In Fig. 27-73,${\mathbf{R}}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\hspace{0.33em}}\mathbf{\Omega }\mathbf{,}{\mathbf{R}}_{\mathbf{2}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\hspace{0.33em}}\mathbf{\Omega }\mathbf{,}{\mathbf{R}}_{\mathbf{3}}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{\hspace{0.33em}}\mathbf{\Omega }\mathbf{,}{\mathbf{C}}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\hspace{0.33em}}\mathbf{\mu F}\mathbf{,}{\mathbf{C}}_{\mathbf{2}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\hspace{0.33em}}\mathbf{\mu F}\mathbf{,}$and the ideal battery has emf $\mathbf{\epsilon }\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{V}$. Assuming that the circuit is in the steady state, what is the total energy stored in the two capacitors?

We know that a voltage drop across a capacitor and resistor drop has the same value. Thus, the voltage drop across the resistor calculated using the loop condition is used to calculate the individual energy stored within the plates of the capacitors. Thus, the total energy is the sum of their energies.

Formulae:

The voltage equation using Ohm’s law,

V=IR (i)

Here I is the current, and R is the resistance.

Kirchhoff’s voltage law,

$\sum _{closedloop}V=0$ (ii)

The energy stored within the capacitor plates,

$U_c=1/2C{\epsilon }^2$ (iii)

In the steady state situation, there is no current going to the capacitors, so the resistors all have the same current. Thus, using equation (i) in the loop rule of equation (ii), the current value can be given s follows:

$$\begin{gathered} \epsilon -i{R}_1-i{R}_2-i{R}_3=0 \\ i=\frac{\epsilon }{R_4+R_2+R_3} \end{gathered}$$

Substitute the values in the above expression, and we get,

$\begin{array}{l}i=\frac{20.0V}{5.0\Omega +10.0\Omega +15.0\Omega }\\ =\frac{2}{3}A\end{array}$

Now, the voltage across the resistor R₁can be given using equation (i) as follows:

$\begin{array}{l}{v}_1=\left(\frac{2}{3}A\right))5.0\Omega )\\ =\frac{10}{3}V\\ \end{array}$

$$\begin{gathered} U_1=\frac{1}{2}\times \left(5.0\times {10}^{-6}F\right){\left(\frac{10}{3}V\right)}^2 \\ =2.78\times {10}^{-5}J \end{gathered}$$

Now, the voltage across the resistor R₂can be given using equation (i) as follows:

$\begin{array}{l}{v}_2=\left(\frac{2}{3}A\right)\left(10.0\Omega \right)\\ =\frac{20}{3}V\end{array}$

The above voltage is also the voltage value across capacitor 2, thus the energy stored within capacitor 2 can be given using equation (iii) as follows:

$$\begin{gathered} U_2=\frac{1}{2}\times \left(10.0\times {10}^{-6}F\right){\left(\frac{20}{3}V\right)}^2 \\ =2.22\times {10}^{-4}J \end{gathered}$$

Thus, the total energy stored within the given capacitors can be given as follows:

$\begin{array}{l}U=u_1+u_2\\ =2.78\times {10}^{-5}J+2.22\times {10}^{-4}J\\ =0.278\times {10}^{-4}J+2.22\times {10}^{-4}J\\ =2.498\times {10}^{-4}J\\ =2.50\times {10}^{-4}J\end{array}$

Hence, the value of the energy is $=2.50\times {10}^{-4}J$.