question_answer

A capacitor of capacitance \[5\mu F\] is connected as shown in the figure. The internal resistance of the cell is\[0.5\Omega \]. The amount of charge on the capacitor plate is              [MP PET 1997]

A)                                                      \[0\mu C\]

B)                    \[5\mu C\]

C)                    \[10\mu C\]

D)                    \[25\mu C\]

Correct Answer: C

Solution :

                   In steady state condition. No current flows through line (1). Hence total current \[i=\frac{2.5}{(1+1+0.5)}=1\,A\]                     Potential difference a cross line (2) = potential difference a cross capacitor                    \[=1\times 2=2\,Volt\] So, charge on capacitor = 5 ´ 2 = 10 mC