Chapter 25, Problem 65SP

Referring to Fig. 25-12, what is the equivalent capacitance between terminals A and B? Are any capacitors removable? Which capacitors are in series? Which are in parallel? [Hint: Redraw the circuit and watch for shorts.]

Fig. 25-12

To determine

The equivalent capacitance between terminals A and B in the circuit given in Fig. 25-12. Also, find the capacitors that are removable, connected in series, and connected in parallel.

Answer to Problem 65SP

Solution:

$9.0\mu \text{F}$, Capacitors $C_1$, $C_3$ , and $C_7$ are removable, $C_4$, $C_5$ are connected in parallel with $C_2$, $C_6$ , and $C_9$ are also connected in parallel.

Explanation of Solution

Given data:

Refer to the circuit given is Fig.25.12.

Formula used:

Write the expression for equivalent capacitance when they are connected in series.

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$

Here, $C_{eq}$ is the equivalent capacitance, $C_1$ and $C_2$ are the individual capacitances.

Write the expression for equivalent capacitance when they are connected in parallel.

$C_{eq}=C_1+C_2$

Explanation:

Consider the following figure 1.

If starting from A to B in the circuit shown in Figure 1, the capacitor $C_1$ is open circuited, the capacitors $C_3$ and $C_7$ are shorted. Therefore, the capacitors $C_1$, $C_3$, and $C_7$ are removable from the circuit.

The capacitors $C_4$ and $C_5$ are connected in parallel.

The expression for equivalent capacitance when they are connected in parallel is,

$C_{e{q}_1}=C_4+C_5$

Here, $C_{e{q}_1}$ is the equivalent capacitance of $C_4$ and $C_5$.

Substitute $3.0\mu \text{F}$ for $C_4$ and $3.0\mu \text{F}$ for $C_5$.

$\begin{array}{c}{C}_{e{q}_1}=3.0\mu \text{F}+3.0\mu \text{F}\\ =6.\text{0 }\mu \text{F}\end{array}$

Figure 2 is a reduced form of figure 1.

From figure 2, the resultantcombination $C_{e{q}_1}$ is connected in series with $C_2$.

The expression for equivalent capacitance, when they are connected in series is,

$\frac{1}{C_{e{q}_2}}=\frac{1}{C_{e{q}_1}}+\frac{1}{C_2}$

Here, $C_{e{q}_2}$ is the equivalent capacitance of $C_{e{q}_1}$ and $C_2$.

Substitute $6.0\mu \text{F }$ for $C_{e{q}_1}$ and $3.0\mu \text{F}$ for $C_2$

$\begin{array}{c}\frac{1}{C_{e{q}_2}}=\frac{1}{6.0\mu \text{F}}+\frac{1}{3.0\mu \text{F}}\\ \frac{1}{C_{e{q}_2}}=\frac{3}{6.0\mu \text{F}}\\ C_{e{q}_2}=2.0\mu \text{F}\end{array}$

The equivalent capacitance $C_{e{q}_2}$ is connected in parallel with $C_6$.

The expression for equivalent capacitance, when capacitors are connected in parallel is,

$C_{e{q}_3}=C_{e{q}_2}+C_6$

Here, $C_{e{q}_3}$ is the equivalent capacitance of $C_{e{q}_2}$ and $C_6$.

Substitute $2.0\mu \text{F}$ for $C_{e{q}_2}$ and $2.0\mu \text{F}$ for $C_6$

$\begin{array}{c}{C}_{e{q}_3}=2.0\mu \text{F}+2.0\mu \text{F}\\ =4.\text{0 }\mu \text{F}\end{array}$

The equivalent capacitance $C_{e{q}_3}$ is connected in parallel with $C_9$.

The expression for equivalent capacitance, when capacitors are connected in parallel is,

$C_{e{q}_4}=C_{e{q}_3}+C_9$

Here, $C_{e{q}_4}$ is the equivalent capacitance of $C_{e{q}_3}$ and $C_9$.

