Chapter 24, Problem 32P

(a)

To determine

The equivalent capacitance of the combination.

Answer to Problem 32P

The equivalent capacitance of the combinationis $15.157\text{μF}$ .

Explanation of Solution

Given:

Two capacitors $\text{4}\text{μF}\text{and}\text{15}\text{μF}$ are connected in series and this series combination is in parallelwith $12\text{μF}$

Formula used:

1. If two capacitors $C_{1}and{C}_2$ are connected in series then their equivalent capacitance will be $\frac{C_1\times C_2}{C_1+C_2}$
2. If two capacitors $C_{1}and{C}_2$ are connected in parallel then their equivalent capacitance will be $C_1+C_2$

Calculation:

To find equivalent capacitance between the terminals

Assume the terminals as ‘a’ and ‘b’.

  

As $4\mu Fand15\mu F$ capacitors are connected in series their equivalent capacitance will be

  $\begin{array}{l}{C}_{eq}=\frac{\text{4μF}\text{×}\text{15}\text{μF}}{\text{4μF}\text{+}\text{15}\text{μF}}\\ \text{=}\text{3}\text{.157}\text{μF}\end{array}$

Now the given circuit can be reduced to

  

Now $3.15\text{7}\text{μF}\text{and}\text{12}\text{μF}$ capacitors are connected in parallel. Therefore, total equivalent capacitance $C_{ab}$ will be :

  $\begin{array}{c}{C}_{ab}=\text{3}\text{.157}\text{μF}\text{+}\text{12}\text{μF}\\ \text{=}\text{15}\text{.157}\text{μF}\end{array}$

Conclusion:

The equivalent capacitance between the terminals ‘a’ and ‘b’ is $15.157\text{μF}$ .

(b)

To determine

The charge stored on positive charged plate of each capacitor.

Answer to Problem 32P

The charge stored on positively charged plate of each capacitor is $q_4=631.57\mu C;q_{15}=631.57\mu Cand{q}_{12}=2400\mu C$ .

Explanation of Solution

Given:

Two capacitors $4\text{μF}\text{and}\text{15}\text{μF}$ are connected in series and this series combination is in parallel with $12\text{μF}$

Formula used:

Charge = Capacitance x voltage

  $\Rightarrow q=c\times v$

Calculation:

The voltage in the given circuit is found out by applying Voltage division rule (as explained in sub part (c)).

  

Therefore, the charge stored on $15\mu F$ capacitor will be

  $\begin{array}{c}q=15\times {10}^{-6}\times 42.105\\ =631.57\text{μ}\text{C}\end{array}$

The charge stored on $4\mu F$ capacitor will be

  $\begin{array}{c}q=4\times {10}^{-6}\times 157.894\\ =631.57\mu C\end{array}$

Similarly, charge on $12\mu F$ capacitor will be

  $\begin{array}{c}q=12\times {10}^{-6}\times 200\\ =2400\mu C\end{array}$

Conclusion:

The charge stored on positively charged plate of each capacitor is $q_4=631.57\mu C;q_{15}=631.57\mu Cand{q}_{12}=2400\mu C$ .

(c)

To determine

The voltage across each capacitor.

Answer to Problem 32P

The voltage across each capacitor is $V_{4\mu \text{F}}=157.89\text{4}\text{V};V_{15\mu \text{F}}=42.105\text{V}$ .

Explanation of Solution

Given:

The supply voltage $V_s\left(t\right)=200V$

Formula used: If two capacitors $C_{1}and{C}_2$ are connected in series, then voltage across each capacitor will be

  $\begin{array}{l}{V}_1=\frac{C_2}{C_1+C_2}{V}_s\left(t\right)\\ V_2=\frac{C_1}{C_1+C_2}{V}_s\left(t\right)\end{array}$

Where, $V_s\left(t\right)$ is the supply voltage.

Calculation:

Thevoltage across $\text{4}\text{μF}\text{and}\text{15}\text{μF}$ series combination will be 200V(applying voltage division rule)

The voltage across $12\text{μF}$ capacitor will be 200V( $\because$ it is directly connected to source)

  

  

The voltage across $4\text{μF}$ is

  $\begin{array}{c}{V}_4=200\times \frac{\text{15μF}}{\text{4μF+15μF}}\\ =157.894V\end{array}$

The voltage across $15\text{μF}$ is

  $\begin{array}{c}{V}_{15}=200\times \frac{4\mu F}{4\mu F+15\mu F}\\ =42.105V\end{array}$

By applying Kirchhoff’s law (Algebraic sum of the voltage drops in a loop is zero),

  $\begin{array}{l}200=157.89+42.105\\ 200\simeq 199.99\text{V}\end{array}$

Conclusion:

The voltage across each capacitor is $V_{4\mu \text{F}}=157.89\text{4}\text{V};V_{15\mu \text{F}}=42.105\text{V}$ .

(d)

To determine

The energy stored in each capacitor.

Answer to Problem 32P

The energy stored in each capacitor is:

  $E_4=0.0498J;E_{15}=0.0132J;E_{12}=0.24J$

Explanation of Solution

Given:

The values of capacitors and supply voltage = 200V

Formula used: The energy stored will be given as:

  $E=\frac{1}{2}\times C\times V^2$

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

Calculation:

The energy stored in $4\text{μF}$ capacitor will be

  $\begin{array}{c}{E}_4=\frac{1}{2}\times 4\times {10}^{-6}\times {\left(157.894\right)}^2\\ =0.0498\text{J}\end{array}$

The energy stored in $15\text{μF}$ capacitor will be

  $\begin{array}{c}{E}_{15}=\frac{1}{2}\times 15\times {10}^{-6}\times {\left(42.105\right)}^2\\ =0.0132\text{J}\end{array}$

The energy stored in $12\text{μF}$ capacitor will be

  $\begin{array}{c}{E}_{12}=\frac{1}{2}\times 12\times {10}^{-6}\times {\left(200\right)}^2\\ =0.24\text{J}\end{array}$

Conclusion:

The energy stored in each capacitor is $E_4=0.0498J;E_{15}=0.0132J;E_{12}=0.24J$ .