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Function of Capacitors in Electrical Circuits
A capacitor is an electrical device that stores electric charge, which possesses energy, and discharges it when needed.
This simulation demonstrates how a capacitor works, how it stores energy when connected to a battery, and discharges all the stored charge at once when connected to a lamp.
This is what happens in a camera flash.
A capacitor functions similarly to a battery, but there are similarities and differences between them.
Capacitor
Battery
Stores charges with energy and discharges them when needed.
Stores charges with energy and discharges them when needed.
Similarity
Discharges charges quickly in one burst.
Discharges charges over a long period.
Difference
Capacitance of a Capacitor
Capacitance is the ability of a capacitor to hold charge.
A capacitor may be uncharged, yet we still say it has capacitance.
This simulation illustrates how a capacitor is charged using a battery and the relationship between the potential difference across the capacitor and its stored charge.
Complete the following table based on the experiment.
Calculating Capacitance
Charge (nC)
Potential Difference (V)
Capacitance (nF)
Capacitance from the Slope of the Charge vs. Voltage Graph
Calculate the slope of the graph. What do you conclude from this experiment?
\[.......................................................\]
\[.......................................................\]
\[.......................................................\]
1
A capacitor has a potential difference of \[12\;V\] across its plates and acquires a charge of \[3\;nC\]. The source voltage is then reduced to \[7\;V\]. Which of the following answers correctly represents the charge and capacitance of the capacitor after the change in source voltage?
\[q_2= 2.25×10^{-9}\;\;C\;\;\;\;,\;\;\;\; c=3.2×10^{-10}\;\;F\;\;\;\;\;\;-C\]
\[q_2= 2×10^{-9}\;\;C\;\;\;\;,\;\;\;\; c=2.8×10^{-10}\;\;F\;\;\;\;\;\;-A\]
\[q_2= 4×10^{-9}\;\;C\;\;\;\;,\;\;\;\; c=5.7×10^{-10}\;\;F\;\;\;\;\;\;-D\]
\[ q_2= 1.75×10^{-9}\;\;C\;\;\;\;,\;\;\;\; c=2.5×10^{-10}\;\;F\;\;\;\;\;\;-B\]
Click here to see the solution method.
Choose the correct answer:
2
Which of the following units is equivalent to the farad?
\[\frac{A^2.S^2 }{kg.m^3}\;\;\;\;\;\;-C\]
\[\frac{A^2.S^3 }{kg.m^2}\;\;\;\;\;\;-A\]
\[\frac{A^2.S^4 }{kg.m^2}\;\;\;\;\;\;-D\]
\[ \frac{A^2.S^2 }{kg.m^3}\;\;\;\;\;\;-B\]
Click here to see the solution method.
Choose the correct answer:
Charging and Discharging a Parallel-Plate Capacitor
In this simulation, a capacitor in its simplest form consists of two metal plates separated by an insulating material. When these plates are connected to a power source:
Process of Charging and Discharging a Capacitor
1. Charging Process
When the circuit is closed:
- Electrons begin to move from the source (battery).
- Charges accumulate on the capacitor plates.
- An electric field is created between the plates.
- The current gradually decreases until it stops completely.
Voltage Across the Capacitor:
\[ V_C(t) = V_0 (1 - e^{\frac {-t}{RC}}) \]
2. Discharging Process
When the circuit is opened:
- The stored charges act as an energy source.
- Current flows in the opposite direction.
- The charge decreases exponentially over time.
- The voltage drops until it reaches zero.
Discharging Equation:
\[ V_C(t) = V_0 e^{\frac {-t}{RC}} \]
Important Notes:
The time constant (τ = RC) determines the speed of the process.
Capacitance (C) determines the amount of charge stored.
Resistance (R) affects the rate of charging/discharging.
Practical Applications:
- Electronic timing circuits.
- Emergency lighting systems.
- Current filters in power supplies.
Important Notes
When a capacitor is connected to a battery, the voltage across the capacitor equals the battery voltage once charging is complete, and the capacitor's voltage will not change unless the battery voltage is altered.
