
Ctotal = (C1 * C2) / (C1 + C2)
= (3 * 6) / (3 + 6) = 18/9 = 2
Option C: 500 milliseconds
1 TC = C * R
1 TC = (1x10^-6) * (100x103)
1 TC = 100 milliseconds
Note that question asks fully charged not 63.2% charged which is 1 Time Constant.
5 TC = 500 milliseconds and fully charged
Option A: 1 second
1 TC = C * R
Need to work out R with Ohms law, R = V/I , R = 100V / 0.025 A, R = 4000 ohms
1 TC = (50x10^-6) F * (4 x10^3) ohms
1 TC = 200 x10^-3 seconds , 1 TC = 0.2 seconds
However, 5 TC is required for full charge
0.2 seconds * 5 = 1 second
Option A = 6.4 Volts
Since Q = VC and Q = It where I is current and t is time
Then It = VC
V = I*t / C, V = (40x10^-6)(4) / (25x10^-6)
= 6.4
Option B: 4 u(micro)F
Remember: Capacitors in series behave like resistors in parallel
1/Ctotal = 1/C1 + 1/C2 + 1/C3 + …
Option B: for the emf to reach 63.2% of maximum
Option A: plate area and inversely proportional to the distance between the plates
Option B: the type of material separating the plates
Note: The capacitance of a capacitor is only dependant upon its physical properties (size and material) and not what you apply to it
Option C: equal to the sum of all the capacitances
Option C: 1x10^-12 farad
Option C: remains equal to the original charging voltage supply
Note: If there is no external circuit, the charge can go nowhere
Option A: zero
Note: If the capacitance is fully charged, it is equal and opposite to the source voltage, so no current flows
Option C: D.C
Option A: 0.4 u(micro)F
Note: capacitors in parallel behave like resistors in series - Total capacitance is equal to the sum of all capacitors
Option B: the charge stored on each is the same
Note: If they were not the same, there would be a current flow between them, until they equalised out
Option C: the product C * V
Remember: Q = C * V
Option B: is fully discharged before removing it from the circuit
Note: Charged capacitors can be lethal, even when charged to say, 1 volt
Option B: the electrostatic energy storing capacity of the capacitor dielectric
Option C: 36 microfarads
Remember: Capacitors in parallel behave like resistors in series. Just add them up
Option A: 8 seconds
1 TC = C * R
Where TC is the Time Constant, C is capacitance in Farads and R is resistance in ohms
Option B: the maximum continuous voltage is can take
Option A: picofarads
Option C: expontential
Option C: small physical size for a large capacity
Option C: heavy / light loads
Electrolytic capacitors are used in circuits of all sizes
Option A: 100 microseconds
1 TC = C * R
1 TC = (20x10^-12) * (1x10^6)
1 TC = 20x10^-6
5 TC ( 5 time constants max charge) = 100 microseconds
Option A: Exponential
Option B: 10,000 pf
103 means 10 followed by 3 zeros. Capacitors are always measured in picofarads
Option A: c = K*A / d
Option A: 36 milliFarad
Option C: 8 seconds
Option B: Same as circuit voltage
Note: The answer to this question depends upon how accurate you want to go. Normal theory is that a fully charged capacitor has an equal (and opposite) voltage to the supply. However, a capacitor is never fully charged (something in the order of 99.99999% charged).
Option C: Large output compared to size required
An Electrolytic Capacitor is used where a large amount of capacitance is required. - Dielectric is very thin layer of oxide - so thin that is it possible to make capacitors with a large value of capacitance for a small physical size.
Option B: 10 nanofarads
Option B: 10 nanofarads
'’Black boys R*** Our Young Girls But Violet Gives Willings’’
Option B: the same as applied voltage
Option B: electrostatic storying capability of the dielectric
Option C: 1 * 10^-6 Farads
Option C: 600 * 10^-12 coulombs
Q = CV, so direct relationship of charge to voltage. 1.5V is 1/4 of 6V, so charge is 1/4 of 2400 picocoulombs.
Option A: Farads per meter
Option B: 4 picofarads
