Question

Five capacitors are connected as shown in the figure.The equivalent capacitance between a and b​

Answer 1

Answer:

Lets Analyse the diagram Step by Step.

First we can see that the two 2 μF capacitors are in series with each other as same current flows through them. Hence Net capacitance of 2μF capacitors are :

⇒ Net Capacitance = ( 2 μF × 2 μF ) / ( 2 μF + 2 μF )

⇒ Net Capacitance = ( 4 μF ) / ( 4 μF ) = 1 μF

Therefore Net Capacitance is 1 μF for the series of 2 μF capacitors.

Now if we observe, the diagonal branch of 1 μF is parallel to the series combination of 2 μF capacitor. So net capacitance of these 2 capacitors would be:

⇒ Net Capacitance = 1 μF + 1 μF = 2 μF

Now if we see there are only 3 capacitors : 1 μF, 2 μF and 2 μF.

The 2μF capacitor is in series with  other 2 μF.

Therefore Net Capacitance would be 1 μF.

Again the 1μF is in parallel connection with the 1μF capacitor we got now.

Therefore Total Capacitance between these 2 is :

⇒ 1 μF + 1 μF = 2 μF

Therefore the total capacitance is 2 μF for the system.