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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

Answer:

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