Question

The potential difference between the points $A$ and $B$ in the following circuit shown in the figure.

• A. $\frac{2}{3}volt$
• B. $\frac{4}{5}volt$
• C. $\frac{8}{9}volt$
• D. $2volt$

Answer 1

The correct option is A $\frac{2}{3}volt$

From the given figure we ca easily find the resultant resistance of circuit,

$⟹\frac{1}{R_{eq}}=\frac{1}{15}+\frac{1}{15}=\frac{2}{15}$

$R_{eq}=\frac{15}{2}\Omega$

Using formula for Ohm's law,

$V=I{R}_{eq}$

$2=I{R}_{eq}$

$⟹I=\frac{4}{15}A$

So, a current of $\frac{4}{15}A$ is passing through circuit.

Using Kirchhoff's law for loop $TRBSuT$

$⟹I_1\left(10\right)+I_1\left(5\right)-2=0$

$⟹I_1=\frac{2}{15}A$

$⟹$ Current through arm $PAQ$

$⟹I-I_1=\frac{4}{15}-\frac{2}{15}=\frac{2}{15}A$

So, potential at point $A$ is given by

$V_A=IR=\frac{2}{15}\times 5=\frac{2}{3}V$

So, potential at point $B$ is given by

$V_B=IR=\frac{2}{15}\times 10=\frac{4}{3}V$

So, potential difference, $V_A-V_B=\frac{4}{3}-\frac{2}{3}=\frac{2}{3}V$