A) 12V
B) 8V
C) 4V
D) zero
Correct Answer: C
Solution :
From the capacitors circuit diagram. \[C=\frac{2\times 4}{2+4}=\frac{4}{3}\mu F\]in one loop \[q=12\times \frac{4}{3}=16\mu C\]across one set of \[2\mu F\] and \[4\mu F\] capacitors. \[{{V}_{p}}=12-\frac{16}{4}=8V\] \[{{V}_{Q}}=12-\frac{16}{2}=4V\] \[\Rightarrow \] \[{{V}_{p}}-{{V}_{Q}}=8-4=4V\]
You need to login to perform this action.
You will be redirected in
3 sec
