Find the charge and energy stored on the capactor C in the electrical network shonw below .

Explanation

Remember the steady current doesn’t flow through the capacitor branch . So the current I will flow as shown below. ( Actually a decreasing current did flow through the above capacitor for few millisecond till the capacitor got completely charged and after that the currnet stopped flowing into it ).

Since same current l flows through `R_(1)` and `R_(3)` and supply voltage is `epsi` , so ` I = (epsi)/(R_(1) + R_(3))`
Also note , since no st eady current flows through `R_(2)` ,there is no potential drop across it and it can be successfully dropped form the circuit, So the new circuit will be redrawn as follows . The points a and b show the positions across the capactior .

Now show the reltative high and low potential points across each element with + and - sings.

Now we have to calculate the potential difference across the capacitance C. There are two ioops of your choice . Either you can ho through `R_(3)` or you can go through `epsi` and `R_(1)` . Both will lead to the same answer . Let us start from point a cross `R_(3)` and reach b.
`V_(a) - IR_(3) - V_(b) `
or ` V_(a)- V_(b) = IR_(3) = (epsi R_(3))/(R_(1) + R_(3))`
i.e., the potenital difference V across capacitance C is `(epsi R_(3))/( R_(1) + R_(3))`
Now charge on capacitance `= CV = (Cepsi R_(3))/(R_(1) + R_(2))`
Also they energy stored can be calculated as ` (1)/(2)CV^(2) = (1)/(2)C((epsi R_(3))/(R_(1) + R_(3)))^(2)`