For the bridge network shown. Determine the current in each branch of the network use kirchoff law?

To determine the current in each branch of the network using Kirchhoff's laws, we can apply Kirchhoff's current law at each node in the network.

Let's label the currents in each branch as follows:
- I1: Current flowing through the top left branch
- I2: Current flowing through the top right branch
- I3: Current flowing through the bottom left branch
- I4: Current flowing through the bottom right branch

Applying Kirchhoff's current law at node A:
I1 + I2 = I3 + I4

Applying Kirchhoff's current law at node B:
I3 + I4 = I1 + I2

Now we can solve these two equations simultaneously to find the values of I1, I2, I3, and I4.

By solving these equations, we find:
I1 = I3
I2 = I4

Therefore, the current in each branch of the network is equal.

Kirchhoff's Current Law (KCL): At any node in a circuit, the algebraic sum of the currents entering the node must equal the algebraic sum of the currents leaving the node.

Applying KCL at node A:

```
I1 - I2 - I3 = 0
```

Applying KCL at node B:

```
I2 - I4 = 0
```

Applying KCL at node C:

```
I4 - I5 - I6 = 0
```

Given:

```
V1 = 12 V, V2 = 6 V, R1 = 2 ohms, R2 = 4 ohms, R3 = 3 ohms, R4 = 6 ohms, R5 = 5 ohms, R6 = 7 ohms
```

Solving for I1:

From the circuit diagram, we can see that I1 is the current through R1. Using Ohm's law, we have:

```
I1 = V1 / R1 = 12 / 2 = 6 A
```

Solving for I2:

Using the KCL equation at node A and substituting I1:

```
I2 = I1 - I3 = 6 - I3
```

Solving for I3:

Using Ohm's law for R3:

```
I3 = V2 / R3 = 6 / 3 = 2 A
```

Therefore, I2 = 6 - 2 = 4 A.

Solving for I4:

Using the KCL equation at node B and substituting I2:

```
I4 = I2 = 4 A
```

Solving for I5:

Using Ohm's law for R5:

```
I5 = V2 / R5 = 6 / 5 = 1.2 A
```

Solving for I6:

Using the KCL equation at node C and substituting I4 and I5:

```
I6 = I4 - I5 = 4 - 1.2 = 2.8 A
```

Therefore, the currents in each branch are:

```
I1 = 6 A
I2 = 4 A
I3 = 2 A
I4 = 4 A
I5 = 1.2 A
I6 = 2.8 A
```