Refer to the figure given below. The values o | vedclass

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(A) Let the central node be $O$ (at $0 	ext{ V}$), the top node be $A$, and the bottom node be $B$. The $5 	ext{ V}$ source is connected between the

Vedclass · Accepted Answer

$I_1 = 2.5 	ext{ A}, I_2 = 1.875 	ext{ A}, I_3 = 1.875 	ext{ A}$

Vedclass · Answer

(A) Let the central node be $O$ (at $0 	ext{ V}$), the top node be $A$, and the bottom node be $B$. The $5 	ext{ V}$ source is connected between the right node and the central vertical branch. The $10 	ext{ V}$ source is connected to the left.Applying nodal analysis at the central node $O$:Let $V_A$ be the potential at the top node and $V_B$ be the potential at the bottom node.Due to symmetry, $V_A = -V_B = V$. The $5 	ext{ V}$ source is connected between the right node $(5 	ext{ V})$ and the central branch. The current $I_2$ flows from the $5 	ext{ V}$ source into the parallel combination of $2 \Omega$ and $2 \Omega$ resistors.Equivalent resistance of the two $2 \Omega$ resistors in series is $4 \Omega$. The $5 	ext{ V}$ source sees $4 \Omega$ in parallel with the rest of the circuit.Solving the nodal equations: $I_3$ is the current through the $10 	ext{ V}$ source, $I_1$ is the current through the $1 \Omega$ resistor, and $I_2$ is the current from the $5 	ext{ V}$ source.Calculations yield $I_1 = 2.5 	ext{ A}$, $I_2 = 1.875 	ext{ A}$, and $I_3 = 1.875 	ext{ A}$.Thus, the correct option is $A$.

Refer to the figure given below. The values of $I_1, I_2$ and $I_3$ are . . . . . . .

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