Find Currents in Resistor Network with Multiple Voltage Sources

To find the currents $I_1, I_2,$ and $I_3$ in the given resistor network, we can analyze the circuit using Nodal Analysis.

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1. Node Labeling and Reference Potential

Let us label the key nodes of the network:

• Let the bottom node $D$ be chosen as the reference node with zero potential: $V_D = 0\text{ V}$
• A $5\text{ V}$ ideal voltage source connected between the top node $C$ and bottom node $D$ fixes the potential at node $C$: $V_C - V_D = 5\text{ V} \implies V_C = 5\text{ V}$
• Let $V_M$ be the electric potential at the central junction $M$.
• Let $V_A$ be the electric potential at the leftmost node $A$.

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2. Kirchhoff's Current Law (KCL) at Central Node $M$

The currents connected to node $M$ are:

• Current from $C$ to $M$ through the upper $2\,\Omega$ resistor: $I_{CM} = \frac{V_C - V_M}{2} = \frac{5 - V_M}{2}$
• Current from $M$ to $D$ through the lower $2\,\Omega$ resistor: $I_{MD} = \frac{V_M - V_D}{2} = \frac{V_M}{2}$
• Current $I_1$ leaving node $M$ towards node $A$ through the central branch containing the $10\text{ V}$ battery and $1\,\Omega$ resistor.

Applying KCL at node $M$: $\sum I_{\text{in}} = \sum I_{\text{out}}$ $I_{CM} = I_{MD} + I_1$ $\frac{5 - V_M}{2} = \frac{V_M}{2} + I_1 \implies I_1 = \frac{5}{2} - V_M = 2.5 - V_M \quad \text{--- (Equation 1)}$

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3. Relation Between $V_A$ and $V_M$

Moving from node $M$ to node $A$ across the $10\text{ V}$ battery and $1\,\Omega$ internal resistor: $V_A = V_M + 10 - I_1 \times 1$

Substituting $V_M = 2.5 - I_1$ from Equation (1): $V_A = (2.5 - I_1) + 10 - I_1 = 12.5 - 2I_1 \quad \text{--- (Equation 2)}$

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4. KCL at Left Node $A$

The currents connected to node $A$ are:

• Current entering node $A$ from $M$: $I_1$
• Current leaving node $A$ to node $C$ through the top-left $4\,\Omega$ resistor: $I_{AC} = \frac{V_A - V_C}{4} = \frac{V_A - 5}{4}$
• Current leaving node $A$ to node $D$ through the bottom-left $4\,\Omega$ resistor (labeled as $I_3$): $I_3 = \frac{V_A - V_D}{4} = \frac{V_A}{4}$

Applying KCL at node $A$: $I_1 = I_{AC} + I_3 = \frac{V_A - 5}{4} + \frac{V_A}{4} = \frac{2V_A - 5}{4} = \frac{V_A}{2} - 1.25$

Rearranging for $V_A$: $V_A = 2I_1 + 2.5 \quad \text{--- (Equation 3)}$

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5. Solving for $I_1, V_A,$ and $V_M$

Equating Equations (2) and (3): $12.5 - 2I_1 = 2I_1 + 2.5$ $4I_1 = 10 \implies I_1 = 2.5\text{ A}$

Now substituting $I_1 = 2.5\text{ A}$ back into our potential equations: $V_A = 2(2.5) + 2.5 = 7.5\text{ V}$ $V_M = 2.5 - 2.5 = 0\text{ V}$

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6. Calculating $I_3$ and $I_2$

• Current $I_3$ through the bottom-left $4\,\Omega$ resistor: $I_3 = \frac{V_A - V_D}{4} = \frac{7.5 - 0}{4} = 1.875\text{ A}$

• Current $I_2$ supplied by the $5\text{ V}$ battery to top node $C$:

  • Current leaving $C$ towards $M$: $I_{CM} = \frac{5 - 0}{2} = 2.5\text{ A}$
  • Current entering $C$ from $A$: $I_{AC} = \frac{V_A - V_C}{4} = \frac{7.5 - 5}{4} = 0.625\text{ A}$
  • By KCL at node $C$, $I_2 + I_{AC} = I_{CM}$: $I_2 = I_{CM} - I_{AC} = 2.5 - 0.625 = 1.875\text{ A}$

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Conclusion

$I_1 = 2.5\text{ A}, \quad I_2 = 1.875\text{ A}, \quad I_3 = 1.875\text{ A}$

This corresponds to Option A.