As per the given circuit, the value of current through the battery will be ______ A.
Numerical value type. Enter your answer.
Answer: 1
The positive terminal of the battery is at the bottom, so current leaves through the $2\ \Omega$ resistor to the bottom-right node and must return to the top-left node.
$D_2$ points from left to right, so it is reverse biased and its branch carries no current.
Path 1: up through the right $10\ \Omega$, then back through $D_1$ (forward biased) and its $6\ \Omega$: $16\ \Omega$.
Path 2: along the diagonal through $6\ \Omega$, $D_3$ (forward biased) and $10\ \Omega$: $16\ \Omega$.
The two paths are in parallel: $8\ \Omega$. Adding $2\ \Omega$: $R = 10\ \Omega$, so $I = \dfrac{10}{10} = 1\ \text{A}$ (ideal diodes).
Solution by Sreeraj P, M.Sc Physics