Current Electricity Ques 52
- In the circuit shown, the potential difference between $A$ and $B$ is
(2019 Main, 11 Jan II)
(a) $3 $ $V$
(b) $1 $ $V$
(c) $6 $ $V$
(d) $2 $ $V$
Show Answer
Answer:
Correct Answer: 52.(d)
Solution:
Formula:
- In the given circuit, let’s assume currents in the arms are $i _1, i _2$ and $i _3$, respectively.
Now, $\quad i _1=\frac{V _1}{R _1}=\frac{1}{1}=1 A$
Similarly, $\quad i _2=\frac{2}{1}=2 A$
and $\quad i _3=\frac{3}{1}=3 A$
Total current in the $\operatorname{arm} D A$ is
$ i=i _1+i _2+i _3=6 A $
As all three resistors between $D$ and $C$ are in parallel.
$\therefore$ Equivalent resistance between terminals $D$ and $C$ is
$ \begin{aligned} & \frac{1}{R _{D C}} =\left(\frac{1}{1}+\frac{1}{1}+\frac{1}{1}\right) \\ \therefore \quad & R _{D C} =\frac{1}{3} \Omega \end{aligned} $
So, potential difference across $D$ and $C$ is
$V _{D C}=i R _{D C}=6 \times \frac{1}{3} $
$\Rightarrow V _{D C}=2 V $
$\text { Now, } V _{A D} \text { and } V _{C B}=0$
(In case of open circuits, $I=0$ )
So, $\quad V _{A B}=V _{A D}+V _{D C}+V _{C B}=V _{D C}$
So, $\quad V _{A B}=2 $ $V$
BETA
