Question

In Fig. $C_1=6.0\mu F,C_2=3.0\mu F,C_3=4.0\mu F\text{, and}{C}_4=8.0\mu F\text{.}$and $C_4=8.0\mu F\text{.}$The capacitor network is connected to an applied potential difference Vab. After the charges on the capacitors have reached their final values, the voltage across C3 is 40.0 V. What are

(a) the voltages across$C_1\text{and}{C}_2$

(b) the voltage across $C_4$.

(c) the voltage $v_{ab}$applied to the network?

Short answer

a) the voltages across $C_1\text{and}{C}_2$is 13.34V and 26.68V respectively.

b) the voltage across $C_{{}_4}$is 30.0 V.

c) the voltage Vab applied to the network is 70.0V.

Given:

$C_1=6.0\mu F,C_2=3.0\mu F,C_3=4.0\mu F,C_4=8.0\mu F$

Capacitors $C_1\text{and}{C}_2$are in series. The equivalent capacitance of 1 and 2$C_{12}$is in parallel with$C_3$.

$C_{123}\text{and}{C}_4\text{are in parallel}{V}_3=400\mathrm{V}$

Step-by-step solution

Calculate the voltage across C1 and C2.

We want to find $V_1\text{and}{V}_2$. In the parallel connection, the voltage does not change so the voltage across $C_1\text{and}{C}_2$is the same for $C_3$so

$V_1+V_2=40.0\mathrm{V}$ (1)

In the series connection, the charge Q is the same for both capacitors $C_1\text{and}{C}_2$so we could obtain the next relation

$\begin{array}{r}{Q}_1=Q_2\\ C_1{V}_1=C_2{V}_2\\ V_2=\frac{C_1}{C_2}{V}_1\\ V_2=\left(\frac{6.0\mu F}{3.0\mu F}\right)V_1\\ V_2=2V_1\end{array}$

Let us plug this expression into equation (1) to get

$\begin{array}{r}{V}_1+V_2=40.0\mathrm{V}\\ V_1+2V_1=40.0\mathrm{V}\\ V_1=13.34\mathrm{V}\end{array}$

Now we can get the value of$V_2$by using the above equation

$V_2=2V_1=2\left(13.34\mathrm{V}\right)=26.68\mathrm{V}$

Calculate the voltage across c4.

Before finding $v_4$we must calculate the equivalent capacitance for $C_1,C_2\text{and}{C}_3$The equivalent capacitance for the three capacitors is $C_{123}$and will be given by

$\begin{array}{r}{C}_{123}=C_{12}+C_3\\ C_{123}=\frac{1}{\left[\frac{1}{C_1}+\frac{1}{C_2}\right]}+C_3\end{array}$

Now let us put our values into this equation to get $c_{123}$

$\begin{array}{r}{C}_{123}=\frac{1}{\left[\frac{1}{C_1}+\frac{1}{C_2}\right]}+C_3\\ C_{123}=\frac{1}{\left[\frac{1}{6.0\mu F}+\frac{1}{3.0\mu F}\right]}+4.0\mu F\\ C_{123}=6.0\mu F\end{array}$

Now we can conclude from the figure that $C_{123}\text{and}{C}_4$are in series, this means the charge is the same for $C_{123}\text{and}{C}_4$Hence, we could get the next relation to finding $v_4$

$\begin{array}{r}{Q}_{123}=Q_4\\ C_{123}{V}_3=C_4{V}_4\\ V_4=\left(\frac{C_{123}}{C_4}\right)V_3\end{array}$

Now we can put our values of $C_{123},C_4\text{and}{V}_3$into the above equation to get $v_4$$where{V}_3=V_{123}=40.0\mathrm{V}$

$\begin{array}{r}{V}_4=\left(\frac{C_{123}}{C_4}\right)V_3\\ V_4=\left(\frac{6.0\mu F}{8.0\mu F}\right)40.0\mathrm{V}\\ V_4=30\mathrm{V}\end{array}$

Calculate the voltage Vab applied to the network.

We want to find the applied voltage across ab. As the capacitances $C_{123}\text{and}{C}_4$are in series, therefore, the applied voltage will be the summation of both voltage

$V_{ab}=V_4+V_{123}=30.0\mathrm{V}+40.0\mathrm{V}=70.0\mathrm{V}$