Question

A battery of emf $10\text{V}$ is connected to a group of resistances as shown in figure. The potential difference $(V_A-V_B)$ between the points $\text{A}$ and $\text{B}$ is

• A. $-2\text{V}$
• B. $2\text{V}$
• C. $5\text{V}$
• D. $\frac{20}{11}\text{V}$

Answer 1

The correct option is B $2\text{V}$

Resistance between upper branch and lower branch in parallel part is same, so equal amount of current flows through them.


Let the total current is $i$.

$\therefore i=\frac{V}{R_{eq}}$

Equivalent resistance of circuit

$R_{eq}=3+\frac{4\times 4}{4+4}=3+2=5\Omega$

$\therefore i=\frac{10}{5}=2\text{A}$

So, current in each branch will be $1\text{A}$.

Now,
$V_C-V_A=1\times 1=1\text{V}.......\left(1\right)$

Also,
$V_C-V_B=1\times 3=3\text{V}........\left(2\right)$

Solving eqs.(1) and (2), we get

$V_A-V_B=3-1=2\text{V}$

Hence, the correct option is $\left(b\right)$.