Let's solve the circuit step by step using **Kirchhoff’s Laws**. We are given: - \( V_1 = 8.00~\text{V} \) - \( V_2 = 5.00~\text{V} \) - \( V_3 = 9.00~\text{V} \) - \( R_1 = 2.00~\Omega \) - \( R_2 = 4.00~\Omega \) - \( R_3 = 1.00~\Omega \) Let: - \( I_1 \) = current through \( R_1 \) (downward) - \( I_2 \) = current through \( R_2 \) (rightward) - \( I_3 \) = current through \( R_3 \) (downward) The direction is as shown in your assumption. --- ## **Step 1: Write Junction Rule (at top node)** **Incoming currents = Outgoing currents:** \[ I_1 = I_2 + I_3 \] --- ## **Step 2: Loop 1 (left loop: \( V_1 \to R_1 \to R_2 \to V_2 \))** Start at the top, go down through \( R_1 \), to the left through \( R_2 \), up through \( V_2 \), and back to the top. \[ -V_1 + I_1 R_1 + I_2 R_2 + V_2 = \] --- ## **Step 3: Loop 2 (right loop: \( V_3 \to R_3 \to R_2 \to V_2 \))** Start at the top, go down through \( R_3 \), left through \( R_2 \), up through \( V_2 \), and back to the top. \[ -V_3 + I_3 R_3 + I_2 R_2 + V_2 = \] --- ## **Step 4: Substitute Values** ### Equation 1 (Junction): \[ I_1 = I_2 + I_3 \tag{1} \] ### Equation 2 (Left loop): \[ -8.00 + 2.00I_1 + 4.00I_2 + 5.00 = \\ 2.00I_1 + 4.00I_2 = 3.00 \tag{2} \] ### Equation 3 (Right loop): \[ -9.00 + 1.00I_3 + 4.00I_2 + 5.00 = \\ 1.00I_3 + 4.00I_2 = 4.00 \tag{3} \] --- ## **Step 5: Solve the System** ### From (1): \( I_1 = I_2 + I_3 \) ### Plug into (2): \[ 2(I_2 + I_3) + 4I_2 = 3 \\ 2I_2 + 2I_3 + 4I_2 = 3 \\ 6I_2 + 2I_3 = 3 \tag{4} \] ### From (3): \[ I_3 + 4I_2 = 4 \implies I_3 = 4 - 4I_2 \tag{5} \] ### Substitute (5) into (4): \[ 6I_2 + 2(4 - 4I_2) = 3 \\ 6I_2 + 8 - 8I_2 = 3 \\ (6I_2 - 8I_2) + 8 = 3 \\ -2I_2 = 3 - 8 \\ -2I_2 = -5 \implies I_2 = \boxed{2.5~\text{A}} \] ### Find \( I_3 \): \[ I_3 = 4 - 4(25) = 4 - 10 = -6~\text{A} \] ### Find \( I_1 \): \[ I_1 = I_2 + I_3 = 2.5 + (-6) = -3.5~\text{A} \] --- ## **Step 6: Directions** - **Positive**: direction as assumed is correct. - **Negative**: actual current is in the **opposite** direction. #### So: - \( I_1 = 3.5~\text{A} \) **upward** (opposite of assumed down) - \( I_2 = 2.5~\text{A} \) **right** (as assumed) - \( I_3 = 6~\text{A} \) **upward** (opposite of assumed down) --- ## **Final Answers** | Current | Magnitude (A) | Direction | |---------|---------------|---------------------| | \( I_1 \) | 3.5 | **Up** (not down) | | \( I_2 \) | 2.5 | **Right** | | \( I_3 \) | 6. | **Up** (not down) | --- ### **Boxed Final Answers** \[ \boxed{I_1 = 3.5~\text{A}~\text{up}} \] \[ \boxed{I_2 = 2.5~\text{A}~\text{right}} \] \[ \boxed{I_3 = 6.~\text{A}~\text{up}} \]