in the circuit shown in fig. determine the branch currents using mesh analysis.

Explanation:

Step1: Define mesh currents

Let the left - hand mesh current be $I_1$ and the right - hand mesh current be $I_2$, both in the clock - wise direction.

Step2: Apply KVL to the left - hand mesh

For the left - hand mesh, starting from the lower left corner and moving clock - wise, we have $42-8I_1 - 4(I_1 - I_2)=0$. Simplifying gives $42-12I_1 + 4I_2 = 0$, or $12I_1-4I_2=42$.

Step3: Apply KVL to the right - hand mesh

For the right - hand mesh, starting from the lower right corner and moving clock - wise, we have $4(I_2 - I_1)-6I_2+10 = 0$. Simplifying gives $4I_2-4I_1-6I_2 + 10 = 0$, or $4I_1+2I_2=10$.

Step4: Solve the system of equations

Multiply the second equation $4I_1+2I_2=10$ by 3, we get $12I_1 + 6I_2=30$. Subtract the first equation $12I_1-4I_2=42$ from it: [ \begin{align*} (12I_1 + 6I_2)-(12I_1-4I_2)&=30 - 42\ 10I_2&=- 12\ I_2&=-1.2A \end{align*} ] Substitute $I_2=-1.2A$ into $4I_1+2I_2=10$: [ \begin{align*} 4I_1+2\times(-1.2)&=10\ 4I_1-2.4&=10\ 4I_1&=12.4\ I_1&=3.1A \end{align*} ]

Step5: Find branch currents

The current through the $8\Omega$ resistor is $I_1 = 3.1A$. The current through the $6\Omega$ resistor is $I_2=-1.2A$. The current through the $4\Omega$ resistor is $I_1 - I_2=3.1-(-1.2)=4.3A$.

Answer:

Current through $8\Omega$ resistor: $3.1A$, Current through $6\Omega$ resistor: $1.2A$ (in the opposite direction of assumed $I_2$), Current through $4\Omega$ resistor: $4.3A$