The power distribution network in a particular apparatus can be modeled by the circuit below.
Figure 1-1
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.
I1 = Vs/(Rw1+RL1) = 4/(1+15) = 0.25A. = Current thru RL1.
VL1 = I1*RL1 = 0.25 * 15 =
I2 = (Vs/(Rw2+RL2) =
VL2 = I2*RL2 =
To determine the node potentials eL1 and eL2, we need to solve the circuit using Kirchhoff's laws and Ohm's law. Here's how you can do it step by step:
1. Assign voltages to the nodes:
- Choose the ground node as the reference point and assign it a voltage of 0V.
- Label the voltage at the node connected to RL1 as eL1.
- Label the voltage at the node connected to RL2 as eL2.
2. Apply Kirchhoff's voltage law (KVL) to the loops:
- Start with the loop on the left side: VS - RL1*(eL1 - 0) - RW1*(eL1 - eL2) = 0.
(The voltage across RL1 is RL1*(eL1 - 0), and the voltage across RW1 is RW1*(eL1 - eL2).)
- Move to the loop on the right side: 0 - RL2*(eL2 - 0) - RW2*(eL2 - eL1) = 0.
(The voltage across RL2 is RL2*(eL2 - 0), and the voltage across RW2 is RW2*(eL2 - eL1).)
3. Simplify the equations:
- Rewriting the first equation: VS - 15*(eL1) - 1*(eL1 - eL2) = 0.
- Rewriting the second equation: -20*(eL2) - 2*(eL2 - eL1) = 0.
4. Solve the equations simultaneously to find eL1 and eL2:
- Rearrange the first equation: VS - 15*(eL1) - (eL1 - eL2) = 0.
Simplify it further: VS - 16*(eL1) + eL2 = 0. ----(equation 1)
- Rearrange the second equation: -20*(eL2) - 2*(eL2 - eL1) = 0.
Simplify it further: 2*(eL1) - 18*(eL2) = 0. ----(equation 2)
To solve these equations, we can use any method, such as substitution or elimination.
5. Substitute the value of eL1 from equation 2 into equation 1:
- Substitute 2*(eL1) from equation 2 into equation 1: VS - 16*(eL1) + eL2 = 0.
Simplify it further: VS - 16*(2*(eL1)) + eL2 = 0.
Rewrite the equation: VS - 32*(eL1) + eL2 = 0. ----(equation 3)
6. Solve equations 3 and 2 simultaneously to find eL1 and eL2:
- Multiply equation 2 by 32: 64*(eL1) - 576*(eL2) = 0. ----(equation 4)
- Subtract equation 3 from equation 4: 64*(eL1) - 576*(eL2) - (VS - 32*(eL1) + eL2) = 0.
Simplify it further: 96*(eL1) - 575*(eL2) - VS = 0.
Now we have two equations:
- 96*(eL1) - 575*(eL2) - VS = 0 ----(equation 5)
- 2*(eL1) - 18*(eL2) = 0 ----(equation 6)
Solve equations 5 and 6 simultaneously to find the values of eL1 and eL2.
7. Substitute the known values (VS = 4.0V, RL1 = 15.0Ω, RL2 = 20.0Ω, RW1 = 1.0Ω, RW2 = 2.0Ω) into the equations and solve for eL1 and eL2.
By following these steps, you can determine the node potentials eL1 and eL2 in volts.