Let's solve the circuit step by step using **phasor analysis** (AC circuit analysis). --- ## **Given:** - \( R = 8\,\Omega \) - \( X_{L1} = 20\,\Omega \) - \( X_{C} = 16\,\Omega \) - \( X_{L2} = 14\,\Omega \) - \( E_1 = 48 \angle ^\circ \) V - \( E_2 = 32 \angle 45^\circ \) V We are to find the **current \( I_L \)** flowing in the circuit. --- ## **Step 1: Draw the Phasor (Impedance) Diagram** The circuit can be simplified to a single loop: \[ E_1 - (R + jX_{L1} - jX_C + jX_{L2})I_L = E_2 \] --- ## **Step 2: Write the Loop Equation (KVL)** Write KVL around the loop: \[ E_1 - I_L(R + jX_{L1} - jX_C + jX_{L2}) = E_2 \] Solve for \( I_L \): \[ E_1 - E_2 = I_L(R + jX_{L1} - jX_C + jX_{L2}) \] --- ## **Step 3: Calculate the Total Impedance** \[ Z_{total} = R + jX_{L1} - jX_C + jX_{L2} \] Substitute values: - \( R = 8 \) - \( X_{L1} = 20 \) - \( X_C = 16 \) - \( X_{L2} = 14 \) \[ Z_{total} = 8 + j20 - j16 + j14 \] \[ = 8 + j(20 - 16 + 14) \] \[ = 8 + j18 \] --- ## **Step 4: Calculate \( E_1 - E_2 \) (in phasor form)** \[ E_1 = 48 \angle ^\circ \] \[ E_2 = 32 \angle 45^\circ \] Convert \( E_2 \) to rectangular form: \[ E_2 = 32(\cos 45^\circ + j\sin 45^\circ) = 32(.7071 + j.7071) = 22.63 + j22.63 \] \( E_1 \) in rectangular form: \( 48 + j \) \[ E_1 - E_2 = (48 - 22.63) + j( - 22.63) = 25.37 - j22.63 \] --- ## **Step 5: Find the Current \( I_L \)** \[ I_L = \frac{E_1 - E_2}{Z_{total}} \] \[ = \frac{25.37 - j22.63}{8 + j18} \] --- ### **Divide Complex Numbers** Let’s write numerator and denominator in polar form: - **Numerator:** \( 25.37 - j22.63 \) - Magnitude: \( \sqrt{25.37^2 + 22.63^2} = \sqrt{644.37 + 511.91} = \sqrt{1156.28} \approx 34 \) - Angle: \( \tan^{-1} \left( \frac{-22.63}{25.37} \right) = \tan^{-1}(-.892) \approx -41.5^\circ \) - So, \( 34 \angle -41.5^\circ \) - **Denominator:** \( 8 + j18 \) - Magnitude: \( \sqrt{8^2 + 18^2} = \sqrt{64 + 324} = \sqrt{388} \approx 19.7 \) - Angle: \( \tan^{-1} \left( \frac{18}{8} \right) = \tan^{-1}(2.25) \approx 66.8^\circ \) - So, \( 19.7 \angle 66.8^\circ \) \[ I_L = \frac{34 \angle -41.5^\circ}{19.7 \angle 66.8^\circ} = 1.73 \angle (-41.5^\circ - 66.8^\circ) \] \[ = 1.73 \angle -108.3^\circ\ \text{A} \] --- ## **Final Answer** \[ \boxed{ I_L = 1.73 \angle -108.3^\circ\ \text{A} } \] --- ### **Summary Table** | Quantity | Value | |:---------------------|:--------------------------| | \( Z_{total} \) | \( 8 + j18 = 19.7 \angle 66.8^\circ\ \Omega \) | | \( E_1 - E_2 \) | \( 34 \angle -41.5^\circ \) V | | \( I_L \) | \( 1.73 \angle -108.3^\circ \) A | If you need voltage across any element or further calculations, let me know!