Question

When two capacitors are connected in parallel the resulting combination has capacitance $10 \, \mu F$. The same capacitors when connected in series results in a capacitance $0.5 \, \mu F$. The respective values of individual capacitors are

• A. $1.9 \, \mu F$ and $0.2 \, \mu F $
• B. $(8 + 2 \sqrt{5} ) \, \mu F$ and $(2 - 2 \sqrt{5} ) \, \mu F $
• C. $(5 + 2 \sqrt{5} ) \, \mu F$ and $(5 - 2 \sqrt{5} ) \, \mu F $
• D. $12 \, \mu F$ and $1.7 \, \mu F $
• E. $5 \, \mu F$ and $2 \, \mu F $

Answer 1

Answer: (C)

When capacitors are connected in parallel then
$C_{ eq }=C_{1}+C_{2}=10\, \mu F$ ...(i)
When capacitors are connected in series then
$C_{ eq }=\frac{C_{1} C_{2}}{C_{1}+C_{2}}=0.5\, \mu F$
or $C_{1} C_{2}=0.5\left(C_{1}+C_{2}\right)$
or $C_{1} C_{2}=5(\mu F )^{2}$ ...(ii)
$\left(\because C_{1}+C_{2}=10\, \mu F \right)$
$\because \left(C_{1}+C_{2}\right)^{2}=C_{1}^{2}+C_{2}^{2}-2 C_{1} C_{2}$ ...(iii)
$\left(C_{1}-C_{2}\right)^{2}=C_{1}^{2}+C_{2}^{2}-2 C_{1} C_{2}$ ...(iv)
Put the value from Eqs. (i) and (ii) in the eq. (iii)
Hence, $C_{1}^{2}+C_{2}^{2}=(10\, \mu)^{2}-2 \times 5\, \mu^{2}=90\, \mu^{2}$
Put the above value in Eq (iv)
$\left(C_{1}-C_{2}\right)^{2}=90\, \mu^{2}-2 \times 5\, \mu^{2}=80\, \mu^{2}$
$\therefore C_{1}-C_{2}=\sqrt{5}\, \mu$ ...(v)
Now from Eqs. (i) and (v),
$C_{1}=\frac{10+4 \sqrt{5}}{2}=(5+2 \sqrt{5})\, \mu F$
$C_{2}=\frac{10-4 \sqrt{5}}{2}=(5-2 \sqrt{5})\, \mu F$