a 3 μf capacitor is connected to a 6 v battery. what is the charge on each plate of the capacitor?

Learning Objectives

By the cease of this section, you will exist able to:

• Explicate how to decide the equivalent capacitance of capacitors in series and in parallel combinations
• Compute the potential difference beyond the plates and the charge on the plates for a capacitor in a network and determine the net capacitance of a network of capacitors

Several capacitors tin can exist continued together to be used in a variety of applications. Multiple connections of capacitors behave equally a unmarried equivalent capacitor. The total capacitance of this equivalent single capacitor depends both on the private capacitors and how they are connected. Capacitors can be arranged in two elementary and common types of connections, known equally series and parallel, for which we can easily calculate the total capacitance. These ii basic combinations, serial and parallel, can also be used equally part of more complex connections.

The Serial Combination of Capacitors

Effigy viii.eleven illustrates a series combination of iii capacitors, arranged in a row within the excursion. As for whatever capacitor, the capacitance of the combination is related to the charge and voltage by using Equation 8.1. When this series combination is connected to a battery with voltage V, each of the capacitors acquires an identical charge Q. To explicate, kickoff note that the charge on the plate connected to the positive terminal of the battery is $+Q$ and the charge on the plate connected to the negative last is $\text{−}Q$ . Charges are then induced on the other plates and then that the sum of the charges on all plates, and the sum of charges on whatsoever pair of capacitor plates, is zero. All the same, the potential drib $V_1=Q\text{/}{C}_{\mathrm{one}}$ on one capacitor may be dissimilar from the potential drop $5_2=Q\text{/}{C}_2$ on another capacitor, because, generally, the capacitors may have different capacitances. The series combination of ii or three capacitors resembles a single capacitor with a smaller capacitance. By and large, any number of capacitors connected in serial is equivalent to one capacitor whose capacitance (chosen the equivalent capacitance) is smaller than the smallest of the capacitances in the serial combination. Charge on this equivalent capacitor is the same as the charge on any capacitor in a series combination: That is, all capacitors of a serial combination have the same accuse. This occurs due to the conservation of charge in the circuit. When a charge Q in a series excursion is removed from a plate of the first capacitor (which we announce as $\text{−}Q$ ), it must be placed on a plate of the second capacitor (which we announce as $+Q),$ so on.

Figure 8.11 (a) Three capacitors are continued in series. The magnitude of the charge on each plate is Q. (b) The network of capacitors in (a) is equivalent to one capacitor that has a smaller capacitance than any of the individual capacitances in (a), and the charge on its plates is Q.

We can discover an expression for the total (equivalent) capacitance by considering the voltages beyond the private capacitors. The potentials beyond capacitors i, two, and three are, respectively, $5_1=Q\text{/}{C}_{\mathrm{one}}$ , $V_2=Q\text{/}{C}_2$ , and ${\mathrm{Five}}_3=Q\text{/}{C}_3$ . These potentials must sum up to the voltage of the battery, giving the following potential balance:

$$V=V_1+V_2+V_3.$$

Potential V is measured across an equivalent capacitor that holds charge Q and has an equivalent capacitance $C_{\text{S}}$ . Inbound the expressions for $V_1$ , ${\mathrm{Five}}_2$ , and $V_3$ , we get

$$\frac{Q}{C_{\text{Southward}}}=\frac{Q}{C_1}+\frac{Q}{C_{\mathrm{ii}}}+\frac{Q}{C_{\mathrm{three}}}.$$

Canceling the charge Q, nosotros obtain an expression containing the equivalent capacitance, $C_{\text{S}}$ , of iii capacitors connected in serial:

$$\frac{1}{C_{\text{Southward}}}=\frac{\mathrm{ane}}{C_{\mathrm{i}}}+\frac{1}{C_2}+\frac{1}{C_3}.$$

This expression can be generalized to any number of capacitors in a series network.

Series Combination

For capacitors connected in a series combination, the reciprocal of the equivalent capacitance is the sum of reciprocals of private capacitances:

$$\frac{1}{C_{\text{Southward}}}=\frac{1}{C_1}+\frac{\mathrm{ane}}{C_2}+\frac{1}{C_{\mathrm{iii}}}+\text{⋯}.$$

8.7

Example 8.four

Equivalent Capacitance of a Series Network

Find the total capacitance for three capacitors continued in serial, given their individual capacitances are $\mathrm{i.000}\mu \text{F}$ , $\mathrm{five.000}\mu \text{F}$ , and $8.000\mu \text{F}$ .

Strategy

Considering there are just three capacitors in this network, we tin can find the equivalent capacitance by using Equation viii.7 with iii terms.

