Question

(a) In Fig. 25-19a, are capacitors 1 and 3 in series? (b) In the samefigure, are capacitors 1 and 2 in parallel? (c) Rank the equivalent capacitances of the four circuits shown in Fig. 25-19, greatest first.

Short answer

a) In Fig.25-19(a), the capacitor 1 and 3 are not in series.

b) In Fig.25-19(a), the capacitor 1 and 2 are in parallel.

c) The equivalent capacitances of the four circuits shown in Fig.25-19 are same.

Step-by-step solution

The given data

The fig.25-19 for all the three capacitors is given.

Understanding the concept of the equivalent capacitance

Using the information from fig.25-19, we can find in fig. 259(a); that capacitors 1 and 3 are in series, and capacitors 1 and 2 are in parallel. Also by using the information from fig.25-19 and formulae for the equivalent capacitance for series and parallel arrangements, we can find the ranks of the equivalent capacitances of the four circuits shown in fig.25-19.

Formulae:

If capacitors are in series, the equivalent capacitance$C_{eq}$is given by,

$\frac{1}{C_{eq}}=\sum _{}\frac{1}{C}$ …(i)

If capacitors are in parallel, the equivalent capacitance$C_{eq}$is given by,

$C_{eq}=\sum _{}C$ …(ii)

(a) Calculation to check whether the capacitors 1 and 3 are in series

In fig.25-19 (a) we consider only $C_1$and $C_2$, then we can see that the capacitor 1 and 2 are in parallel but $C_{12}$and $C_3$are in series.

Therefore, the capacitors 1 and 3 are not in series.

(b) Calculation to check whether the capacitors 1 and 2 are in parallel

As stated in part , we can see that the capacitors 1 and 2 are in parallel.

(c) Ranking the circuits with their equivalent capacitance

Let’s calculate the equivalent capacitance of all the circuits to compare them.

Circuit (a)

From the above diagram, we can see that the capacitors$C_1$and $C_2$are in parallel, as they are connected between the two common points at A and B andlocalid="1661493749423" $C_3$is in series with these two capacitors.

Find the equivalent capacitancelocalid="1661493755515" $C_{12}$oflocalid="1661493764078" $C_1$andlocalid="1661493767805" $C_2$first.

localid="1661493771006" $C_{12}=C_1+C_2$

localid="1661493776093" $C_{12}$is in series with localid="1661493784923" $C_3$. So, the equivalent capacitance of the entire circuit is,

localid="1661493789257" $$\begin{gathered} \frac{1}{C_{eq}}=\frac{1}{C_{12}}+\frac{1}{C_3} \\ =\frac{1}{C_1+C_2}+\frac{1}{C_3} \\ =\frac{C_1+C_2+C_3}{\left(C_1+C_2\right).C_3} \\ {C}_{eq}=\frac{\left(C_1+C_2\right).C_2}{C_1+C_2+C_3} \end{gathered}$$

Circuit (b)

From the above diagram, we can see that the capacitorslocalid="1661493794425" $C_1$andlocalid="1661493798640" $C_2$are in parallel, as they are connected between the two common points at A and B andlocalid="1661493803482" $C_3$is in series with these two capacitors.

Find the equivalent capacitancelocalid="1661493817944" $C_{12}$oflocalid="1661493822264" $C_1$andlocalid="1661493827991" $C_2$first.

localid="1661493833052" $C_{12}=C_1+C_2$

localid="1661493837187" $C_{12}$is in series withlocalid="1661493841057" $C_3$. So, the equivalent capacitance of the entire circuit is,

localid="1661493844789" $$\begin{gathered} \frac{1}{C_{eq}}=\frac{1}{C_{12}}+\frac{1}{C_3} \\ =\frac{1}{C_1+C_2}+\frac{1}{C_3} \\ =\frac{C_1+C_2+C_3}{\left(C_1+C_2\right).C_3} \\ {C}_{eq}=\frac{\left(C_1+C_2\right).C_2}{C_1+C_2+C_3} \end{gathered}$$

Circuit (c)

From the above diagram, we can see that the capacitors and are in parallel, as they are connected between the two common points at A and B and is in series with these two capacitors.

Find the equivalent capacitance of and first.

is in series with . So, the equivalent capacitance of the entire circuit is,

localid="1661493850267" $$\begin{gathered} \frac{1}{C_{eq}}=\frac{1}{C_{12}}+\frac{1}{C_3} \\ =\frac{1}{C_1+C_2}+\frac{1}{C_3} \\ =\frac{C_1+C_2+C_3}{\left(C_1+C_2\right).C_3} \\ {C}_{eq}=\frac{\left(C_1+C_2\right).C_2}{C_1+C_2+C_3} \end{gathered}$$

Circuit (d)

From the above diagram, we can see that the capacitors$C_1$and$C_2$are in parallel, as they are connected between the two common points at A and B and$C_3$is in series with these two capacitors.

Find the equivalent capacitancelocalid="1661493857338" $C_{12}$oflocalid="1661493868162" $C_1$andlocalid="1661493876675" $C_2$first.

localid="1661493886060" $C_{12}=C_1+C_2$

is in series with . So, the equivalent capacitance of the entire circuit is,

localid="1661493891061" $$\begin{gathered} \frac{1}{C_{eq}}=\frac{1}{C_{12}}+\frac{1}{C_3} \\ =\frac{1}{C_1+C_2}+\frac{1}{C_3} \\ =\frac{C_1+C_2+C_3}{\left(C_1+C_2\right).C_3} \\ {C}_{eq}=\frac{\left(C_1+C_2\right).C_2}{C_1+C_2+C_3} \end{gathered}$$

From the above calculations of equivalent capacitance of circuits, (a), (b), (c) and (d), we can see that the equivalent capacitance of all the circuit is same. So, we can rank them at the same level.