How do I solve for this? I already did the first part:?

Part 1 Procedure (Basic Network): My R1 = 75 ohms, R2 = 83 ohms, R3 = 22 ohms, R4 = 51 Ohms, E1 = 3 volts, E2 = 1.49 volts

Part 2 Procedure (Basic Network): I1 = +15 mA, I2 = +10 mA, I3 = -23 mA?

Part 3 Procedure (Modified Network): I1 = 33 mA, I2 = -21 mA, I3 = -12 mA

Data Part I: Verification of Kirchhoff?s Laws

1) ?I_i = 0

At Junction G

I₁ + I₂ + I₃ = 15.0 mA + 10.0 mA – 23.0 mA = 0 mA

2) ?V_i = 0?

At Loop a-k-j-g-c-a_??

V₁ + V₂ + V₃ + V_4? = +1.25 V -+0.9 V + 0.55 V +-2.7 =? 0.0 V

Part II: Basic Network

3) Set up 3 equations with 3 unknowns (the currents) for the Basic Network: one for junction g, and the other two for loops a-k-j-g-c-a and i-g-c-f-i.? You should include two loops equations and the current equation in terms of e?s, I?s and R?s.

4) Solve these equations by a method of your choice, but show clearly your method and include most of the steps in your calculation.

WARNING😕 Don?t use your measured values of currents in any of your calculations. You will use the recorded values of resistances and the emf’s in order to determine the theoretical values of the currents.

NOTE: It is useful to solve your equations algebraically without inserting values for the resistances and emf’s. ?Then it will be easier to calculate the currents for Part III: the Modified Network.? For example, display the formula for I1, in the following form:

5) Display your results in Table 1 as shown. The discrepancies should be stated as the difference between the calculated and measured values, not in the % difference.

Part III: Modified Network

6) Set up 3 equations with 3 unknowns for the modified network, as in (3) above.

Hint: You should now appreciate the importance of solving the equations algebraically, as in (3), because by reversing the sign of ?2 and setting R4 = 0 in (3) you automatically have the solution for the Modified Network.

7) Solve the modified equations.