Discussion Forum



1)

Two capacitors when connected in series have a capacitance of 3 μF and when connected in parallel have a capacitance of 16 μF. Their individual capacities are


A) 1 µF, 2 µF

B) 6 µF, 2 µF

C) 12 µF, 4 µF

D) 3 µF, 6 µF

Answer:

Option C

Explanation:

$C_{s}= \frac{C_{1}C_{2}}{C_{1}+C_{2}} =3$

$C_{p}=C_{1}+C_{2} = 16 \therefore C_{1}C_{2}=48$

$C_{1}-C_{2} = \sqrt{(C_{1}+C_{2})^{2}-4C_{1}C_{2}}$ $= \sqrt{16^{2}-4\times48}$=8

C1+C2 = 16μF

C1-C2=8μF

2C1 =  24μF = C1 = 12μF

 

.'. C2 = 48/12 = 4μF