Question

Determine the current in branches AB, AC, and BC of the network shown in the figure.
\includegraphics[width=0.5\linewidth]{22.png}

Hint:

When solving circuits with multiple loops and nodes, it is useful to clearly define all current directions and check consistency of signs in your KVL and KCL equations.

Answer 1

1. Given Equations:
For closed loop ADCA:
For closed loop ADCA:
\[ 7I_1 - 6I_2 - 2I_3 = 10 \quad (i)} \] For closed loop ABCA:
\[ I_1 + 6I_2 + 2I_3 = 10 \quad (ii)} \] For closed loop BCDED:
\[ 2I_1 - 4I_2 - 4I_3 = -5 \quad (iii)} \] 2. Solve Equations (i) and (ii):
Add Equations (i) and (ii):
\[ 8I_1 = 20 \implies I_1 = 2.5 \, A} \] 3. Substitute \( I_1 = 2.5 \, A} \) into Equation (ii):
\[ 6I_2 + 2I_3 = 7.5 \quad (iv)} \] 4. Substitute \( I_1 = 2.5 \, A} \) into Equation (iii):
\[ I_2 + I_3 = 2.5 \quad (v)} \] 5. Solve Equations (iv) and (v):
From Equation (v): \( I_3 = 2.5 - I_2 \). Substitute into Equation (iv):
\[ 6I_2 + 2(2.5 - I_2) = 7.5 \] \[ 4I_2 = 2.5 \implies I_2 = 0.625 \, A} \] Then, \( I_3 = 2.5 - 0.625 = 1.875 \, A} \). 6. Verify the Results:
Current in branch AB: \( I_2 = 0.625 \, A} \),
Current in branch AC: \( I_1 = 2.5 \, A} \),
Current in branch BC: \( I_2 + I_3 = 2.5 \, A} \).
Final Answer:
Current in branch \( AB \): \( I_2 = 0.625 \, A} \),
Current in branch \( AC \): \( I_1 = 2.5 \, A} \),
Current in branch \( BC \): \( I_2 + I_3 = 2.5 \, A} \).