Question

Two condensers A and B each having slabs of dielectric constant $K=2$ are connected in series. When they are connected across 230 V supply, potential across A is 130 V and that across B is 100 V. If the dielectric in the condenser of smaller capacitance is replaced by one for which $K=5$ what will be the values of potential difference across them?

Answer 1

In series potential difference distributes in value ratio capacitance.

$\therefore \frac{V_A}{V_B}=\frac{C_2}{C_1}$

or $\frac{130}{100}=\frac{C_2}{C_1}$

or $C_1=\frac{C_2}{1.3}=C_A$

K is made 2.5 times. Therefore $C_1$ will also become 2.5 times.

$C_1^{\prime }=2.5C_1=\frac{2.5C_2}{1.3}$

or $\frac{C_1^{\prime }}{C_2}=\frac{25}{13}$

Now, $\frac{V_A^{\prime }}{V_B^{\prime }}=\frac{C_2}{C_1^{\prime }}=\frac{13}{25}$

or $V_A^{\prime }=\left(\frac{13}{13+25}\right)\left(230\right)=78.68V$

$V_B^{\prime }=230.V_B^{\prime }=151.32V$