Answer:
0, 0, 0.0024 A
Step-by-step explanation:
Let's analyze what happens to this circuit when current starts flowing.
First of all, we observe that the branch 3 (in parallel with branches 1 and 2) has zero resistance: this means that when the current reaches the node, all the current will flow through the branch 3. This also means that no current will flow across resistors R1 and R2: so, we can already say that
![I_1=0](https://img.qammunity.org/2021/formulas/physics/middle-school/f6qs3pri0w877mnw22glsg8tmk1cfwanhu.png)
![I_2=0](https://img.qammunity.org/2021/formulas/physics/middle-school/brbzc02ftgoj3rlya5vywvyzl3evhvb5ky.png)
Now we have to determine the current through the branch 3, This is equal to the current passing through resistor R3. Since there are no other "active" resistors in the circuit, we can directly use Ohm's law:
![V=R_3I_3](https://img.qammunity.org/2021/formulas/physics/middle-school/jvn2f2fgzgkqwjb00w8ycv9ru6bj2rc07y.png)
where:
V = 12 V is the voltage of the battery, which corresponds to the potential difference across R3
is the resistance
Therefore, solving for I3, we find
![I_3=(V)/(R_3)=(12)/(5000)=0.0024 A](https://img.qammunity.org/2021/formulas/physics/middle-school/dd4tkv9imuhsabusll1iw0disqu4i3zjm1.png)