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) pF. The capacitances of the capacitors are, respectively:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.2 pF, 0.3 pF and 0.4 pF
Let the capacitances be 2x, 3x, 4x
Parallel combination:
$C_p = 2x + 3x + 4x = 9x$
Series combination:
$\frac{1}{C_s} = \frac{1}{2x} + \frac{1}{3x} + \frac{1}{4x} = \frac{6 + 4 + 3}{12x} = \frac{13}{12x}$
$C_s = \frac{12x}{13}$
Given: $C_p - C_s = \frac{21}{26}$ pF
$9x - \frac{12x}{13} = \frac{21}{26}$
$\frac{117x - 12x}{13} = \frac{21}{26}$
$\frac{105x}{13} = \frac{21}{26}$
$x = \frac{21}{26} \cdot \frac{13}{105} = \frac{273}{2730} = 0.1\ \text{pF}$
Capacitances:
2x = 0.2 pF, 3x = 0.3 pF, 4x = 0.4 pF
Capacitances = 0.2 pF, 0.3 pF, 0.4 pF