Answer:
The equivalent capacitance will be
Step-by-step explanation:
We have given two capacitance
![C_=10\mu F\ and\ C_2=20\mu F](https://img.qammunity.org/2020/formulas/physics/college/qky0unfb08759xppukocjd359ft5e44u1j.png)
They are connected in parallel
So equivalent capacitance
![C=C_1+C_2=10+20=30\mu F](https://img.qammunity.org/2020/formulas/physics/college/bjsl6sxpox2vec0j17os7eln0g39a8vpmh.png)
This equivalent capacitance is now connected in series with
![30\mu F](https://img.qammunity.org/2020/formulas/physics/college/xo5oti9cmelakbf168whp3100s2rlpz8xg.png)
In series combination of capacitors the equivalent capacitance is given by
![(1)/(C)=(1)/(30)+(1)/(30)](https://img.qammunity.org/2020/formulas/physics/college/582mkf6utl44lo4lw7iv31c7hx5p6qvtb9.png)
![C=(30)/(2)=15\mu F](https://img.qammunity.org/2020/formulas/physics/college/vzimtbfixjsv389z8cjnqpp96fg3at2cxn.png)
So the equivalent capacitance will be