Question

In the given circuit the potential at point $E$ is:

• A. Zero
• B. $-8V$
• C. $-\frac{4}{3}V$
• D. $\frac{4}{3}V$

Answer 1

Answer: (D)

Given$,$

$e.m.f.=8V.$

$R₁=5\Omega$

$R₂=1\Omega$

From the Figure we can observe that the Resistors are in series$.$

$\therefore R=R₁+R₂$

$=5+1$

$=6\Omega$

Using Ohm's law$,$

$V=I\times R$

$\Rightarrow I=V/R$

$\Rightarrow I=8/6$

$\therefore I=4/3A.$

$\because$ Current remains same in the Series$,$

$\therefore$ Current flowing through the $1\Omega$ Resistors is $4/3A.$

Again Using the Ohm's law$,$

$V=I\times R$

$V=4/3\times 1$

$\therefore V=4/3V.$

Hence,

option $\left(D\right)$ is correct answer.