8.3: Capacitores em série e em paralelo
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- 184619
Ao final desta seção, você poderá:
- Explique como determinar a capacitância equivalente de capacitores em série e em combinações paralelas
- Calcule a diferença de potencial entre as placas e a carga nas placas para um capacitor em uma rede e determine a capacitância líquida de uma rede de capacitores
Vários capacitores podem ser conectados juntos para serem usados em uma variedade de aplicações. Várias conexões de capacitores se comportam como um único capacitor equivalente. A capacitância total desse capacitor único equivalente depende dos capacitores individuais e de como eles estão conectados. Os capacitores podem ser organizados em dois tipos simples e comuns de conexões, conhecidas como séries e paralelas, para as quais podemos calcular facilmente a capacitância total. Essas duas combinações básicas, série e paralela, também podem ser usadas como parte de conexões mais complexas.
A combinação de capacitores em série
\(\PageIndex{1}\)A figura ilustra uma combinação em série de três capacitores, dispostos em uma linha dentro do circuito. Como em qualquer capacitor, a capacitância da combinação está relacionada à carga e à tensão:
\[ C=\dfrac{Q}{V}.\]
Quando essa combinação em série é conectada a uma bateria com tensão V, cada um dos capacitores adquire uma carga Q idêntica. Para explicar, primeiro observe que a carga na placa conectada ao terminal positivo da bateria é\(+Q\) e a carga na placa conectada ao terminal negativo é\(-Q\). As cargas são então induzidas nas outras placas para que a soma das cargas em todas as placas e a soma das cargas em qualquer par de placas de capacitores seja zero. No entanto, a queda de potencial\(V_1 = Q/C_1\) em um capacitor pode ser diferente da queda de potencial\(V_2 = Q/C_2\) em outro capacitor, porque, geralmente, os capacitores podem ter capacitâncias diferentes. A combinação em série de dois ou três capacitores se assemelha a um único capacitor com uma capacitância menor. Geralmente, qualquer número de capacitores conectados em série é equivalente a um capacitor cuja capacitância (chamada de capacitância equivalente) é menor do que a menor das capacitâncias na combinação em série. A carga neste capacitor equivalente é a mesma de qualquer capacitor em uma combinação em série: ou seja, todos os capacitores de uma combinação em série têm a mesma carga. Isso ocorre devido à conservação da carga no circuito. Quando uma carga Q em um circuito em série é removida de uma placa do primeiro capacitor (que denotamos como\(-Q\)), ela deve ser colocada em uma placa do segundo capacitor (que denotamos como\(+Q\)) e assim por diante.
We can find an expression for the total (equivalent) capacitance by considering the voltages across the individual capacitors. The potentials across capacitors 1, 2, and 3 are, respectively, \(V_1 = Q/C_1\), \(V_2 = Q/C_2\), and \(V_3 = Q/C_3\). 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 \(C_S\). Entering the expressions for \(V_1\), \(V_2\), and \(V_3\), we get
\[\dfrac{Q}{C_S} = \dfrac{Q}{C_1} + \dfrac{Q}{C_2} + \dfrac{Q}{C_3}.\]
Canceling the charge Q, we obtain an expression containing the equivalent capacitance, \(C_S\), of three capacitors connected in series:
\[\dfrac{1}{C_S} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3}.\]
This expression can be generalized to any number of capacitors in a series network.
For capacitors connected in a series combination, the reciprocal of the equivalent capacitance is the sum of reciprocals of individual capacitances:
\[\dfrac{1}{C_S} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3} + \dots \label{capseries}\]
Find the total capacitance for three capacitors connected in series, given their individual capacitances are \(1.000 \mu F\), \(5.000 \mu F\), and \(8.000 \mu F\).
Strategy
Because there are only three capacitors in this network, we can find the equivalent capacitance by using Equation \ref{capseries} with three terms.
Solution
We enter the given capacitances into Equation \ref{capseries}:
\[ \begin{align*} \dfrac{1}{C_S} &= \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3} \\[4pt] &= \dfrac{1}{1.000 \mu F} + \dfrac{1}{5.000 \mu F} + \dfrac{1}{8.000 \mu F} \\[4pt] &= \dfrac{1.325}{\mu F}.\end{align*} \]
Now we invert this result and obtain
\[ \begin{align*} C_S &= \dfrac{\mu F}{1.325} \\[4pt] &= 0.755 \mu F.\end{align*} \nonumber\]
Significance
Note that in a series network of capacitors, the equivalent capacitance is always 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 \(\PageIndex{2a}\). 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 store a different charge. To find the equivalent capacitance \(C_p\) 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-hand side of this equation, we use the relation \(Q = C_pV\), which holds for the entire network. On the right-hand side of the equation, we use the relations \(Q_1 = C_1 V\), \(Q_2 = C_2V\), and \(Q_3 = C_3V\) for the three capacitors in the network. In this way we obtain
\[C_pV = C_1V + C_2V + C_3V.\]
This equation, when simplified, is the expression for the equivalent capacitance of the parallel network of three capacitors:
\[C_p = C_1 + C_2 + C_3.\]
This expression is easily generalized to any number of capacitors connected in parallel in the network.
For capacitors connected in a parallel combination, the equivalent (net) capacitance is the sum of all individual capacitances in the network,
\[C_p = C_1 + C_2 + C_3 + ... \label{capparallel}\]
Find the net capacitance for three capacitors connected in parallel, given their individual capacitances are \(1.0 \mu F\), \(5.0 \mu F\), and \(8.0 \mu F\).
