Question 216 of 380

In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point $B$ with respect to the point $A$ is:

A$+2V$B$-2V$C$–1V$D$+1V$

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Answer: (D) $+1V$

Using kirchhoff's law in the circuit $ACDB$, we get
Voltage at $B$= voltage at $A$ + voltage of battery of $1V$ + potential difference across $2\Omega$ resistance - voltage of battery of $2V$
$\Rightarrow {V}_{A}+1+2(1)-2={V}_{B}$
Now potential at point $B$ with respect to point $A$ is given by,
$\Rightarrow {V}_{B}-{V}_{A}=1V$

JEE Main Physics practice question on Current Electricity (Kirchoffs Laws), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.