Let's analyze and solve the circuit step-by-step. --- ## Given Data: - \(V_T = 54\,V\) - \(R_1 = 1.2\,k\Omega = 120\,\Omega\) - \(R_2 = 1.5\,k\Omega = 150\,\Omega\) - \(R_3 = 1\,k\Omega = 100\,\Omega\) --- ## Step 1: Find equivalent resistance for the parallel resistors \(R_2\) and \(R_3\) \[ \frac{1}{R_{23}} = \frac{1}{R_2} + \frac{1}{R_3} \] \[ \frac{1}{R_{23}} = \frac{1}{150} + \frac{1}{100} \] \[ \frac{1}{R_{23}} = \frac{2}{300} + \frac{3}{300} = \frac{5}{300} \] \[ R_{23} = \frac{300}{5} = 600\,\Omega \] --- ## Step 2: Find the total resistance \(R_T\) The total resistance seen by the source: \[ R_T = R_1 + R_{23} = 120\,\Omega + 600\,\Omega = 180\,\Omega \] --- ## Step 3: Find the total current \(I_T\) Using Ohm's Law: \[ I_T = \frac{V_T}{R_T} = \frac{54\,V}{180\,\Omega} = .03\,A = 30\,mA \] --- ## Step 4: Find the current through \(R_2\) and \(R_3\) Since \(R_2\) and \(R_3\) are in parallel, the current splits: \[ I_2 = \frac{R_3}{R_2 + R_3} \times I_T \] or directly using voltage division: **Voltage across \(R_2\) and \(R_3\):** \[ V_{R_{23}} = I_T \times R_{23} = .03\,A \times 600\,\Omega = 18\,V \] **Current \(I_2\) through \(R_2\):** \[ I_2 = \frac{V_{R_{23}}}{R_2} = \frac{18\,V}{150\,\Omega} = .012\,A = 12\,mA \] **Current \(I_3\) through \(R_3\):** \[ I_3 = \frac{V_{R_{23}}}{R_3} = \frac{18\,V}{100\,\Omega} = .018\,A = 18\,mA \] --- ## **Summary of Results:** | Quantity | Value | |------------|--------| | \( R_T \) | 180 Ω | | \( I_T \) | 30 mA | | \( I_2 \) | 12 mA | | \( I_3 \) | 18 mA | --- Let me know if you'd like me to clarify further steps or solve for \(R_I\), \(V_1, V_2, V_3\), or \(I_1, I_2, I_3\) specifically.