Q. The equivalent capacitance of three capacitors of capacitance $C_{1},C_{2}$ and $C_{3}$ are connected in parallel is 12 units and product $C_{1}.C_{2}.C_{3} = 48$ unit. When the capacitors $C_{1}$ and $C_{2}$ are connected in parallel, the equivalent capacitance is 6 units. Then the capacitances are
A) 2, 3, 7
B) 1, 5, 2, 5, 8
C) 1, 5, 6
D) 4, 2, 6
Explanation
(d) $C_{1} + C_{2} + C_{3} = 12$ …(i)
$C_{1}C_{2}C_{3} = 48$ …(ii)
$C_{1} + C_{2} = 6$ …(iii)
From equation (i) and (iii)
$C_{3} = 6$ …(iv)
From equation (ii) and (iv) $C_{1}C_{2} = 8$
Also $\left( C_{1} - C_{2} \right)^{2} = \left( C_{1} + C_{2} \right)^{2} - 4C_{1}C_{2}$
$\left( C_{1} - C_{2} \right)^{2} = (6)^{2} - 4 \times 8 = 4$
$\Rightarrow C_{1} - C_{2} = 2$ …(v)
On solving (iii) and (v) $C_{1} = 4,\ C_{2} = 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.