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 $KVL$ in loop $(1)$,
$-5+10I_1+10\left(I_1-I_2\right)=0$
$20I_1-10I_2=5..(i)$
By $KVL$ in loop$(2)$
$10I_2+3+10\left(I_2-I_1\right)=0$
$-10I_1+20I_2=-3$
$-20I_1+40I_2=-6.....(ii)$
Solving Eqs. (i) and (ii), we get
$30I_2=-1$
$I_2=\frac{-1}{30}A$
From Eq. (i), $20I_1-10\left(\frac{-1}{30}\right)=5\Rightarrow I_1=\frac{7}{30}$ A
$\therefore$ Current in branch $AB,I_{AB}=I_1-I_2=\frac{7}{30}-\left(\frac{-1}{30}\right)$
$I_{AB}=\frac{8}{30}A$
$\therefore$ Voltage across a terminal $AB$
$V_{AB}=I_{AB}\times R_{AB}=\frac{8}{30}\times 10=\frac{8}{3}V$