Solution

We enter the given capacitances into Equation eight.seven:

$$\begin{array}{ccc}\hfill \frac{1}{C_{\text{S}}}& =\hfill & \frac{1}{C_{\mathrm{ane}}}+\frac{1}{C_2}+\frac{1}{C_3}\hfill \\ & =\hfill & \frac{1}{1.000\mu \text{F}}+\frac{1}{\mathrm{five.000}\mu \text{F}}+\frac{\mathrm{one}}{\mathrm{viii.000}\mu \text{F}}\hfill \\ \hfill \frac{\mathrm{i}}{C_{\text{S}}}& =\hfill & \frac{1.325}{\mu \text{F}}.\hfill \end{array}$$

Now nosotros capsize this result and obtain $C_{\text{S}}=\frac{\mu \text{F}}{\mathrm{one.325}}=0.755\mu \text{F}\text{.}$

Significance

Annotation that in a series network of capacitors, the equivalent capacitance is e'er less than the smallest individual capacitance in the network.

The Parallel Combination of Capacitors

A parallel combination of three capacitors, with one plate of each capacitor connected to one side of the circuit and the other plate connected to the other side, is illustrated in Figure 8.12(a). Since the capacitors are connected in parallel, they all have the same voltage V across their plates. However, each capacitor in the parallel network may shop a different charge. To detect the equivalent capacitance $C_{\text{P}}$ of the parallel network, we note that the total charge Q stored by the network is the sum of all the individual charges:

$$Q=Q_1+Q_2+Q_3.$$

On the left-paw side of this equation, we use the relation $Q=C_{\text{P}}V$ , which holds for the entire network. On the right-hand side of the equation, we use the relations $Q_{\mathrm{one}}=C_{1}V,Q_2=C_{2}V,$ and $Q_3=C_3\mathrm{Five}$ for the three capacitors in the network. In this mode we obtain

$$C_{\text{P}}V=C_{1}V+C_{2}5+C_{\mathrm{three}}\mathrm{Five}.$$

This equation, when simplified, is the expression for the equivalent capacitance of the parallel network of three capacitors:

$$C_{\text{P}}=C_{\mathrm{one}}+C_2+C_3.$$

This expression is easily generalized to whatever number of capacitors continued in parallel in the network.

Parallel Combination

For capacitors connected in a parallel combination, the equivalent (net) capacitance is the sum of all individual capacitances in the network,

$$C_{\text{P}}=C_1+C_{\mathrm{ii}}+C_{\mathrm{iii}}+\text{⋯}.$$

8.8

Effigy eight.12 (a) Three capacitors are connected in parallel. Each capacitor is connected directly to the battery. (b) The charge on the equivalent capacitor is the sum of the charges on the individual capacitors.

Case 8.5

Equivalent Capacitance of a Parallel Network

Discover the net capacitance for three capacitors connected in parallel, given their private capacitances are $1.0\mu \text{F},5.0\mu \text{F},\text{and}\mathrm{viii.0}\mu \text{F}\text{.}$

Strategy

Because in that location are only three capacitors in this network, we tin find the equivalent capacitance by using Equation 8.8 with three terms.

Solution

Entering the given capacitances into Equation viii.eight yields

$$\begin{array}{ccc}\hfill C_{\text{P}}& =\hfill & C_1+C_2+C_3=1.0\mu \text{F}+\mathrm{v.0}\mu \text{F}+\mathrm{eight.0}\mu \text{F}\hfill \\ \hfill C_{\text{P}}& =\hfill & \mathrm{xiv.0}\mu \text{F}\text{.}\hfill \end{array}$$

Significance

Notation that in a parallel network of capacitors, the equivalent capacitance is always larger than any of the individual capacitances in the network.

Capacitor networks are normally some combination of series and parallel connections, every bit shown in Figure 8.thirteen. To find the internet capacitance of such combinations, we identify parts that contain only series or simply parallel connections, and notice their equivalent capacitances. We repeat this process until we can determine the equivalent capacitance of the unabridged network. The post-obit example illustrates this process.

Figure viii.13 (a) This circuit contains both series and parallel connections of capacitors. (b) $C_1$ and $C_{\mathrm{ii}}$ are in series; their equivalent capacitance is $C_{\text{S}}.$ (c) The equivalent capacitance $C_{\text{Due south}}$ is connected in parallel with $C_3.$ Thus, the equivalent capacitance of the entire network is the sum of $C_{\text{South}}$ and $C_3.$

Instance viii.six

Equivalent Capacitance of a Network

Observe the total capacitance of the combination of capacitors shown in Figure eight.13. Assume the capacitances are known to three decimal places $(C_{\mathrm{ane}}=\mathrm{one.000}\mu \text{F},C_2=\mathrm{v.000}\mu \text{F,}$ $C_3=8.000\mu \text{F})\text{.}$ Round your answer to 3 decimal places.