If a capacitor is connected to a battery and then disconnected, the charge on the capacitor remains constant unless discharge occurs.
When we refer to the charge on a capacitor, we mean the charge on one of the plates, not the combined charge of both plates.
Spherical Capacitor
Cylindrical Capacitor
Parallel-Plate Capacitor
Two concentric spheres, the outer sphere is negatively charged, and the inner sphere is positively charged.
Two concentric cylinders, the outer cylinder is negatively charged, and the inner cylinder is positively charged.
Two parallel plates with a small separation. One plate is positively charged, and the other is negatively charged.
\[C=4\pi\epsilon_0\frac{r_1 \cdot r_2}{r_2 - r_1}\]
\[C=\frac{2\pi\epsilon_0 L}{\ln\frac{r_2}{r_1}}\]
\[C=\epsilon_0 \frac{A}{d}\]
3
A student studied the factors affecting the capacitance of a parallel-plate capacitor and changed the distance between the plates while keeping other factors constant. The relationship between the capacitance and the inverse of the distance between the plates was plotted, resulting in the following graph. The common area between the plates equals:
\[A= 6 \;\;cm^2\;\;\;\;\;\;-C\]
\[A= 2 \;\;cm^2\;\;\;\;\;\;-A\]
\[A= 9 \;\;cm^2\;\;\;\;\;\;-D\]
\[A= 4 \;\;cm^2\;\;\;\;\;\;-B\]
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Parallel Connection of Capacitors
Series Connection of Capacitors
Each capacitor has two plates, one connected to the positive terminal and the other to the negative terminal. The voltage across each capacitor equals the battery voltage. \[∆V=V_1=V_2\] Each capacitor stores charge according to its capacitance, so the total charge is distributed across the capacitors. \[q_{tot}=q_1+q_2\] \[C_{eq} \cdot ∆V = C_1 \cdot V_1 + C_2 \cdot V_2\] \[C_{eq}=C_1 + C_2\]
The plate connected to the positive terminal of the battery is positively charged, and the other plate is negatively charged by induction. All capacitors carry the same charge. \[q_{tot} = q_1 = q_2\] The battery voltage is distributed across the capacitors until charging is complete. The capacitor with the larger capacitance requires a lower voltage to charge. \[∆V=V_1+V_2\] \[\frac{q_{tot}}{C_{eq}} = \frac{q_1}{C_1} + \frac{q_2}{C_2}\] \[\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\]
For series connection:
\[\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ... + \frac{1}{C_n}\]
Where:
\[C_{total}\] Total capacitance (Farads)
\[C_1, C_2, C_3, ...\] Capacitance of each individual capacitor
✵ Practical Example:
If we have two capacitors with capacitances \[C_1=4\;μF, C_2=6\;μF\] connected in series:
\[\frac{1}{C_{total}} = \frac{1}{4} + \frac{1}{6} = \frac{3 + 2}{12} = \frac{5}{12}\]
\[C_{total} = \frac{12}{5} = 2.4μF\]
✵ Important Notes:
- The total capacitance in a series connection is always less than the smallest capacitance in the circuit.
- This type of connection is used to withstand high electrical voltages.
For parallel connection:
\[C_{total} = C_1 + C_2 + C_3 + ... + C_n\]
Where:
\[C_{total}\] Total capacitance (Farads)
\[C_1, C_2, C_3, ...\] Capacitance of each individual capacitor
Practical Example:
If we have three capacitors with capacitances \[C_1=4\;μF, C_2=6\;μF, C_3=2\;μF\] connected in parallel:
\[C_{total} = C_1 + C_2 + C_3 = 4μF + 6μF + 2μF = 12μF\]
✵ Important Notes:
- The total capacitance in a parallel connection is always greater than any individual capacitance in the circuit.
- This type of connection is used in energy storage systems in electronic devices.
- This type of connection is used in timing and filtering circuits.
- This type of connection is used in solar power systems.
