In the circuit shown in the figure, if potential, at point A is taken to be zero, the potential at point B is |
A) \[-1\text{ }V\]
B) \[+2\text{ }V\]
C) \[-2\text{ }V\]
D) \[+1\text{ }V\]
Correct Answer: D
Solution :
[d] By KVL along path ACDB |
\[{{V}_{A}}+1+(1)(2)-2={{V}_{B}}\] |
\[0+1={{V}_{B}}\] |
\[{{V}_{B}}=1\,V\] |
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