A) \[{{V}_{A}}=4\,V;{{V}_{D}}=9\,V\]
B) \[{{V}_{A}}=3\,V;{{V}_{D}}=4\,V\]
C) \[{{V}_{A}}=9\,V;{{V}_{D}}=3\,V\]
D) \[{{V}_{A}}=4\,V;{{V}_{D}}=3\,V\]
Correct Answer: D
Solution :
(d) \[{{V}_{A}}=4\,V;{{V}_{D}}=3\,V\] \[{{V}_{A}}-{{V}_{B}}=2\times 2=4\,V\] \[\therefore {{V}_{A}}-0=4\,V\Rightarrow {{V}_{A}}=4\,V\] \[[\because \,\,{{V}_{B}}=0]\] As point D is connected to positive terminal of battery of emf \[3\text{ }V\] and \[{{V}_{B}}=0,\therefore \,\,{{V}_{D}}=3V\]
You need to login to perform this action.
You will be redirected in
3 sec