Solved Example
A set of capacitors is connected as shown in the figure:
Calculate the equivalent capacitance.
\[C_{23} = C_2 + C_3 = 1 \mu F + 1 \mu F = 2 \mu F\]
\[C_{eq} = \left[\frac{1}{C_1} + \frac{1}{C_{23}}\right]^{-1} = \left[\frac{1}{2 \mu F} + \frac{1}{2 \mu F}\right]^{-1} = 1 \mu F\]
Calculate the charge and voltage across each capacitor.
\[C_{eq} = \frac{q_{tot}}{∆V}\]
\[q_{tot} = C_{eq} \cdot ∆V = 1 \times 10^{-6} \times 10 = 1 \times 10^{-5} C\]
\[q_{tot} = q_1 = q_{23} = 1 \times 10^{-5} C\]
\[∆V_1 = \frac{q_1}{C_1} = \frac{1 \times 10^{-5}}{2 \times 10^{-6}} = 5 V\]
\[∆V_{23} = ∆V_2 = ∆V_3 = 5 V\]
\[q_2 = C_2 \cdot ∆V_2 = 1 \times 10^{-6} \times 5 = 0.5 \times 10^{-5} C\]
\[q_3 = C_3 \cdot ∆V_3 = 1 \times 10^{-6} \times 5 = 0.5 \times 10^{-5} C\]
Energy Stored in a Capacitor
When charging a capacitor, work is done, and this work is converted into energy stored in the capacitor.
\[dW = ∆V \cdot dq\]\[\int dW = \int ∆V \cdot dq = \int \frac{q}{C} \cdot dq = \frac{1}{C} \int q \cdot dq = \frac{1}{2} \frac{q^2}{C}\]\[W = U = \frac{1}{2} \frac{q^2}{C}\]
Since \[C = \frac{q}{∆V}, U = \frac{1}{2} q \cdot ∆V\]And since
\[q = C \cdot ∆V, U = \frac{1}{2} C \cdot ∆V^2\]
\[U = \frac{1}{2} \frac{q^2}{C} = \frac{1}{2} q \cdot ∆V = \frac{1}{2} C \cdot ∆V^2\]
The work done in charging the capacitor is equal to the area under the curve.
Energy density: This is the energy stored per unit volume.
\[u = \frac{1}{2} \epsilon_0 E^2\]
\[u = \frac{U}{V} = \frac{\frac{1}{2} C \cdot ∆V^2}{A \cdot d} = \frac{\frac{\epsilon_0 A}{d} \cdot ∆V^2}{2 A \cdot d} = \frac{1}{2} \epsilon_0 \left(\frac{∆V}{d}\right)^2\]
\[u = \frac{1}{2} \epsilon_0 E^2\]
Capacitors and Dielectrics
When a dielectric material is placed between the plates of a capacitor, the capacitance increases by a factor (K), called the dielectric constant, which varies depending on the material. The capacitance of a parallel-plate capacitor becomes:
\[C = K \cdot C_0 = K \cdot \epsilon_0 \frac{A}{d}\]
To explain the reason for the increase in capacitance, there are two cases:
The capacitor is connected to a battery
When the dielectric material is inserted,
the material becomes polarized, creating an electric field that opposes the capacitor's field.
This leads to a decrease in the voltage across the capacitor, making
the battery's voltage higher. As a result, negative charges begin to move from the plate
connected to the positive terminal to the negative terminal, increasing the charge
(the charge increases) and thus (the field increases)
until the capacitor's voltage equals the battery's voltage again, meaning
(the voltage ultimately remains constant)
\[c=\frac{q}{∆𝑉}\]
The charge increased while the voltage remained constant, indicating (an increase in capacitance).
The capacitor is charged and disconnected from the battery.
When the dielectric material is inserted,
the material becomes polarized, creating an electric field that opposes the capacitor's field.
This leads to a decrease in the voltage across the capacitor,
meaning the voltage (decreases).
Since the capacitor is disconnected from the battery,
the charge (remains constant),
and the field (decreases) due to the reduction in voltage.
\[c=\frac{q}{∆𝑉}\]
The voltage decreased while the charge remained constant, indicating (an increase in capacitance).
A capacitor is an electrical device that stores electric charge, which possesses energy, and discharges it when needed.
