Question

Calculate the voltage across $AB$ terminals in the given circuit,

• A. $\frac{3}{8} V$
• B. $\frac{8}{3} V$
• C. $\frac{3}{2} V$
• D. $\frac{2}{3} V$

Answer 1

Answer: (B)

Given, circuit diagram is shown in the figure, If $I_{1}$ and $I_{2}$ be the loop currents in loop ( 1 ) and loop $(2)$, respectively then

By $K V L$ in loop $( 1)$,
$-5+10 I_{1}+10\left(I_{1}-I_{2}\right)=0 $
$20 I_{1}-10 I_{2}=5\,..(i)$
By $K V L$ in loop$(2)$
$10 I_{2}+3+10\left(I_{2}-I_{1}\right)=0$
$-10 I_{1}+20 I_{2}=-3 $
$-20 I_{1}+40 I_{2}=-6\,.....(ii)$
Solving Eqs. (i) and (ii), we get
$30 I_{2}=-1$
$I_{2}=\frac{-1}{30} A$
From Eq. (i), $20 I_{1}-10\left(\frac{-1}{30}\right)=5 \Rightarrow I_{1}=\frac{7}{30}$ A
$\therefore $ Current in branch $A B, I_{A B}=I_{1}-I_{2}=\frac{7}{30}-\left(\frac{-1}{30}\right)$
$I_{A B}=\frac{8}{30}\,A$
$\therefore $ Voltage across a terminal $A B$
$V_{A B}=I_{A B} \times R_{A B}=\frac{8}{30} \times 10=\frac{8}{3} V$