A capacitor of capacitance `1muF` is connected in parallel with a resistance `10Omega` and the combination is connected across the terminals of a battery of EMF 50 V and internal resistance `1Omega`. The potential difference across the capacitor at steady state is (in Volt)

Options

A. (a)`(50)/(11)`
B. (b)`(100)/(11)`
C. (c)`(250)/(11)`
D. (d)`(500)/(11)`

Answer

D

Explanation

At steady state, there is no current through the capacitor and hence it can be removed from the circuit while analysing it. So, current through the circuit
`i=(50)/(10+1)=(50)/(11)A`
The potential difference across the capacitor is equal to the potential difference across the `10Omega` resistance therefore, `V=iR=((50)/(11))(10)=(500)/(11)`Volt