This simulation demonstrates how a capacitor works, how it stores energy when connected to a battery, and discharges all the stored charge at once when connected to a lamp.
This is what happens in a camera flash.
A capacitor functions similarly to a battery, but there are similarities and differences between them.
Capacitor |
Battery |
|
Stores charges with energy and discharges them when needed. |
Stores charges with energy and discharges them when needed. |
Similarity |
Discharges charges quickly in one burst. |
Discharges charges over a long period. |
Difference |
Capacitance of a Capacitor
Capacitance is the ability of a capacitor to hold charge.
A capacitor may be uncharged, yet we still say it has capacitance.
This simulation illustrates how a capacitor is charged using a battery and the relationship between the potential difference across the capacitor and its stored charge.
Calculating Capacitance
| Charge (nC) | Potential Difference (V) | Capacitance (nF) |
|---|---|---|
Capacitance from the Slope of the Charge vs. Voltage Graph
Calculate the slope of the graph. What do you conclude from this experiment?
\[.......................................................\] \[.......................................................\] \[.......................................................\]
A capacitor has a potential difference of \[12\;V\] across its plates and acquires a charge of \[3\;nC\]. The source voltage is then reduced to \[7\;V\]. Which of the following answers correctly represents the charge and capacitance of the capacitor after the change in source voltage?
\[q_2= 2.25×10^{-9}\;\;C\;\;\;\;,\;\;\;\; c=3.2×10^{-10}\;\;F\;\;\;\;\;\;-C\] |
\[q_2= 2×10^{-9}\;\;C\;\;\;\;,\;\;\;\; c=2.8×10^{-10}\;\;F\;\;\;\;\;\;-A\] |
\[q_2= 4×10^{-9}\;\;C\;\;\;\;,\;\;\;\; c=5.7×10^{-10}\;\;F\;\;\;\;\;\;-D\] |
\[ q_2= 1.75×10^{-9}\;\;C\;\;\;\;,\;\;\;\; c=2.5×10^{-10}\;\;F\;\;\;\;\;\;-B\] |
Click here to see the solution method.
Choose the correct answer:
2
Which of the following units is equivalent to the farad?
\[\frac{A^2.S^2 }{kg.m^3}\;\;\;\;\;\;-C\]
\[\frac{A^2.S^3 }{kg.m^2}\;\;\;\;\;\;-A\]
\[\frac{A^2.S^4 }{kg.m^2}\;\;\;\;\;\;-D\]
\[ \frac{A^2.S^2 }{kg.m^3}\;\;\;\;\;\;-B\]
Click here to see the solution method.
Choose the correct answer:
Which of the following units is equivalent to the farad?
\[\frac{A^2.S^2 }{kg.m^3}\;\;\;\;\;\;-C\] |
\[\frac{A^2.S^3 }{kg.m^2}\;\;\;\;\;\;-A\] |
\[\frac{A^2.S^4 }{kg.m^2}\;\;\;\;\;\;-D\] |
\[ \frac{A^2.S^2 }{kg.m^3}\;\;\;\;\;\;-B\] |
Choose the correct answer:
Process of Charging and Discharging a Capacitor
1. Charging Process
When the circuit is closed:
- Electrons begin to move from the source (battery).
- Charges accumulate on the capacitor plates.
- An electric field is created between the plates.
- The current gradually decreases until it stops completely.
Voltage Across the Capacitor:
\[ V_C(t) = V_0 (1 - e^{\frac {-t}{RC}}) \]2. Discharging Process
When the circuit is opened:
- The stored charges act as an energy source.
- Current flows in the opposite direction.
- The charge decreases exponentially over time.
- The voltage drops until it reaches zero.
Discharging Equation:
\[ V_C(t) = V_0 e^{\frac {-t}{RC}} \]Important Notes:
Practical Applications:
- Electronic timing circuits.
- Emergency lighting systems.
- Current filters in power supplies.
Important Notes
When a capacitor is connected to a battery, the voltage across the capacitor equals the battery voltage once charging is complete, and the capacitor's voltage will not change unless the battery voltage is altered.
