6.3 Series and Parallel Capacitors

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6.3 Series and Parallel Capacitors

We know from resisti ve circuits that the series-parallel combination is a powerful tool for reducing circuits. This technique can be e xtended to series-parallel connections of capacitors, which are sometimes encoun tered. We desire to replace these capacitors by a single equivalent capacitor Ceq.

In order to obtain the equi valent capacitor Ceq of N capacitors in parallel, consider the circuit in Fig. 6.14(a). The equi valent circuit is in Fig. 6.14(b). Note that the capacitors ha ve the same voltage v across them. Applying KCL to Fig. 6.14(a),

$$i = i_1 + i_2 + i_3 + \dots + i_N \tag{6.11}$$

But ik = Ck dv∕dt. Hence,

$$i = C_1 \frac{dv}{dt} + C_2 \frac{dv}{dt} + C_3 \frac{dv}{dt} + \dots + C_N \frac{dv}{dt}$$

=

$$\left(\sum_{k=1}^{N} C_k\right) \frac{dv}{dt} = C_{eq} \frac{dv}{dt}$$

(6.12)

where

$$C_{\text{eq}} = C_1 + C_2 + C_3 + \dots + C_N \tag{6.13}$$

The equivalent capacitance of N parallel-connected capacitors is the sum of the individual capacitances.

We observe that capacitors in parallel combine in the same manner as resistors in series.

We no w obtain Ceq of N capacitors connected in series by comparing the circuit in Fig. 6.15(a) with the equi valent circuit in Fig. 6.15(b). Note that the same current i flows (and consequently the same charge) through the capacitors. Applying KVL to the loop in Fig. 6.15(a),

$$v = v_1 + v_2 + v_3 + \dots + v_N \tag{6.14}$$

But vk=___1 Ck ∫ t0 t i(τ) dτ+vk(t0). Therefore,

$$v = \frac{1}{C_1} \int_{t_0}^t i(\tau) d\tau + v_1(t_0) + \frac{1}{C_2} \int_{t_0}^t i(\tau) d\tau + v_2(t_0)$$

• $\cdots + \frac{1}{C_N} \int_{t_0}^t i(\tau) d\tau + v_N(t_0)$
  = $\left(\frac{1}{\tau_0} + \frac{1}{\tau_0} + \cdots + \frac{1}{\tau_N}\right) \int_0^t i(\tau) d\tau + v_1(t_0) + v_2$

$$= \left(\frac{1}{C_1} + \frac{1}{C_2} + \dots + \frac{1}{C_N}\right) \int_{t_0}^t i(\tau) \, d\tau + v_1(t_0) + v_2(t_0) + \dots + v_N(t_0)$$ $$=\frac{1}{C_{\text{eq}}}\int_{t_0}^t i(\tau)\,d\tau + v(t_0)$$

where

$$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots + \frac{1}{C_N}$$

(6.16)

Figure 6.14

(a) Parallel-connected N capacitors, (b) equivalent circuit for the parallel capacitors.

(a)

(6.15) Figure 6.15

(a) Series-connected N capacitors, (b) equivalent circuit for the series capacitor.

The initial voltage v(t0) across Ceq is required by KVL to be the sum of the capacitor voltages at t0. Or according to Eq. (6.15),

$$v(t_0) = v_1(t_0) + v_2(t_0) + \dots + v_N(t_0)$$

Thus, according to Eq. (6.16),

The equivalent capacitance of series-connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitances.

Note that capacitors in series combine in the same manner as resistors in parallel. For N = 2 (i.e., two capacitors in series), Eq. (6.16) becomes

___1 Ceq = ___1 C1 + ___1 C2

or

$$C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} \tag{6.17}$$

Example 6.6 Find the equi valent capacitance seen between terminals a and b of the circuit in Fig. 6.16.

Solution:

The 20- μF and 5- μF capacitors are in series; their equivalent capaci tance is

$$\frac{20 \times 5}{20 + 5} = 4 \ \mu\text{F}$$

This 4-μF capacitor is in parallel with the 6- μF and 20- μF capacitors; their combined capacitance is

$$4 + 6 + 20 = 30 \,\mu\text{F}$$

This 30-μF capacitor is in series with the 60- μF capacitor. Hence, the equivalent capacitance for the entire circuit is

$$C_{\text{eq}} = \frac{30 \times 60}{30 + 60} = 20 \,\mu\text{F}$$

Find the equivalent capacitance seen at the terminals of the circuit in Practice Problem 6.6 Fig. 6.17.

Answer: 40 μF.

For Practice Prob. 6.6.

For the circuit in Fig. 6.18, find the voltage across each capacitor. Example 6.7

Solution:

We first find the equivalent capacitance Ceq, shown in Fig. 6.19. The two parallel capacitors in Fig. 6.18 can be combined to get 40 + 20 = 60 mF. This 60-mF capacitor is in series with the 20-mF and 30-mF capacitors. Thus,

$$C_{\text{eq}} = \frac{1}{\frac{1}{60} + \frac{1}{30} + \frac{1}{20}} \text{ mF} = 10 \text{ mF}$$

The total charge is

$$q = C_{\text{eq}}v = 10 \times 10^{-3} \times 30 = 0.3 \text{ C}$$

This is the charge on the 20-mF and 30-mF capacitors, because they are in series with the 30-V source. (A crude way to see this is to imagine that charge acts like current, since i = dq∕dt.) Therefore,

$$v_1 = \frac{q}{C_1} = \frac{0.3}{20 \times 10^{-3}} = 15 \text{ V}$$

$v_2 = \frac{q}{C_2} = \frac{0.3}{30 \times 10^{-3}} = 10 \text{ V}$

Having determined v1 and v2, we now use KVL to determine v3 by

$$v_3 = 30 - v_1 - v_2 = 5$$

V

Alternatively, since the 40-mF and 20-mF capacitors are in parallel, they have the same voltage v3 and their combined capacitance is 40 + 20 = 60 mF. This combined capacitance is in series with the 20-mF and 30-mF capacitors and consequently has the same charge on it. Hence,

$$v_3 = \frac{q}{60 \text{ mF}} = \frac{0.3}{60 \times 10^{-3}} = 5 \text{ V}$$

Find the voltage across each of the capacitors in Fig. 6.20. Practice Problem 6.7

Answer:

$$v_1 = 75 \text{ V}, v_2 = 75 \text{ V}, v_3 = 25 \text{ V}, v_4 = 50 \text{ V}.$$

Figure 6.18 For Example 6.7.

Figure 6.19 Equivalent circuit for Fig. 6.18.

Figure 6.20 For Practice Prob. 6.7.

Figure 6.21 Typical form of an inductor.

In view of Eq. (6.18), for an inductor to have voltage across its terminals, its current must vary with time. Hence, ^v = 0 for constant current through the inductor.

Figure 6.22

Various types of inductors: (a) solenoidal wound inductor, (b) toroidal inductor, (c) axial lead inductor.

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