Strategy
Because there are only three capacitors in this network, we can find the equivalent capacitance by using Equation \ref{capparallel} with three terms.
Solution
Entering the given capacitances into Equation \ref{capparallel} yields
\[\begin{align*} C_p &= C_1 + C_2 + C_3 \\[4pt] &= 1.0 \mu F + 5.0 \mu F + 8.0 \mu F \\[4pt] &= 14.0 \mu F. \end{align*}\]
Significance
Note 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 usually some combination of series and parallel connections, as shown in Figure \(\PageIndex{3}\). To find the net capacitance of such combinations, we identify parts that contain only series or only parallel connections, and find their equivalent capacitances. We repeat this process until we can determine the equivalent capacitance of the entire network. The following example illustrates this process.
Find the total capacitance of the combination of capacitors shown in Figure \(\PageIndex{3}\). Assume the capacitances are known to three decimal places (\(C_1 = 1.000 \mu F, C_2 = 5.000 \mu F, C_3 = 8.000 \mu F\)). Round your answer to three decimal places.
Strategy
We first identify which capacitors are in series and which are in parallel. Capacitors \(C_1\) and \(C_2\) are in series. Their combination, labeled \(C_S\) is in parallel with \(C_3\).
Solution
Since \(C_1\) and \(C_2\) are in series, their equivalent capacitance \(C_S\) is obtained with Equation \ref{capseries}:
\[\begin{align*} \dfrac{1}{C_S} &= \dfrac{1}{C_1} + \dfrac{1}{C_2} \\[4pt] &= \dfrac{1}{1.000 \mu F} + \dfrac{1}{5.000 \mu F} \\[4pt] &= \dfrac{1.200}{\mu F} \end{align*}\]
Therefor
\[ C_S = 0.833 \mu F. \nonumber\]
Capacitance \(C_S\) is connected in parallel with the third capacitance \(C_3\), so we use Equation \ref{capparallel} find the equivalent capacitance C of the entire network:
\[\begin{align*} C &= C_S + C_3 \\[4pt] &= 0.833 \mu F + 8.000 \mu F \\[4pt] &= 8.833 \mu F. \end{align*}\]
Determine the net capacitance C of the capacitor combination shown in Figure \(\PageIndex{4}\) when the capacitances are \(C_1 = 12.0 \mu F, C_2 = 2.0 \mu F\), and \(C_3 = 4.0 \mu F\). When a 12.0-V potential difference is maintained across the combination, find the charge and the voltage across each capacitor.
Strategy We first compute the net capacitance \(C_{23}\) of the parallel connection \(C_2\) and \(C_3\). Then C is the net capacitance of the series connection \(C_1\) and \(C_{23}\). We use the relation \(C = Q/V\) to find the charges \(Q_1, Q_2\), and \(Q_3\), and the voltages \(V_1, V_2\), and \(V_3\) across capacitors 1, 2, and 3, respectively.
Solution The equivalent capacitance for \(C_2\) and \(C_3\) is
\[C_{23} = C_2 + C_3 = 2.0 \mu F + 4.0 \mu F = 6.0 \mu F.\]
The entire three-capacitor combination is equivalent to two capacitors in series,
\[\dfrac{1}{C} = \dfrac{1}{12.0 \mu F} + \dfrac{1}{6.0 \mu F} = \dfrac{1}{4.0 \mu F} \Rightarrow C = 4.0 \mu F.\]
Consider the equivalent two-capacitor combination in Figure \(\PageIndex{2b}\). Since the capacitors are in series, they have the same charge, \(Q_1 = Q_{23}\). Also, the capacitors share the 12.0-V potential difference, so
\[12.0 V = V_1 + V_{23} = \dfrac{Q_1}{C_1} + \dfrac{Q_{23}}{C_{23}} = \dfrac{Q_1}{12.0 \mu F} + \dfrac{Q_1}{6.0 \mu F} \Rightarrow Q_1 = 48.0 \mu C.\]
Now the potential difference across capacitor 1 is
\[V_1 = \dfrac{Q_1}{C_1} = \dfrac{48.0 \mu C}{12.0 \mu F} = 4.0 V.\]
Because capacitors 2 and 3 are connected in parallel, they are at the same potential difference:
\[V_2 = V_3 = 12.0 V - 4.0 V = 8.0 V.\]
Hence, the charges on these two capacitors are, respectively,
\[Q_2 = C_2V_2 = (2.0 \mu F)(8.0 V) = 16.0 \mu C,\]
\[Q_3 = C_3V_3 = (4.0 \mu F)(8.0 V) = 32.0 \mu C.\]
Significance As expected, the net charge on the parallel combination of \(C_2\) and \(C_3\) is \(Q_{23} = Q_2 + Q_3 = 48.0 \mu C.\)
Determine the net capacitance C of each network of capacitors shown below. Assume that \(C_1 = 1.0 pF, C_2 = 2.0 pF, C_3 = 4.0 pF\), and \(C_4 = 5.0 pF\). Find the charge on each capacitor, assuming there is a potential difference of 12.0 V across each network.
- Responda a
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\(C = 0.86 pF, Q_1 = 10 pC, Q_2 = 3.4 pC, Q_3 = 6.8 pC\)
- Resposta b
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\(C = 2.3 pF, Q_1 = 12 pC, Q_2 = Q_3 = 16 pC\)
- Resposta c
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\(C = 2.3 pF, Q_1 = 9.0 pC, Q_2 = 18 pC, Q_3 = 12 pC, Q_4 = 15 pC\)