If a capacitor is connected to a battery and then disconnected, the charge on the capacitor remains constant unless discharge occurs.
When we refer to the charge on a capacitor, we mean the charge on one of the plates, not the combined charge of both plates.
Spherical Capacitor |
Cylindrical Capacitor |
Parallel-Plate Capacitor |
Two concentric spheres, the outer sphere is negatively charged, and the inner sphere is positively charged. |
Two concentric cylinders, the outer cylinder is negatively charged, and the inner cylinder is positively charged. |
Two parallel plates with a small separation. One plate is positively charged, and the other is negatively charged. |
\[C=4\pi\epsilon_0\frac{r_1 \cdot r_2}{r_2 - r_1}\] |
\[C=\frac{2\pi\epsilon_0 L}{\ln\frac{r_2}{r_1}}\] |
\[C=\epsilon_0 \frac{A}{d}\] |
A student studied the factors affecting the capacitance of a parallel-plate capacitor and changed the distance between the plates while keeping other factors constant. The relationship between the capacitance and the inverse of the distance between the plates was plotted, resulting in the following graph. The common area between the plates equals:
\[A= 6 \;\;cm^2\;\;\;\;\;\;-C\] |
\[A= 2 \;\;cm^2\;\;\;\;\;\;-A\] |
\[A= 9 \;\;cm^2\;\;\;\;\;\;-D\] |
\[A= 4 \;\;cm^2\;\;\;\;\;\;-B\] |
Click here to show the solutionChoose the correct answer
Parallel Connection of Capacitors |
Series Connection of Capacitors |
Each capacitor has two plates, one connected to the positive terminal and the other to the negative terminal. The voltage across each capacitor equals the battery voltage. \[∆V=V_1=V_2\] Each capacitor stores charge according to its capacitance, so the total charge is distributed across the capacitors. \[q_{tot}=q_1+q_2\] \[C_{eq} \cdot ∆V = C_1 \cdot V_1 + C_2 \cdot V_2\] \[C_{eq}=C_1 + C_2\] |
The plate connected to the positive terminal of the battery is positively charged, and the other plate is negatively charged by induction. All capacitors carry the same charge. \[q_{tot} = q_1 = q_2\] The battery voltage is distributed across the capacitors until charging is complete. The capacitor with the larger capacitance requires a lower voltage to charge. \[∆V=V_1+V_2\] \[\frac{q_{tot}}{C_{eq}} = \frac{q_1}{C_1} + \frac{q_2}{C_2}\] \[\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\] |
For series connection: \[\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ... + \frac{1}{C_n}\]
Where:
-
\[C_{total}\] Total capacitance (Farads)
\[C_1, C_2, C_3, ...\] Capacitance of each individual capacitor
✵ Practical Example:
If we have two capacitors with capacitances \[C_1=4\;μF, C_2=6\;μF\] connected in series:
\[\frac{1}{C_{total}} = \frac{1}{4} + \frac{1}{6} = \frac{3 + 2}{12} = \frac{5}{12}\] \[C_{total} = \frac{12}{5} = 2.4μF\]✵ Important Notes:
- The total capacitance in a series connection is always less than the smallest capacitance in the circuit.
- This type of connection is used to withstand high electrical voltages.
For parallel connection: \[C_{total} = C_1 + C_2 + C_3 + ... + C_n\]
Where:
-
\[C_{total}\] Total capacitance (Farads)
\[C_1, C_2, C_3, ...\] Capacitance of each individual capacitor
Practical Example:
If we have three capacitors with capacitances \[C_1=4\;μF, C_2=6\;μF, C_3=2\;μF\] connected in parallel:
\[C_{total} = C_1 + C_2 + C_3 = 4μF + 6μF + 2μF = 12μF\]✵ Important Notes:
- The total capacitance in a parallel connection is always greater than any individual capacitance in the circuit.
- This type of connection is used in energy storage systems in electronic devices.
- This type of connection is used in timing and filtering circuits.
- This type of connection is used in solar power systems.
