Chapter 24, Problem 33P

(a)

To determine

To Calculate:The equivalent capacitance of twocapacitor in series.

Answer to Problem 33P

  $C_{eq}=\frac{C_1{C}_2}{C_2+C_1}$

Explanation of Solution

Given information:

Two capacitors $C_1$ and $C_2$

Formula used:

The charge stored in a capacitor:

  $Q=CV$

Where, C is the capacitance and V is the potential.

Calculation:As the capacitors are connected in series. So, charge on each capacitor is Q.

Let equivalent capacitance $=C_{eq}$

This series combination is connected to a voltage source V.

So, voltage across $C_1$ and $C_2$ is $V_1=\frac{Q}{C_1}$ and $V_2=\frac{Q}{C_2}$

Total voltage $V=V_1+V_2$

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

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

  $\frac{1}{C_{eq}}=\frac{C_2+C_1}{C_1{C}_2}$

  $C_{eq}=\frac{C_1{C}_2}{C_2+C_1}$

(b)

To determine

To Proof: $C_{eq}$ must be less than $C_1$ and $C_2$ .

Answer to Problem 33P

  $C_{eq}<C_2$ and $C_{eq}<C_1$

Explanation of Solution

Given information:

From part (a), equivalent capacitance of two capacitors is $C_{eq}=\frac{C_1{C}_2}{C_2+C_1}$

Calculation:Dividing numerator and denominator by $C_1$

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

As $\frac{C_2}{C_1}+1$ >1

So $C_{eq}<C_2$

Dividing numerator and denominator by $C_2$

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

As $\frac{C_1}{C_2}+1>1$

So $C_{eq}<C_1$

So $C_{eq}$ must be less than $C_1$ and $C_2$ and hence must be less than the smaller of two values.

(c)

To determine

To Calculate:The equivalent capacitance of three capacitors in series.

Answer to Problem 33P

  $C_{eq}=\frac{C_1{C}_2{C}_3}{C_2{C}_3+C_1{C}_3+C_1{C}_2}$

Explanation of Solution

Given information:

The three capacitors $C_1$ , $C_2$ and $C_3$

Formula used:

The charge stored in a capacitor:

  $Q=CV$

Where, C is the capacitance and V is the potential.

Calculation: As the capacitors are connected in series. So, charge on each capacitor is Q.

Let equivalent capacitance $=C_{eq}$

This series combination is connected to a voltage source V.

So, the voltage across $C_1,C_2\text{and}{\text{ C}}_{\text{3}}$ is $V_1=\frac{Q}{C_1}$

  $V_2=\frac{Q}{C_2}$ and $V_3=\frac{Q}{C_3}$ respectively

Total voltage, $V=V_1+V_2$ + $V_3$

  $\frac{Q}{C_{eq}}=\frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3}$

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

  $\frac{1}{C_{eq}}=\frac{C_2{C}_3+C_1{C}_3+C_1{C}_2}{C_1{C}_2{C}_3}$

  $C_{eq}=\frac{C_1{C}_2{C}_3}{C_2{C}_3+C_1{C}_3+C_1{C}_2}$

(d)

To determine

To Calculate: $C_{eq}$ must be less than $C_1$ , $C_2$ and $C_3$ .

Answer to Problem 33P

  $C_{eq}<C_2$ , $C_{eq}<C_1$ and $C_{eq}<C_3$

Explanation of Solution

Given information:

Equivalent capacitance of three capacitors is $C_{eq}=\frac{C_1{C}_2{C}_3}{C_2{C}_3+C_1{C}_3+C_1{C}_2}$ .

Calculation: $C_{eq}=\left(\frac{C_1{C}_2}{C_2{C}_3+C_1{C}_3+C_1{C}_2}\right)C_3$

Dividing numerator and denominator by $C_1{C}_2$

  $C_{eq}=\left(\frac{1}{\frac{C_3}{C_1}+\frac{C_3}{C_2}+1}\right)C_3$

As $\frac{C_3}{C_1}+\frac{C_3}{C_2}+1>1$

So $C_{eq}<C_3$

  $C_{eq}$ can also be written as $C_{eq}=\left(\frac{C_1{C}_3}{C_2{C}_3+C_1{C}_3+C_1{C}_2}\right)C_2$

Dividing numerator and denominator by $C_1{C}_3$

  $C_{eq}=\left(\frac{1}{\frac{C_2}{C_1}+1+\frac{C_2}{C_3}}\right)C_2$

As $\frac{C_2}{C_1}+1+\frac{C_2}{C_3}>1$

So $C_{eq}$2

  $C_{eq}$ can also be written as

  $C_{eq}=(\frac{C_3{C}_2}{C_2{C}_3+C_1{C}_3+C_1{C}_2})C_1$

Dividing numerator and denominator by $C_2{C}_3$ , we have

  $C_{eq}=\left(\frac{1}{1+\frac{C_1}{C_2}+\frac{C_1}{C_3}}\right)C_1$

As $1+\frac{C_1}{C_2}+\frac{C_1}{C_3}$

So $C_{eq}<C_1$

So, $C_{eq}$ must be less than $C_1$ , $C_2$ and C ₃ and hence must be less than the smaller of three values.