Question

Two parallel-plate capacitors, each of capacitance $40\mu F$, are connected in series. The space between the plates of one capacitor is filled with a dielectric of dielectric constant $K=3$, then the equivalent capacitance of the combination is

• A. $30\mu F$
• B. $120\mu F$
• C. $40\mu F$
• D. $160\mu F$

Answer 1

The correct option is A $30\mu F$

From given,
Let $C_1$ and $C_2$ be the given capacitance $40\mu F$ and $C_2^{\prime }$ be the final after insertion of dielectric.
$C_2=40\mu F$
$C_2^{\prime }=K{C}_2$
$C_2^{\prime }=3\times 40\mu F=120\mu F$
Now
$C_{eq}=\frac{C_1{C}_2^{{}^{\prime }}}{C_1+C_2^{{}^{\prime }}}$
$C_{eq}=\frac{40\times 120}{40+120}=30\mu F$