A) \[4\mu C\]
B) \[6\mu C\]
C) \[1\mu C\]
D) Zero
Correct Answer: B
Solution :
In steady state the capacitor is fully charged and is treated as open circuit, so no current flows through branching containing capacitor in steady state. So, the circuit can be redrawn as The potential difference across capacitor in steady state =V-6-V=-G Volt The negative sign indicates that left plate is of negative potential charge \[=CV=1\times 6=6\mu C\]You need to login to perform this action.
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