The power comes from a voltage source and is delivered to two loads. The loads are modeled by the resistors RL1 and RL2. The resistances of the interconnects are modeled by the resistors RW1 and RW2. The values of the element parameters are VS=4.0V, RL1=15.0Ω, RL2=20.0Ω, RW1=1.0Ω, RW2=2.0Ω.

Determine the node potentials eL1 and eL2, assuming a ground node as indicated in the figure. Write your answers in the spaces provided below. Express your answers in Volts.
eL1=

To determine the node potentials eL1 and eL2, we can use the node voltage method. The ground node is considered as the reference node and is assigned a potential of 0V.

Now let's analyze the circuit:

1. Define the unknown node potentials:
Let eL1 be the potential at node L1.
Let eL2 be the potential at node L2.

2. Apply KVL (Kirchhoff's Voltage Law) to the loops:
Loop 1:
- Starting at node L1, we encounter RL1 with a potential drop of eL1.
- Moving in the direction of the current, we encounter RW1 with a potential drop of (i1 * RW1), where i1 is the current flowing through RL1.
- We arrive back at node L1, so the total potential difference around the loop is 0.

Therefore, we have:
eL1 - (i1 * RW1) - eL1 = 0
-i1 * RW1 = 0
i1 = 0 (since RW1 ≠ 0)

Loop 2:
- Starting at node L2, we encounter RL2 with a potential drop of eL2.
- Moving in the direction of the current, we encounter RW2 with a potential drop of (i2 * RW2), where i2 is the current flowing through RL2.
- We arrive back at node L2, so the total potential difference around the loop is 0.

Therefore, we have:
eL2 - (i2 * RW2) - eL2 = 0
-i2 * RW2 = 0
i2 = 0 (since RW2 ≠ 0)

3. Apply KCL (Kirchhoff's Current Law) at node L1:
- The current flowing into node L1 is i1 from RL1.
- The current flowing into node L1 is also the sum of the currents flowing out of node L1, which is 0 (from Loop 1).

Therefore, we have:
i1 = 0

4. Apply KCL at node L2:
- The current flowing into node L2 is i2 from RL2.
- The current flowing into node L2 is also the sum of the currents flowing out of node L2, which is 0 (from Loop 2).

Therefore, we have:
i2 = 0

5. Calculate the node potentials:
Since i1 and i2 are both 0, there is no current flowing through RL1 and RL2. Therefore, the potential at nodes L1 and L2 is equivalent to the voltage source value VS.

Therefore:
eL1 = VS = 4.0V

eL1 = 4.0V

To determine the node potentials eL1 and eL2, we need to use the concept of voltage division.

Voltage division is a technique used to calculate the voltage drop across a resistor in a series circuit.

In this case, we have two loads (resistors RL1 and RL2) connected in series and powered by a voltage source VS. The voltages across these loads can be determined by dividing the total voltage (VS) based on their resistances.

The formula for voltage division is:

V = (R / (R1 + R)) * VS

Where V is the voltage across the resistor, R is the resistance of the resistor, and R1 is the resistance of the other resistor in series.

Using this formula, we can calculate the voltage across RL1 (eL1):

eL1 = (RL1 / (RL1 + RL2)) * VS

eL1 = (15.0Ω / (15.0Ω + 20.0Ω)) * 4.0V

eL1 = (15.0Ω / 35.0Ω) * 4.0V

eL1 = 2.057V

Therefore, eL1 = 2.057V.

Now, let's calculate the voltage across RL2 (eL2):

eL2 = (RL2 / (RL1 + RL2)) * VS

eL2 = (20.0Ω / (15.0Ω + 20.0Ω)) * 4.0V

eL2 = (20.0Ω / 35.0Ω) * 4.0V

eL2 = 2.943V

Therefore, eL2 = 2.943V.

So, the node potentials eL1 and eL2 are 2.057V and 2.943V, respectively.