Substitute $4.0\mu \text{F}$ for $C_{e{q}_3}$ and $2.0\mu \text{F}$ for $C_9$

$\begin{array}{c}{C}_{e{q}_4}=4.0\text{ pF}+2.0\text{ pF}\\ =6.\text{0 pF}\end{array}$

The equivalent capacitance $C_{e{q}_4}$ is connected in series with $C_8$.

The expression for equivalent capacitance, when capacitors are connected in series is,

$\frac{1}{C_{e{q}_5}}=\frac{1}{C_{e{q}_4}}+\frac{1}{C_8}$

Here, $C_{e{q}_5}$ is the equivalent capacitance of $C_{e{q}_4}$ and $C_8$.

Substitute $6.0\mu \text{F }$ for $C_{e{q}_4}$ and $3.0\mu \text{F}$ for $C_8$

$\begin{array}{c}\frac{1}{C_{e{q}_5}}=\frac{1}{6.0\mu \text{F}}+\frac{1}{3.0\mu \text{F}}\\ \frac{1}{C_{e{q}_5}}=\frac{3}{6.0\mu \text{F}}\\ C_{e{q}_5}=2.0\mu \text{F}\end{array}$

The equivalent capacitance $C_{e{q}_5}$ is connected in parallel with $C_{11}$.

The expression for equivalent capacitance, when capacitors are connected in parallel is,

$C_{e{q}_6}=C_{e{q}_5}+C_{11}$

Here, $C_{e{q}_6}$ is the equivalent capacitance of $C_{e{q}_5}$ and $C_{11}$.

Substitute $2.0\mu \text{F}$ for $C_{e{q}_5}$ and $2.0\mu \text{F}$ for $C_{11}$

$\begin{array}{c}{C}_{e{q}_6}=2.0\text{ pF}+2.0\text{ pF}\\ =4.\text{0 pF}\end{array}$

The equivalent capacitance $C_{e{q}_6}$ is connected in series with $C_{10}$.

The expression for equivalent capacitance, when capacitors are connected in series is,

$\frac{1}{C_{e{q}_7}}=\frac{1}{C_{e{q}_6}}+\frac{1}{C_{10}}$

Here, $C_{e{q}_7}$ is the equivalent capacitance of $C_{e{q}_6}$ and $C_{10}$.

Substitute $4.0\mu \text{F }$ for $C_{e{q}_6}$ and $12.0\mu \text{F}$ for $C_{10}$

$\begin{array}{c}\frac{1}{C_{e{q}_7}}=\frac{1}{4.0\mu \text{F}}+\frac{1}{12.0\mu \text{F}}\\ \frac{1}{C_{e{q}_7}}=\frac{4}{12.0\mu \text{F}}\\ C_{e{q}_7}=3.0\mu \text{F}\end{array}$

The equivalent capacitance $C_{e{q}_7}$ is connected in parallel with $C_{12}$.

The expression for equivalent capacitance, when capacitors are connected in parallel is,

$C_{AB}=C_{e{q}_7}+C_{12}$

Here, $C_{AB}$ is the equivalent capacitance of $C_{e{q}_7}$ and $C_{12}$.

Substitute $3.0\mu \text{F}$ for $C_{e{q}_7}$ and $6.0\mu \text{F}$ for $C_{12}$

$\begin{array}{c}{C}_{AB}=3.0\mu \text{F}+6.0\mu \text{F}\\ =9.\text{0 }\mu \text{F}\end{array}$

Thus, the equivalent capacitance between terminals A and B in the given circuit is $C_{AB}=9.0\mu \text{F}$.

Conclusion:

The equivalent capacitance between terminals A and B in the given circuit is $9.0\mu \text{F}$. The capacitors $C_1$, $C_3$ and $C_7$ are removable, $C_4$, $C_5$ are connected in parallel with $C_2$, $C_6$ , and $C_9$ are also connected in parallel.