Chapter 24, Problem 34P

(a)

To determine

The equivalent capacitance between the terminals.

Answer to Problem 34P

The equivalent capacitance between the terminals is $0.2419\text{μF}$ .

Explanation of Solution

Given:

Two capacitors $1.00\text{μF}\text{and}\text{0}\text{.250}\text{μF}$ are connected in parallel and this parallel combination is in series with $0.300\text{μF}$

Formula used:

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

Calculation:

To find equivalent capacitance between the terminals

Assume the terminals as ‘a’ and ‘b’.

  

As $1.00\text{μF }\text{and}\text{0}\text{.250}\text{μF}$ capacitors are connected in parallel their equivalent capacitance will be

  $\begin{array}{c}{C}_{eq}=\text{1}\text{.00}\text{μF}\text{+}\text{0}\text{.}\text{250}\text{μF}\\ \text{=}\text{1}\text{.}\text{250}\text{μF}\end{array}$

Now the given circuit can be reduced to

  

Now $0.300\text{μF}\text{and}\text{1}\text{.250}\text{μF}$ capacitors are connected in series. Therefore, total equivalent capacitance $C_{ab}$ will be:

  $\begin{array}{l}{C}_{ab}=\frac{\text{0}\text{.300}\text{μF}\text{×}\text{1}\text{.250}\text{μF}}{\text{0}\text{.300}\text{μF}\text{+}\text{1}\text{.250}\text{μF}}\\ \text{=}\text{0}\text{.2419}\text{μF}\\ =\text{0}\text{.242μF}\end{array}$

Conclusion:

The equivalent capacitance between the terminals ‘a’ and ‘b’ is $0.2419\text{μF}$ .

(b)

To determine

The charge stored on each capacitor

Answer to Problem 34P

The charge stored on each capacitor is $q_{0.3\mu F}=2.419\text{μC};q_{1\mu F}=1.935\text{μC}\text{and}{\text{q}}_{0.25\mu F}=0.483\text{μC}$ .

Explanation of Solution

Given:

The value of capacitors are given and two capacitors $1.00\mu Fand0.250\mu F$ are connected in parallel and this parallel combination is in series with $0.300\mu F$

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 $\text{0}\text{.3μF}$ capacitor will be

  $\begin{array}{c}q=0.3\times {10}^{-6}\times 8.064\\ =2.419\text{μ}\text{C}\end{array}$

The charge stored on $1\text{μF}$ capacitor will be

  $\begin{array}{c}q=1\times {10}^{-6}\times 1.935\\ =1.935\text{μ}\text{C}\end{array}$

Similarly, charge on $0.25\text{μF}$ capacitor will be

  $\begin{array}{l}q=0.25\times {10}^{-6}\times 1.935\\ q=0.483\text{μ}\text{C}\end{array}$

Conclusion:

The charge stored on each capacitor is $q_{0.3\mu F}=2.419\text{μC};q_{1\mu F}=1.935\text{μC}\text{and}{\text{q}}_{0.25\mu F}=0.483\text{μC}$

(c)

To determine

The voltage across each capacitor

Answer to Problem 34P

The voltage across each capacitor is ${\text{V}}_{\text{0}\text{.250}\text{μF}}\text{=}\text{8}\text{.064}\text{V}\text{;}{\text{V}}_{\text{1μF}}\text{=}{\text{V}}_{\text{0}\text{.250}\text{μF}}\text{=1}\text{.935}\text{V}$ .

Explanation of Solution

Given:

The supply voltage $V_s\left(t\right)=10V$

Formula used: If two capacitors $C_{1}and{C}_2$ are connected in series then voltage across each capacitor will be $V_1=\frac{C_2}{C_1+C_2}{V}_s\left(t\right)\&V_2=\frac{C_1}{C_1+C_2}{V}_s\left(t\right);where{V}_s\left(t\right)=\text{supply}voltage$

Calculation:

To find the Voltage across each capacitor:

  

The given circuit is reduced as:

  

The elements in the given circuit are connected in series combination. Then applying voltage division rule to find the voltage across each capacitor

Voltage across $0.300\text{μF}$ capacitor will be

  $\begin{array}{c}{V}_{0.300\mu \text{F}}=\text{10}\text{×}\frac{\text{1}\text{.250}\text{μF}}{\text{0}\text{.300}\text{μF}\text{+}\text{1}\text{.250}\text{μF}}\\ \text{=}\text{8}\text{.064}\text{V}\end{array}$

Voltage across $1.250\text{μF}$ capacitor will be

As $1\text{μF}$ and $0.250\text{μF}$ capacitors are connected in parallel, the voltage will be same i.e., 1.935V

  $\therefore V_{1\mu \text{F}}=V_{0.250\mu \text{F}}=1.935\text{V}$

Conclusion:

The voltage across each capacitor is ${\text{V}}_{\text{0}\text{.250}\text{μF}}\text{=}\text{8}\text{.064}\text{V}\text{;}{\text{V}}_{\text{1μF}}\text{=}{\text{V}}_{\text{0}\text{.250}\text{μF}}\text{=1}\text{.935}\text{V}$ .

(d)

To determine

The total stored energy in the circuit

Answer to Problem 34P

  $12.1\times {10}^{-6}\text{J}$

Explanation of Solution

Given:

The values of capacitors and supply voltage = 10V

Formula used: The total stored energy will be given as:

  $E_{total}=\frac{1}{2}\times C_{total}\times V^2$

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

Calculation:

From part (a) $C_{total}=C_{ab}=0.242\mu F$

  $\begin{array}{l}\therefore E_T=\frac{1}{2}\times 0.242\times {10}^{-6}\times {\left(10\right)}^2\\ =12.1\mu J\end{array}$

Conclusion:

The total energy stored in the circuit is $12.1\times {10}^{-6}\text{J}$ .