In the circuit shown in the figure, if potential at point $A$ is taken to be zero, the potential at point…
In the circuit shown in the figure, if potential at point $A$ is taken to be zero, the potential at point $B$ is
- $-1 \mathrm{~V}$
- $+2 \mathrm{~V}$
- $-2 \mathrm{~V}$
- $+1 \mathrm{~V}$
Solution
By KVL along path $A C D B$
$\begin{gathered}
V_A+1+(1)(2)-2=V_B \\
0+1=V_B \\
V_B=1 \mathrm{~V}
\end{gathered}$
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Asked in: NEET 2011 (Mains)
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