1. **State the problem:** We are given currents $I = 5$ A entering node $a$ and $I_2 = 4$ A flowing through resistor $R_2$. We need to determine currents $I_1$, $I_3$, $I_4$, and $I_5$ in the given circuit using Kirchhoff’s Current Law (KCL). 2. **Analyze node $a$:** At node $a$, the current $I$ splits into $I_1$ (top branch) and $I_2$ (bottom branch). Using KCL, the sum of currents flowing out of node $a$ equals the current flowing in: $$I = I_1 + I_2$$ Given $I = 5$ A and $I_2 = 4$ A, solve for $I_1$: $$I_1 = I - I_2 = 5 - 4 = 1\ \text{A}$$ 3. **Currents at nodes $b$ and $c$ through resistors $R_3$ and $R_4$ respectively:** By the problem statement, current $I_3$ flows through $R_3$ downstream of node $b$, and current $I_4$ flows through $R_4$ downstream of node $c$. Since the nodes $b$ and $c$ lead into $R_3$ and $R_4$ with no additional branching, the current flowing into each resistor equals the current in its respective branch: $$I_3 = I_1 = 1\ \text{A}$$ $$I_4 = I_2 = 4\ \text{A}$$ 4. **At node $d$, currents recombine:** The currents $I_3$ and $I_4$ combine to form current $I_5$ flowing through resistor $R_5$: $$I_5 = I_3 + I_4 = 1 + 4 = 5\ \text{A}$$ **Final answers:** $$I_1 = 1\ A,\ I_3 = 1\ A,\ I_4 = 4\ A,\ I_5 = 5\ A$$