Physics - Current Electricity Question with Solution | TestHub

Each pair on the right side of the circuit is in parallel connection, hence it can be replaced by a resistance of $\frac{r}{2}$ as shown.

Now, $r\&\frac{r}{2}$ in each branch on the right side is in series, hence combined resistance value will be, $\frac{3r}{2}$. Upper branch having resistance $\frac{3r}{2}$ is in parallel with another $\frac{3r}{2}$, hence it can be written as $\frac{3r}{4}$.

Further, simplified circuit will have net resistance of $\frac{7r}{8}$.

Therefore, $R_{eq}=\frac{7r}{8}+r=\frac{7}{8}+1=\frac{15}{8}\Omega$

Now, current $I=\frac{3}{\frac{15}{8}}=\frac{8}{5}\text{A}$

Hence, value of $a=8$.