Physics - Current Electricity Question with Solution | TestHub

The current of $9\text{ A}$ entering at point P splits into two parallel paths to reach point R.

Path 1 is along P $\to$ Q $\to$ R. The equivalent resistance of this path is $R_1=3\Omega +3\Omega =6\Omega$. The current through this path is $I_1$.

Path 2 is directly from P $\to$ R. The resistance of this path is $R_2=3\Omega$. The current through this path is $I_2$.

Since the two paths are in parallel, the potential difference across them is equal:

$V_{PR}=I_1{R}_1=I_2{R}_2$

$I_1(6)=I_2(3)$

$I_2=2I_1$

According to Kirchhoff's current law at junction P, the total current is the sum of the currents in the two branches:

$I_1+I_2=9$

Substituting $I_2=2I_1$ into the above equation:

$I_1+2I_1=9$

$3I_1=9\Rightarrow I_1=3\text{ A}$

Now, substituting the value of $I_1$ to find $I_2$:

$I_2=2(3)=6\text{ A}$

Therefore, the currents $I_1$ and $I_2$ are $3\text{ A}$ and $6\text{ A}$ respectively.

Answer: $3A,6A$