Question

10 identical capacitors are connected as shown. The capacitance of each capacitor is 30 F. The equivalent capacitance between A and B is (12 × n) F. Find n.

Answer 1

15/14

Answer 2

To find the equivalent capacitance between points A and B, we simplify the circuit step-by-step from the rightmost end towards A. Let the capacitance of each identical capacitor be $C = 30 \mu F$.

The circuit diagram can be represented with nodes as follows:

Step 1: Simplify the branch Y-S-B

Capacitors $C_9$ and $C_{10}$ are in series between nodes Y and B, as node S is only connected to $C_9$ and $C_{10}$.

The equivalent capacitance $C_{YSB}$ is: $C_{YSB} = \frac{C_9 \times C_{10}}{C_9 + C_{10}} = \frac{C \times C}{C + C} = \frac{C^2}{2C} = \frac{C}{2}$

This $C_{YSB}$ is connected between Y and B.

Step 2: Simplify the branch Y-Z-B

Capacitors $C_3$ and $C_4$ are in series between nodes Y and B, as node Z is only connected to $C_3$ and $C_4$.

The equivalent capacitance $C_{YZB}$ is: $C_{YZB} = \frac{C_3 \times C_4}{C_3 + C_4} = \frac{C \times C}{C + C} = \frac{C^2}{2C} = \frac{C}{2}$

This $C_{YZB}$ is connected between Y and B.

Step 3: Combine effective capacitances at node Y

Now, from node Y, there are two effective capacitors connected to B: $C_{YSB}$ and $C_{YZB}$. These are in parallel.

The effective capacitance between Y and B, $C_{YB\_eff}$, is: $C_{YB\_eff} = C_{YSB} + C_{YZB} = \frac{C}{2} + \frac{C}{2} = C$

So, we can replace the Y-S-B and Y-Z-B branches with a single capacitor $C_{YB\_eff} = C$ between Y and B.

Step 4: Simplify the branch X-P-Q-R-B

This branch can be simplified by successive series combinations from right to left:

• C7 and C8 are in series between Q and B: $C_{QRB} = \frac{C_7 \times C_8}{C_7 + C_8} = \frac{C \times C}{C + C} = \frac{C}{2}$ This $C_{QRB}$ is connected between Q and B.
• C6 and $C_{QRB}$ are in series between P and B: $C_{PQB} = \frac{C_6 \times C_{QRB}}{C_6 + C_{QRB}} = \frac{C \times (C/2)}{C + C/2} = \frac{C^2/2}{3C/2} = \frac{C}{3}$ This $C_{PQB}$ is connected between P and B.
• C5 and $C_{PQB}$ are in series between X and B: $C_{XPB} = \frac{C_5 \times C_{PQB}}{C_5 + C_{PQB}} = \frac{C \times (C/3)}{C + C/3} = \frac{C^2/3}{4C/3} = \frac{C}{4}$ This $C_{XPB}$ is connected between X and B.

Step 5: Combine effective capacitances at node X

Now, from node X, there are two main paths to B:

• Path 1: Through C2 and $C_{YB\_eff}$ $C_2$ and $C_{YB\_eff}$ are in series. $C_{XYB\_path1} = \frac{C_2 \times C_{YB\_eff}}{C_2 + C_{YB\_eff}} = \frac{C \times C}{C + C} = \frac{C}{2}$ This $C_{XYB\_path1}$ is connected between X and B.
• Path 2: Through $C_{XPB}$ This is simply $C_{XPB} = C/4$.

These two effective capacitors, $C_{XYB\_path1}$ and $C_{XPB}$, are in parallel between X and B.

The effective capacitance between X and B, $C_{XB\_eff}$, is: $C_{XB\_eff} = C_{XYB\_path1} + C_{XPB} = \frac{C}{2} + \frac{C}{4} = \frac{2C}{4} + \frac{C}{4} = \frac{3C}{4}$

So, we can replace the entire network between X and B with a single capacitor $C_{XB\_eff} = 3C/4$.

Step 6: Final equivalent capacitance between A and B

Capacitor $C_1$ and $C_{XB\_eff}$ are in series between A and B.

The equivalent capacitance $C_{AB}$ is: $C_{AB} = \frac{C_1 \times C_{XB\_eff}}{C_1 + C_{XB\_eff}} = \frac{C \times (3C/4)}{C + 3C/4} = \frac{3C^2/4}{7C/4} = \frac{3C}{7}$

Step 7: Calculate the value of n

Given $C = 30 \mu F$. $C_{AB} = \frac{3 \times 30 \mu F}{7} = \frac{90}{7} \mu F$.

The problem states that the equivalent capacitance between A and B is $(12 \times n) \mu F$.

So, $\frac{90}{7} = 12 \times n$. $n = \frac{90}{7 \times 12} = \frac{90}{84}$.

Simplify the fraction: $n = \frac{90 \div 6}{84 \div 6} = \frac{15}{14}$.

The final answer is 15/14.

Explanation of the solution:

The circuit is a complex ladder network. It is simplified by identifying series and parallel combinations starting from the rightmost end and progressively moving towards the input terminals A and B.

1. Capacitors C9 and C10 are in series (Y-S-B), giving $C/2$.
2. Capacitors C3 and C4 are in series (Y-Z-B), giving $C/2$.
3. These two effective capacitors ($C/2$ each) are in parallel between Y and B, resulting in $C_{YB\_eff} = C/2 + C/2 = C$.
4. The branch X-P-Q-R-B is simplified by successive series combinations:
   • C7 and C8 in series (Q-R-B) give $C/2$.
   • C6 and this $C/2$ in series (P-Q-B) give $C/3$.
   • C5 and this $C/3$ in series (X-P-B) give $C/4$. So, $C_{XPB} = C/4$.
5. Now, from node X to B, there are two parallel paths:
   • Path via C2 and $C_{YB\_eff}$ (which is C). C2 and $C_{YB\_eff}$ are in series, giving $C/2$.
   • Path via $C_{XPB}$ (which is $C/4$). These two paths are in parallel, so $C_{XB\_eff} = C/2 + C/4 = 3C/4$.
6. Finally, C1 and $C_{XB\_eff}$ are in series between A and B. $C_{AB} = \frac{C \times (3C/4)}{C + 3C/4} = \frac{3C}{7}$.
7. Substitute $C = 30 \mu F$ to get $C_{AB} = \frac{3 \times 30}{7} = \frac{90}{7} \mu F$.
8. Equate this to $(12 \times n) \mu F$ to find $n$: $\frac{90}{7} = 12n \implies n = \frac{90}{84} = \frac{15}{14}$.