The capacitances of three capacitors are in the ratio 2 : 3 : 4. Their equivalent capacitance in parallel combination is more than the equivalent capacitance in series combination by \( (21/26)\,\mathrm{pF} \). The capacitances of the capacitors are, respectively:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 0.2\,\mathrm{pF} \), \( 0.3\,\mathrm{pF} \) and \( 0.4\,\mathrm{pF} \)
Capacitors in parallel add, \( C_{p} = C_{1}+C_{2}+C_{3} \), while in series the reciprocals add. Let the capacitances be \( 2x \), \( 3x \) and \( 4x \).
\( \displaystyle C_{p} = 2x + 3x + 4x = 9x \)
\( \displaystyle \frac{1}{C_{s}} = \frac{1}{2x} + \frac{1}{3x} + \frac{1}{4x} = \frac{6+4+3}{12x} = \frac{13}{12x} \;\Rightarrow\; C_{s} = \frac{12x}{13} \)
\( \displaystyle C_{p} - C_{s} = 9x - \frac{12x}{13} = \frac{105x}{13} = \frac{21}{26}\,\mathrm{pF} \)
\( \displaystyle x = \frac{21}{26} \times \frac{13}{105} = 0.1\,\mathrm{pF} \)
So the capacitances are \( 0.2\,\mathrm{pF} \), \( 0.3\,\mathrm{pF} \) and \( 0.4\,\mathrm{pF} \).