Question

A, B and C are three capacitors. If A and B are connected in series, the effective capacitance is \(6\mu F\). If B and C are connected in series, the effective capacitance is \(4\mu F\). If A and C are connected in series, the effective capacitance is \(3\mu F\). If these three capacitors are connected in parallel, the effective capacitance is:

• A. \(26\mu F\)
• B. \(36\mu F\)
• C. \(13\mu F\)
• D. \(32.4\mu F\)

Hint:

For capacitor series equations, substitute \[ x=\frac1A,\quad y=\frac1B,\quad z=\frac1C \] to convert nonlinear equations into linear ones.

Answer 1

The Correct Option is B

Concept: For capacitors in series, \[ \frac1{C_s} = \frac1{C_1} + \frac1{C_2}. \] Given, \[ \frac{AB}{A+B}=6, \] \[ \frac{BC}{B+C}=4, \] \[ \frac{AC}{A+C}=3. \]

Step 1: Convert into reciprocal form.
\[ \frac1A+\frac1B=\frac16. \] \[ \frac1B+\frac1C=\frac14. \] \[ \frac1A+\frac1C=\frac13. \] Let \[ x=\frac1A,\quad y=\frac1B,\quad z=\frac1C. \] Then \[ x+y=\frac16, \] \[ y+z=\frac14, \] \[ x+z=\frac13. \]

Step 2: Solve equations.
Adding, \[ 2(x+y+z) = \frac16+\frac14+\frac13. \] \[ = \frac34. \] \[ x+y+z=\frac38. \] Hence \[ x=\frac18, \quad y=\frac1{24}, \quad z=\frac14. \] Therefore \[ A=8, \quad B=24, \quad C=4. \]

Step 3: Parallel combination.
\[ C_{eq} = A+B+C. \] \[ = 8+24+4. \] \[ = 36\mu F. \]