Question

Each of the capacitors shown in figure (31-E6) has a capacitance of 2 µF. find the equivalent capacitance of the assembly between the points A and B. Suppose, a battery of emf 60 volts is connected between A and B. Find the potential difference appearing on the individual capacitors.
Figure

Answer 1

The potential difference appearing on the individual capacitors is 20 Volts

Explanation:

Step 1:

There are three rows of capacitors connected in parallel as shown as above.  

For each row, the equivalent capacitance is given by

$1 / \mathrm{Cr}=1 / 2+1 / 2+1 / 2$

$\mathrm{Cr}=23 \mu \mathrm{F}$

$1 / \mathrm{Cr}=1 / 2+1 / 2+1 / 2$

$\mathrm{Cr}=2 / 3 \mu \mathrm{F}$

Step 2:

As three rows are connected in parallel, their equivalent capacitance is given by

$\mathrm{Ceq}=\mathrm{Cr}+\mathrm{Cr}+\mathrm{Cr}$

$2 / 3+2 / 3+2 / 3$

$2 / 3+2 / 3+2 / 3=2 \mu \mathrm{F}$

• The voltage drop across each capacitor is equal to the 60 Volts.

• From the above we can say that all capacitors have the same capacitance in each row hence the potential difference across their plates is the same.

• Therefore, the potential difference across each capacitor = 20 Volts.