Solved Example
Calculate the equivalent capacitance.
\[C_{23} = C_2 + C_3 = 1 \mu F + 1 \mu F = 2 \mu F\] \[C_{eq} = \left[\frac{1}{C_1} + \frac{1}{C_{23}}\right]^{-1} = \left[\frac{1}{2 \mu F} + \frac{1}{2 \mu F}\right]^{-1} = 1 \mu F\]
Calculate the charge and voltage across each capacitor. \[C_{eq} = \frac{q_{tot}}{∆V}\] \[q_{tot} = C_{eq} \cdot ∆V = 1 \times 10^{-6} \times 10 = 1 \times 10^{-5} C\] \[q_{tot} = q_1 = q_{23} = 1 \times 10^{-5} C\] \[∆V_1 = \frac{q_1}{C_1} = \frac{1 \times 10^{-5}}{2 \times 10^{-6}} = 5 V\] \[∆V_{23} = ∆V_2 = ∆V_3 = 5 V\] \[q_2 = C_2 \cdot ∆V_2 = 1 \times 10^{-6} \times 5 = 0.5 \times 10^{-5} C\] \[q_3 = C_3 \cdot ∆V_3 = 1 \times 10^{-6} \times 5 = 0.5 \times 10^{-5} C\]
Energy Stored in a Capacitor
When charging a capacitor, work is done, and this work is converted into energy stored in the capacitor.
\[dW = ∆V \cdot dq\]\[\int dW = \int ∆V \cdot dq = \int \frac{q}{C} \cdot dq = \frac{1}{C} \int q \cdot dq = \frac{1}{2} \frac{q^2}{C}\]\[W = U = \frac{1}{2} \frac{q^2}{C}\] Since \[C = \frac{q}{∆V}, U = \frac{1}{2} q \cdot ∆V\]And since \[q = C \cdot ∆V, U = \frac{1}{2} C \cdot ∆V^2\] \[U = \frac{1}{2} \frac{q^2}{C} = \frac{1}{2} q \cdot ∆V = \frac{1}{2} C \cdot ∆V^2\]
The work done in charging the capacitor is equal to the area under the curve.
Energy density: This is the energy stored per unit volume.
\[u = \frac{1}{2} \epsilon_0 E^2\] \[u = \frac{U}{V} = \frac{\frac{1}{2} C \cdot ∆V^2}{A \cdot d} = \frac{\frac{\epsilon_0 A}{d} \cdot ∆V^2}{2 A \cdot d} = \frac{1}{2} \epsilon_0 \left(\frac{∆V}{d}\right)^2\] \[u = \frac{1}{2} \epsilon_0 E^2\]
Capacitors and Dielectrics
When a dielectric material is placed between the plates of a capacitor, the capacitance increases by a factor (K), called the dielectric constant, which varies depending on the material. The capacitance of a parallel-plate capacitor becomes:
\[C = K \cdot C_0 = K \cdot \epsilon_0 \frac{A}{d}\]
To explain the reason for the increase in capacitance, there are two cases:
The capacitor is connected to a battery
When the dielectric material is inserted,
the material becomes polarized, creating an electric field that opposes the capacitor's field.
This leads to a decrease in the voltage across the capacitor, making
the battery's voltage higher. As a result, negative charges begin to move from the plate
connected to the positive terminal to the negative terminal, increasing the charge
(the charge increases) and thus (the field increases)
until the capacitor's voltage equals the battery's voltage again, meaning
(the voltage ultimately remains constant)
\[c=\frac{q}{∆𝑉}\]
The charge increased while the voltage remained constant, indicating (an increase in capacitance).
The capacitor is charged and disconnected from the battery.
When the dielectric material is inserted,
the material becomes polarized, creating an electric field that opposes the capacitor's field.
This leads to a decrease in the voltage across the capacitor,
meaning the voltage (decreases).
Since the capacitor is disconnected from the battery,
the charge (remains constant),
and the field (decreases) due to the reduction in voltage.
\[c=\frac{q}{∆𝑉}\]
The voltage decreased while the charge remained constant, indicating (an increase in capacitance).
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