Question

The capacitive network shown is assembled with initially uncharged capacitors. A potential difference, Vab = +100 V, is applied across the network. The switch S in the network is kept open throughout.
The potential difference Vcd, across the open switch S, in SI units, is?
Answer is +40

Answer 1

Concepts and reason

The concept required to solve this problem is that the potential difference between two point is the sum of all the potential drops across it.

In this problem first divide the whole circuit in to two halves with two potential difference, then find the find the potential difference across each part using the formula for equivalent capacitance finally perform the arithmetic to find the potential difference across CD.

Fundamentals

Capacitance is given as follows:

$C = \frac{Q}{V}$

Here, C is the capacitance, Q is the charge, and V is the voltage drop across the capacitor.

Equivalent capacitance in series connection:

$\frac{1}{{{C_{net}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}$

Here, ${C_{net}}$ is the equivalent capacitance in the series connection, ${C_1}$ and ${C_2}$ are the two capacitance connected in series arrangement.

Equivalent capacitance in parallel connection:

${C_{net}} = {C_1} + {C_2}$

Here, ${C_{net}}$ is the equivalent capacitance in the parallel connection, ${C_1}$ and ${C_2}$ are the two capacitance connected in parallel arrangement.

Kirchhoff’s loop rule:

This law states that the sum of potential drop around the loop must be zero.

$\sum V = 0$

Here, $\sum V$ represents the algebraic sum of all the potential drop around the loop.

The diagram of the given circuit is as follows:

Consider the closed loop starting from the point a passing through c and b (leave d) and then coming back to a.

In this loop if we consider the potential across ac as ${V_{ac}}$ , potential across cb as ${V_{cb}}$ , and potential across ab as ${V_{ab}}$ . Then using Kirchhoff’s law,

${V_{ab}} = {V_{ac}} + {V_{cb}}$

Along the loop acba,

Here, $9{\rm{ \mu F}}$ and $15{\rm{ \mu F}}$ are in parallel so their equivalent capacitance is calculated using the formula ${C_{net}} = {C_1} + {C_2}$ .

Substitute $9{\rm{ \mu F}}$ for ${C_1}$ and $15{\rm{ \mu F}}$ for ${C_2}$ in the equation ${C_{net}} = {C_1} + {C_2}$ .

$\begin{array}{c}\\{C_{net}} = 9{\rm{ \mu F}} + 15{\rm{ \mu F}}\\\\ = 24{\rm{ \mu F}}\\\end{array}$

Suppose ${Q_1}$ charge is flowing through the loop acba.

The voltage drop across ac is calculated using the formula ${V_{ac}} = \frac{{{Q_1}}}{{{C_{ac}}}}$ .

Here, ${C_{ac}}$ is the equivalent capacitance across ac which is equal to ${\rm{24 \mu F}}$ .

Substitute ${\rm{24 \mu F}}$ for ${C_{ac}}$ in the equation ${V_{ac}} = \frac{{{Q_1}}}{{{C_{ac}}}}$ .

${V_{ac}} = \frac{{{Q_1}}}{{24{\rm{ \mu F}}}}$ .

Now, calculate the potential across cb as follows:

As ${\rm{16 \mu F}}$ is in series with ${\rm{24 \mu F}}$ (the equivalent capacitance of $9{\rm{ \mu F}}$ and $15{\rm{ \mu F}}$ ) so same charge will be stored on ${\rm{16 \mu F}}$ capacitor.

The voltage drop across cb is calculated using the formula ${V_{cb}} = \frac{{{Q_1}}}{{{C_{cb}}}}$ .

Here, ${C_{cb}}$ is the equivalent capacitance across cb which is equal to ${\rm{16 \mu F}}$ .

Substitute ${\rm{16 \mu F}}$ for ${C_{cb}}$ in the equation ${V_{cb}} = \frac{{{Q_1}}}{{{C_{cb}}}}$ .

${V_{cb}} = \frac{{{Q_1}}}{{{\rm{16 \mu F}}}}$

Now, substitute $\frac{{{Q_1}}}{{24{\rm{ \mu F}}}}$ for ${V_{ac}}$ , $\frac{{{Q_1}}}{{{\rm{16 \mu F}}}}$ for ${V_{cb}}$ , and +100 V for ${V_{ab}}$ in the equation $\begin{array}{c}\\{V_{ab}} = {V_{ac}} + {V_{cb}}\\\\100{\rm{ V}} = \frac{{{Q_1}}}{{24{\rm{ \mu F}}}} + \frac{{{Q_1}}}{{{\rm{16 \mu F}}}}\\\\{Q_1} = 960{\rm{ \mu C}}\\\end{array}$

Along the loop adba,

Here, ${\rm{4 \mu F}}$ , ${\rm{8 \mu F}}$ , and ${\rm{12 \mu F}}$ are in parallel so their equivalent capacitance is calculated using the formula ${C_{net}} = {C_1} + {C_2} + {C_3}$ .

