Physics - Current Electricity Question with Solution | TestHub

Let the potential at node $D$ be $0\text{ V}$.

The potentials at nodes $A$ and $C$ are $V_A=15\text{ V}$ and $V_C=30\text{ V}$ respectively.

Let the potential at node $B$ be $V_B$. Applying Kirchhoff's Current Law (KCL) at node $B$:

$\frac{V_B-V_A}{6}+\frac{V_B-V_D}{3}+\frac{V_B-V_C}{3}=0$

Substituting the values of $V_A$, $V_C$, and $V_D$:

$\frac{V_B-15}{6}+\frac{V_B-0}{3}+\frac{V_B-30}{3}=0$

Multiplying the entire equation by $6$:

$(V_B-15)+2V_B+2(V_B-30)=0$

$V_B-15+2V_B+2V_B-60=0$

$5V_B-75=0$

$5V_B=75\Rightarrow V_B=15\text{ V}$

The current through the branch $BD$ is:

$I_{BD}=\frac{V_B-V_D}{3}=\frac{15-0}{3}=5\text{ A}$

Answer: $5$ A