Electrostatics Question 580
A capacitor of capacitance $ 5\mu F $ is connected. The internal resistance of the cell is $ 0.5\Omega $. The amount of charge on the capacitor plate is [MP PET 1997]
Options:
A) $ 0,\mu\text{C}$
B) $ 5\mu C $
C) $ 10\mu C $
D) $ 25\mu C $
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Answer:
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)}=1A $
Potential difference across line (2) = potential difference across capacitor $ =1\times 2=2Volt $
So, charge on capacitor = 5×2 = 10 mC
BETA
