1. **Stating the problem:** We need to find the currents $I_{R1}$, $I_{R2}$, and $I_{R3}$ in the circuit using Norton's theorem. 2. **Understanding Norton's theorem:** Norton's theorem states that any linear electrical network with voltage sources and resistors can be replaced at terminals by an equivalent current source $I_{no}$ in parallel with a Norton resistance $R_{no}$. 3. **Step A: Short circuit $R3$ to find $I_{no}$ and $R_{no}$** - Short $R3$ means replace $R3$ with a wire (0 Ohm). - Calculate the Norton current $I_{no}$ as the current through the shorted $R3$. - Calculate the Norton resistance $R_{no}$ by turning off all independent voltage sources (replace voltage sources with short circuits) and calculating the equivalent resistance seen from the terminals where $R3$ was connected. 4. **Calculate $I_{no}$:** - With $R3$ shorted, the circuit reduces to two voltage sources $E1=10V$ and $E2=6V$ connected through $R1=2\Omega$ and $R2=6\Omega$. - The voltage difference across $R1$ and $R2$ is $E1 - E2 = 10 - 6 = 4V$. - The total resistance in series is $R1 + R2 = 2 + 6 = 8\Omega$. - So, $I_{no} = \frac{4V}{8\Omega} = 0.5A$ flowing from $E1$ side to $E2$ side. 5. **Calculate $R_{no}$:** - Turn off voltage sources: replace $E1$ and $E2$ with short circuits. - The resistors $R1=2\Omega$ and $R2=6\Omega$ are in parallel from the terminals where $R3$ was connected. - Calculate equivalent resistance: $$ R_{no} = \frac{R1 \times R2}{R1 + R2} = \frac{2 \times 6}{2 + 6} = \frac{12}{8} = 1.5\Omega $$ 6. **Norton equivalent circuit:** - Current source $I_{no} = 0.5A$ in parallel with $R_{no} = 1.5\Omega$. - Reconnect $R3 = 3\Omega$ in parallel with $R_{no}$. 7. **Calculate total current through $R3$ branch:** - Total resistance in parallel: $$ R_{total} = \frac{R_{no} \times R3}{R_{no} + R3} = \frac{1.5 \times 3}{1.5 + 3} = \frac{4.5}{4.5} = 1\Omega $$ - The Norton current $I_{no} = 0.5A$ splits between $R_{no}$ and $R3$ inversely proportional to their resistances. 8. **Calculate current through $R3$ ($I_{R3}$):** $$ I_{R3} = I_{no} \times \frac{R_{no}}{R_{no} + R3} = 0.5 \times \frac{1.5}{1.5 + 3} = 0.5 \times \frac{1.5}{4.5} = 0.5 \times \frac{1}{3} = 0.1667A $$ 9. **Calculate current through $R1$ ($I_{R1}$) and $R2$ ($I_{R2}$):** - The current through $R1$ and $R2$ is the same as the current through $R_{no}$ branch: $$ I_{R1} = I_{R2} = I_{no} - I_{R3} = 0.5 - 0.1667 = 0.3333A $$ 10. **Summary of currents:** - $I_{R1} = 0.3333A$ - $I_{R2} = 0.3333A$ - $I_{R3} = 0.1667A$ These currents are consistent with the circuit and Norton's theorem.