Substitute ${\rm{4 \mu F}}$ for ${C_1}$ , ${\rm{8 \mu F}}$ for ${C_2}$ , ${\rm{12 \mu F}}$ for ${C_3}$ in the equation ${C_{net}} = {C_1} + {C_2} + {C_3}$ .

$\begin{array}{c}\\{C_{net}} = 4{\rm{ \mu F}} + 8{\rm{ \mu F}} + 12{\rm{ \mu F}}\\\\ = 24{\rm{ \mu F}}\\\end{array}$

Suppose ${Q_2}$ charge is flowing through the loop adba.

The voltage drop across db is calculated using the formula ${V_{db}} = \frac{{{Q_2}}}{{{C_{db}}}}$ .

Here, ${C_{db}}$ is the equivalent capacitance across db which is equal to ${\rm{24 \mu F}}$ .

Substitute ${\rm{24 \mu F}}$ for ${C_{db}}$ in the equation ${V_{db}} = \frac{{{Q_2}}}{{{C_{db}}}}$ .

${V_{db}} = \frac{{{Q_2}}}{{24{\rm{ \mu F}}}}$ .

Now, calculate the potential across ad as follows:

As ${\rm{6 \mu F}}$ is in series with ${\rm{24 \mu F}}$ (the equivalent capacitance of ${\rm{4 \mu F}}$ , ${\rm{8 \mu F}}$ , and ${\rm{12 \mu F}}$ ) so same charge will be stored on ${\rm{6 \mu F}}$ capacitor.

The voltage drop across ad is calculated using the formula ${V_{ad}} = \frac{{{Q_2}}}{{{C_{ad}}}}$ .

Here, ${C_{ad}}$ is the equivalent capacitance across ad which is equal to ${\rm{6 \mu F}}$ .

Substitute ${\rm{6 \mu F}}$ for ${C_{ad}}$ in the equation ${V_{ad}} = \frac{{{Q_2}}}{{{C_{ad}}}}$ .

${V_{ad}} = \frac{{{Q_2}}}{{{\rm{6 \mu F}}}}$

Now, substitute $\frac{{{Q_2}}}{{{\rm{6 \mu F}}}}$ for ${V_{ad}}$ , $\frac{{{Q_2}}}{{{\rm{24 \mu F}}}}$ for ${V_{db}}$ , and +100 V for ${V_{ab}}$ in the equation $\begin{array}{c}\\{V_{ab}} = {V_{ad}} + {V_{db}}\\\\100{\rm{ V}} = \frac{{{Q_2}}}{{{\rm{6 \mu F}}}} + \frac{{{Q_2}}}{{{\rm{24 \mu F}}}}\\\\{Q_2} = 480{\rm{ \mu C}}\\\end{array}$

Along the loop bcdb,

The voltage drop across cd is calculated using the equation ${V_{bc}} = {V_{cd}} + {V_{db}}$ .

The equation ${V_{bc}} = {V_{cb}} + {V_{db}}$ can be simplified as,

${V_{bc}} = {V_{cd}} + {V_{db}}$

Substitute $\frac{{{Q_1}}}{{16{\rm{ \mu F}}}}$ for ${V_{bc}}$ and $\frac{{{Q_2}}}{{{\rm{24 \mu F}}}}$ for ${V_{db}}$ in the above equation.

$\frac{{{Q_1}}}{{16{\rm{ \mu F}}}} = {V_{cd}} + \frac{{{Q_2}}}{{{\rm{24 \mu F}}}}$

Rewrite the above expression for ${V_{cd}}$ .

${V_{cd}} = \frac{{{Q_1}}}{{16{\rm{ \mu F}}}} - \frac{{{Q_2}}}{{{\rm{24 \mu F}}}}$

Now substitute $960{\rm{ \mu F}}$ for ${Q_1}$ and $480{\rm{ \mu F}}$ for ${Q_2}$ in the equation ${V_{cd}} = \frac{{{Q_1}}}{{16{\rm{ \mu F}}}} - \frac{{{Q_2}}}{{{\rm{24 \mu F}}}}$ .

$\begin{array}{c}\\{V_{cd}} = \frac{{960{\rm{ \mu C}}}}{{16{\rm{ \mu F}}}} - \frac{{480{\rm{ \mu C}}}}{{{\rm{24 \mu F}}}}\\\\ = 40{\rm{ V}}\\\end{array}$

The voltage drop across cd is ${V_{cd}}$ is 40 V.

Ans:

The voltage drop across cd is ${V_{cd}} = 40{\rm{ V}}$ .