Let's solve the circuit step by step. --- # Given Data: | Resistors | Resistance | |-------------|--------------| | \( R_1 \) | 1.2 kΩ | | \( R_2 \) | 1.5 kΩ | | \( R_3 \) | 1 kΩ | | Voltage Source | \( V_T \) | 54 V | --- # Step 1: Find the equivalent resistance \( R_T \) **Note**: \( R_T \) is usually the total equivalent resistance seen from the source. ### Since the resistors \( R_2 \) and \( R_3 \) are in parallel: \[ \frac{1}{R_{23}} = \frac{1}{R_2} + \frac{1}{R_3} \] \[ \frac{1}{R_{23}} = \frac{1}{150} + \frac{1}{100} \quad \text{(values in ohms)} \] \[ \frac{1}{R_{23}} = \frac{2}{300} + \frac{3}{300} = \frac{5}{300} \] \[ R_{23} = \frac{300}{5} = 600\, \Omega \] ### Now, \( R_{23} \) is in series with \( R_1 \): \[ R_T = R_1 + R_{23} = 120\, \Omega + 600\, \Omega = 180\, \Omega \] --- # **Part a: \( R_T \)** \[ \boxed{ R_T = 1.8\, \text{k}\Omega } \] --- # Step 2: Find the total current \( I_T \) The total current from the source: \[ I_T = \frac{V_T}{R_T} = \frac{54\,V}{180\, \Omega} \] \[ I_T = .03\, \text{A} = 30\, \text{mA} \] --- # **Part b: \( I_T \)** \[ \boxed{ I_T = 30\, \text{mA} } \] --- # Step 3: Find the voltages across \( R_1 \), \( R_2 \), and \( R_3 \) Since \( R_2 \) and \( R_3 \) are in parallel, the voltage across them is the same and equal to the voltage across \( R_{23} \): \[ V_{R_{23}} = I_{R_{23}} \times R_{23} \] But first, find the current through \( R_1 \): \[ I_{R_1} = \frac{V_{R_1}}{R_1} \] Total current: \[ I_T = I_{R_1} + I_{R_{23}} \] Since the total current splits between \( R_2 \) and \( R_3 \), the voltage across the parallel combination: \[ V_{R_{23}} = V_{R_2} = V_{R_3} \] And: \[ V_{R_{23}} = V_T - V_{R_1} \] But because the resistors are in series with the total: \[ V_{R_1} = I_T \times R_1 = .03\,A \times 120\, \Omega = 36\,V \] Hence, \[ V_{R_{23}} = V_T - V_{R_1} = 54\,V - 36\,V = 18\,V \] --- # Part c: Voltages \( V_1, V_2, V_3 \) - \( V_1 \): Voltage across \( R_1 \): **36 V** - \( V_2 \) and \( V_3 \): Voltage across \( R_2 \) and \( R_3 \): \[ V_2 = V_3 = V_{R_{23}} = 18\,V \] --- # Step 4: Find the currents \( I_2 \) and \( I_3 \) - Current through \( R_2 \): \[ I_2 = \frac{V_{R_2}}{R_2} = \frac{18\,V}{150\, \Omega} = .012\,A = 12\,mA \] - Current through \( R_3 \): \[ I_3 = \frac{V_{R_3}}{R_3} = \frac{18\,V}{100\, \Omega} = .018\,A = 18\,mA \] --- # **Part d: Currents \( I_2 \) and \( I_3 \)** \[ \boxed{ I_2 = 12\, \text{mA}, \quad I_3 = 18\, \text{mA} } \] --- # **Summary:** | Part | Result | |---------|--------| | a. \( R_T \) | 1.8 kΩ | | b. \( I_T \) | 30 mA | | c. \( V_1, V_2, V_3 \) | 36 V, 18 V, 18 V | | d. \( I_2, I_3 \) | 12 mA, 18 mA | Let me know if you'd like additional details or clarification!