Question

In Fig. P24.59, each capacitance C₁ , $\mathbf{6}\mathbf{.}\mathbf{9}\mathbf{\mu F}$and each capacitance ${\mathbf{C}}_{\mathbf{2}}$is $\mathbf{4}\mathbf{.}\mathbf{6}\mathbf{\mu F}$. (a) Compute the equivalent capacitance of the network between points a and b. (b) Compute the charge on each of the three capacitors nearest a and b when $\mathbf{}{\mathbf{V}}_{\mathbf{ab}}\mathbf{=}\mathbf{420}\mathbf{V}$. (c) With 420 V across a and b, compute $\mathbf{}{\mathbf{V}}_{\mathbf{cd}}$.

Short answer

(a) $C_t=2.3\mu F$

(b) $\begin{array}{l}Q=Q_4=9.66\times {10}^{a}C,\\ Q_2=6.44\times {10}^{-4}C.\end{array}$

(c) $V_{cd}=46.67V.$

Step-by-step solution

parameters.

$\begin{array}{l}{C}_1=6.9\mu F\\ C_2=4.6\mu F\end{array}$

(a) Computing the equivalent capacitance of the network.

The 3 capacitors$C_1$ are in series so the equivalent$C_1\text{'}$ is given by

$\frac{1}{C_1\text{'}}=\frac{1}{C_1}+\frac{1}{C_1}=\frac{1}{C_1}=\frac{1}{C_1}\Rightarrow C_1\text{'}=\frac{C_1}{3}=\frac{6.9}{3}=2.3\mu F$

The capacitance$C_1\text{'},C_2$ are in parallel so the equivalent$C_{12}\text{'}$ is given by

$C_{12}\text{'}=C_1\text{'}+C_2=2.3+4.6=6.9\mu F$

The 2 capacitorsrole="math" localid="1664269686387" $C_1\text{'},C_2$ are in series so the equivalent$C_{12}\text{'}$ is given by

$\begin{array}{l}\frac{1}{C_3}=\frac{1}{C_1}+\frac{1}{C_1}+\frac{1}{C_{12}\text{'}}==\frac{3}{6.9}\\ C_3=\frac{6.9}{3}=2.3\mu F\end{array}$

The capacitorsrole="math" localid="1664269567298" $C_3,C_2$ are in parallel so the equivalent $C_{23}$is given by

$C_{23}=C_3+C_2=2.3+4.6=6.9\mu F$

The 2 capacitors are in series so the equivalent$C_t$ is given by

$\begin{array}{l}\frac{1}{C_3}=\frac{1}{C_1}+\frac{1}{C_1}+\frac{1}{C_{23}}=\frac{3}{6.9}\\ C_t=\frac{6.9}{3}=2.3\mu F\end{array}$

(b) Computing the charge of each of the three capacitors.

Looking to fig (4) the charge flows through series capacitors is the same and is given by

$Q=V_{ab}{C}_t=420\times 2.3\times {10}^{-6}=9.66\times {10}^{-4}C$

So the voltage on each in fig (4) is given by

$V_1=\frac{Q}{C_1}=\frac{9.66\times {10}^{-4}}{6.9\times {10}^{-6}}=140V$

So the voltage on $V_{23}$is given by

$V_{23}=V_{ab}-2V_2=420-2\times 140=140V$

The voltage across the capacitors on parallel is the same so the charge on $C_2$ is given by

$Q_2=V_{23}{C}_2=140\times 4.6\times {10}^{-6}=6.44\times {10}^{-4}C$.

(c) Computing the Vcd .

Looking to fig (2) knowing the voltage across the capacitor$C_{12}\text{'}$ , to get that, the charge flow should be known before.

The charge flow in$C_{12}\text{'}$ is given by

$Q_{12}\text{'}=Q-Q_2=\left(9.66-6.44\right)\times {10}^{-4}=3.22\times {10}^{-4}$.

So the voltage V_cd is given by

$V_{cd}=\frac{Q_{12}\text{'}}{C_{12}\text{'}}\frac{3.22x{10}^{-4}}{6.9x{10}^{-6}}=46.67V$