Strategy

We start identify which capacitors are in serial and which are in parallel. Capacitors $C_1$ and $C_2$ are in series. Their combination, labeled $C_{\text{S}},$ is in parallel with $C_3.$

Solution

Since $C_{\mathrm{ane}}\text{and}{C}_{\mathrm{two}}$ are in series, their equivalent capacitance $C_{\text{Southward}}$ is obtained with Equation 8.7:

$$\frac{1}{C_{\text{S}}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{\mathrm{ane}}{\mathrm{ane.000}\mu \text{F}}+\frac{1}{5.000\mu \text{F}}=\frac{1.200}{\mu \text{F}}\Rightarrow C_{\text{S}}=0.833\mu \text{F}\text{.}$$

Capacitance $C_{\text{Due south}}$ is connected in parallel with the third capacitance $C_3$ , so nosotros use Equation 8.8 to find the equivalent capacitance C of the entire network:

$$C=C_{\text{South}}+C_3=0.833\mu \text{F}+8.000\mu \text{F}=\mathrm{eight.833}\mu \text{F}\text{.}$$

Example 8.7

Network of Capacitors

Determine the cyberspace capacitance C of the capacitor combination shown in Figure 8.14 when the capacitances are $C_{\mathrm{one}}=12.0\mu \text{F},C_2=2.0\mu \text{F},$ and $C_3=\mathrm{four.0}\mu \text{F}$ . When a 12.0-V potential difference is maintained beyond the combination, find the accuse and the voltage across each capacitor.

Figure 8.14 (a) A capacitor combination. (b) An equivalent two-capacitor combination.

Strategy

We beginning compute the net capacitance $C_{23}$ of the parallel connection $C_2$ and $C_3$ . So C is the net capacitance of the serial connexion $C_{\mathrm{one}}$ and $C_{23}$ . We employ the relation $C=Q\text{/}V$ to find the charges $Q_1$ , $Q_2$ , and $Q_3$ , and the voltages $5_1$ , $V_2$ , and ${\mathrm{Five}}_3$ , across capacitors 1, 2, and iii, respectively.

Solution

The equivalent capacitance for $C_2$ and $C_3$ is

$$C_{23}=C_2+C_{\mathrm{iii}}=2.0\mu \text{F}+\mathrm{iv.0}\mu \text{F}=6.0\mu \text{F}\text{.}$$

The unabridged iii-capacitor combination is equivalent to two capacitors in series,

$$\frac{\mathrm{one}}{C}=\frac{1}{12.0\mu \text{F}}+\frac{1}{\mathrm{vi.0}\mu \text{F}}=\frac{1}{4.0\mu \text{F}}\Rightarrow C=4.0\mu \text{F}\text{.}$$

Consider the equivalent two-capacitor combination in Effigy viii.xiv(b). Since the capacitors are in series, they accept the same charge, $Q_1=Q_{23}$ . Also, the capacitors share the 12.0-Five potential difference, so

$$12.0\text{V}={\mathrm{Five}}_{\mathrm{ane}}+V_{23}=\frac{Q_1}{C_{\mathrm{ane}}}+\frac{Q_{23}}{C_{23}}=\frac{Q_{\mathrm{i}}}{12.0\mu \text{F}}+\frac{Q_{\mathrm{one}}}{6.0\mu \text{F}}\Rightarrow Q_1=48.0\mu \text{C}.$$

Now the potential difference across capacitor i is

$$5_1=\frac{Q_{\mathrm{ane}}}{C_{\mathrm{one}}}=\frac{48.0\mu \text{C}}{12.0\mu \text{F}}=\mathrm{iv.0}\text{V}\text{.}$$

Because capacitors 2 and 3 are connected in parallel, they are at the same potential departure:

$$V_{\mathrm{ii}}={\mathrm{Five}}_3=12.0\text{Five}-4.0\text{V}=8.0\text{V}.$$

Hence, the charges on these two capacitors are, respectively,

$$\begin{array}{l}{Q}_2=C_2{5}_2=(2.0\mu \text{F})(8.0\text{Five})=\mathrm{xvi.0}\mu \text{C,}\hfill \\ Q_3=C_3{V}_3=(4.0\mu \text{F})(\mathrm{eight.0}\text{V})=32.0\mu \text{C}\text{.}\hfill \end{array}$$

Significance

As expected, the cyberspace charge on the parallel combination of $C_2$ and $C_3$ is $Q_{23}=Q_{\mathrm{two}}+Q_{\mathrm{three}}=48.0\mu \text{C}\text{.}$

Cheque Your Understanding 8.v

Check Your Agreement Determine the cyberspace capacitance C of each network of capacitors shown below. Assume that $C_1=1.0\text{pF}$ , $C_2=\mathrm{two.0}\text{pF}$ , $C_3=\mathrm{iv.0}\text{pF}$ , and $C_{\mathrm{iv}}=\mathrm{five.0}\text{pF}$ . Observe the charge on each capacitor, assuming there is a potential divergence of 12.0 V across each network.

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Source: https://openstax.org/books/university-physics-volume-2/pages/8-2-capacitors-in-series-and-in-parallel

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