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)
$\left(\because C_1+C_2=10\mu F\right)$
$\because {\left(C_1+C_2\right)}^2=C_1^2+C_2^2-2C_1{C}_2$ ...(iii)
${\left(C_1-C_2\right)}^2=C_1^2+C_2^2-2C_1{C}_2$ ...(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),
$C_1=\frac{10+4\sqrt{5}}{2}=(5+2\sqrt{5})\mu F$
$C_2=\frac{10-4\sqrt{5}}{2}=(5-2\sqrt{5})\mu F$