In the following circuit, the magnitude of current $$I_1$$, is ______ A.

Let's use Nodal Analysis to find the currents in the circuit.

1. Set Reference Node:

Let the bottom wire (connected to the negative terminal of the $$2\text{ V}$$ battery and the bottom of the lower $$1\ \Omega$$ resistor) be at potential $$0\text{ V}$$.

2. Determine Fixed Node Potentials:

The top wire connected to the positive terminal of the $$2\text{ V}$$ battery is at potential:

$$$V_{\text{top}} = 2\text{ V}$$$

Let the central junction between the two $$1\ \Omega$$ resistors be node $$C$$ at potential $$V_C$$.

The middle branch has a $$5\text{ V}$$ battery with its positive terminal connected to node $$C$$. Therefore, the left node $$L$$ between the $$5\text{ V}$$ battery and the middle $$1\ \Omega$$ resistor has potential $$V_L = V_C - 5\text{ V}$$.

3. Apply Nodal Analysis at Node C:

The current entering node $$C$$ from the top branch is $$\frac{V_{\text{top}} - V_C}{1} = \frac{2 - V_C}{1}$$.

The current leaving node $$C$$ through the bottom $$1\ \Omega$$ resistor is $$\frac{V_C - 0}{1} = V_C$$.

The current leaving node $$C$$ through the middle branch (towards the left wire) is $$I_2$$.

The left vertical wire connects the top $$2\ \Omega$$ resistor, middle $$1\ \Omega$$ resistor, and bottom $$2\ \Omega$$ resistor. Let the potential of this left vertical wire be $$V_0$$.

Using KCL at the left vertical wire node:

$$$\frac{V_{\text{top}} - V_0}{2} + \frac{V_L - V_0}{1} + \frac{0 - V_0}{2} = 0$$$

$$$\frac{2 - V_0}{2} + (V_C - 5 - V_0) - \frac{V_0}{2} = 0$$$

$$$1 - \frac{V_0}{2} + V_C - 5 - V_0 - \frac{V_0}{2} = 0$$$

$$$V_C - 4 - 2V_0 = 0 \implies V_0 = \frac{V_C - 4}{2}$$$

4. Apply KCL at Node C:

$$$\frac{2 - V_C}{1} = \frac{V_C - 0}{1} + \frac{V_C - (V_0 + 5)}{1}$$$

$$$2 - V_C = V_C + V_C - 5 - V_0$$$

$$$2 - V_C = 2V_C - 5 - \left(\frac{V_C - 4}{2}\right)$$$

Multiply the entire equation by 2:

$$$4 - 2V_C = 4V_C - 10 - V_C + 4$$$

$$$4 - 2V_C = 3V_C - 6$$$

$$$5V_C = 10 \implies V_C = 2\text{ V}$$$

5. Calculate Current $$I_1$$:

The current $$I_1$$ flows upward through the $$2\text{ V}$$ battery. By KCL at the top node ($$V_{\text{top}} = 2\text{ V}$$):

The current flowing into the top $$2\ \Omega$$ resistor is:

$$$I_3 = \frac{V_{\text{top}} - V_0}{2}$$$

Since $$V_C = 2\text{ V}$$, we have $$V_0 = \frac{2 - 4}{2} = -1\text{ V}$$.

$$$I_3 = \frac{2 - (-1)}{2} = \frac{3}{2} = 1.5\text{ A}$$$

The current flowing downward through the upper $$1\ \Omega$$ resistor is:

$$$I_{\text{down}} = \frac{V_{\text{top}} - V_C}{1} = \frac{2 - 2}{1} = 0\text{ A}$$$

Therefore, by KCL at the top node:

$$$I_1 = I_3 + I_{\text{down}} = 1.5\text{ A} + 0\text{ A} = 1.5\text{ A}$$$

The magnitude of current $$I_1$$ is 1.5 A (or $$\frac{3}{2}$$ A).