Question

Equivalent capacitance between A and B is nC. Write n

Answer 1

1.625

Answer 2

To find the equivalent capacitance between A and B, we can simplify the circuit step-by-step.

1. Simplify the leftmost series branch:
   The four capacitors on the left (A-Q, Q-P, P-Z, Z-B) are connected in series. Each has capacitance C.
   The equivalent capacitance of these four capacitors in series is:
   $C_{series1} = \frac{1}{\frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C}} = \frac{1}{\frac{4}{C}} = \frac{C}{4}$

2. Simplify the bottom-right series branch:
   The two capacitors at the bottom (B-Y, Y-D) are connected in series. Each has capacitance C.
   The equivalent capacitance of these two capacitors in series is:
   $C_{series2} = \frac{1}{\frac{1}{C} + \frac{1}{C}} = \frac{1}{\frac{2}{C}} = \frac{C}{2}$
   This $C_{series2}$ is connected between points B and D.

3. Simplify the parallel combination between B and D:
   The capacitor $C_{BD}$ (between B and D, with capacitance C) is in parallel with the equivalent capacitance $C_{series2} = C/2$ (between B and D).
   The equivalent capacitance between B and D is:
   $C_{BD,eff} = C_{BD} + C_{series2} = C + \frac{C}{2} = \frac{3C}{2}$

4. Redraw the simplified circuit:
   Now, the circuit looks like this:

   • A direct capacitor $C_{AB}$ (vertical C) between A and B.
   • The $C_{series1} = C/4$ between A and B.
   • A branch from A to X (C), then X to D (C), then D to B (with $C_{BD,eff} = 3C/2$).

   This is a network of capacitors. The direct $C_{AB}$ and $C_{series1}$ are in parallel, so their equivalent is $C + C/4 = 5C/4$.
   Let's call the remaining part $C_{AXDB}$. This part is connected in parallel with the $5C/4$ equivalent.

   We need to find the equivalent capacitance of the A-X-D-B part. This is a three-capacitor network: $C_{AX}=C$, $C_{XD}=C$, $C_{DB,eff}=3C/2$.
   This is a simple series connection if we consider the path A-X-D-B. The equivalent capacitance for this path is:
   $C_{AXDB} = \frac{1}{\frac{1}{C_{AX}} + \frac{1}{C_{XD}} + \frac{1}{C_{DB,eff}}} = \frac{1}{\frac{1}{C} + \frac{1}{C} + \frac{1}{3C/2}} = \frac{1}{\frac{1}{C} + \frac{1}{C} + \frac{2}{3C}}$
   $C_{AXDB} = \frac{1}{\frac{3+3+2}{3C}} = \frac{1}{\frac{8}{3C}} = \frac{3C}{8}$

   Now, all three parts are in parallel between A and B:

   1. The direct capacitor $C_{AB} = C$.
   2. The series branch $C_{series1} = C/4$.
   3. The series branch $C_{AXDB} = 3C/8$.

   The total equivalent capacitance $C_{eq}$ between A and B is the sum of these parallel capacitances:
   $C_{eq} = C + C_{series1} + C_{AXDB}$
   $C_{eq} = C + \frac{C}{4} + \frac{3C}{8}$
   To add these, find a common denominator, which is 8:
   $C_{eq} = \frac{8C}{8} + \frac{2C}{8} + \frac{3C}{8}$
   $C_{eq} = \frac{(8 + 2 + 3)C}{8} = \frac{13C}{8}$

The problem states that the equivalent capacitance is $nC$.
Comparing $\frac{13C}{8}$ with $nC$, we get $n = \frac{13}{8}$.

The value of n is $\frac{13}{8}$ or $1.625$.