1.

A capacitor of capacitaance `5muF` is connected as shown in the figure.The internal resistane of the cvell is `0.5omega` The amount of charge on the cap[acitor plates is A. `80muC`B. `40muC`C. `20muC`D. `10muC`

Answer» Correct Answer - d
In steady state, there will be no current in the capacitro branch .Net resistacne of the circuit `R=1+1+0.5=2.5omega`
Current drawn from the cell `i=(V)/(R )=(2.5)/(2.5)=1A`
Potential drop across two parallel bracnhes
`V=E-ir=2.5-1xx0.5=2.5-0.5=2.0V`
So, charge on the capacito plates
`q=CV=5xx2=10muC`


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