Question 129 of 380

All resistances in figure are $1\Omega$ each. The value of current '$I$' is $\dfrac{a}{5}A$. The value of $a$ is _____ .

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Answer: 8

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

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

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

Therefore, ${R}_{eq}=\dfrac{7r}{8}+r=\dfrac{7}{8}+1=\dfrac{15}{8}\Omega$
Now, current $I=\dfrac{3}{\dfrac{15}{8}}=\dfrac{8}{5}A$
Hence, value of $a=8$.

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