1. **State the problem:** We need to find the voltages $V_1$ and $V_2$ in the given AC circuit with complex impedances and voltage sources. 2. **Identify components and variables:** - Voltage source $V_s = 75\angle 0^\circ$ V - Resistor $R_1 = 4\ \Omega$ - Voltage source $100\angle 60^\circ$ V between $V_1$ and $V_2$ - Inductive reactance $j4\ \Omega$ - Capacitive reactance $-j1\ \Omega$ - Resistor $R_2 = 2\ \Omega$ 3. **Write node voltage equations:** Let the ground be at the bottom node. At node $V_1$: $$\frac{V_1 - 75}{4} + \frac{V_1 - V_2 + 100\angle 60^\circ}{j4 - j1} = 0$$ At node $V_2$: $$\frac{V_2 - V_1 - 100\angle 60^\circ}{j4 - j1} + \frac{V_2}{2} = 0$$ 4. **Simplify impedances:** $$j4 - j1 = j(4 - 1) = j3$$ 5. **Rewrite equations:** $$\frac{V_1 - 75}{4} + \frac{V_1 - V_2 + 100\angle 60^\circ}{j3} = 0$$ $$\frac{V_2 - V_1 - 100\angle 60^\circ}{j3} + \frac{V_2}{2} = 0$$ 6. **Multiply both sides to clear denominators:** Multiply first equation by $12j$ (LCM of denominators $4$ and $j3$): $$12j \times \frac{V_1 - 75}{4} + 12j \times \frac{V_1 - V_2 + 100\angle 60^\circ}{j3} = 0$$ Simplify: $$3j(V_1 - 75) + 4(V_1 - V_2 + 100\angle 60^\circ) = 0$$ Multiply second equation by $6j$ (LCM of denominators $j3$ and $2$): $$6j \times \frac{V_2 - V_1 - 100\angle 60^\circ}{j3} + 6j \times \frac{V_2}{2} = 0$$ Simplify: $$2(V_2 - V_1 - 100\angle 60^\circ) + 3j V_2 = 0$$ 7. **Expand and rearrange first equation:** $$3j V_1 - 225j + 4 V_1 - 4 V_2 + 400\angle 60^\circ = 0$$ Group terms: $$(4 + 3j) V_1 - 4 V_2 = 225j - 400\angle 60^\circ$$ 8. **Expand and rearrange second equation:** $$2 V_2 - 2 V_1 - 200\angle 60^\circ + 3j V_2 = 0$$ Group terms: $$(-2) V_1 + (2 + 3j) V_2 = 200\angle 60^\circ$$ 9. **Write system of equations:** $$\begin{cases} (4 + 3j) V_1 - 4 V_2 = 225j - 400\angle 60^\circ \\ -2 V_1 + (2 + 3j) V_2 = 200\angle 60^\circ \end{cases}$$ 10. **Convert $400\angle 60^\circ$ and $200\angle 60^\circ$ to rectangular form:** $$400\angle 60^\circ = 400(\cos 60^\circ + j \sin 60^\circ) = 400(0.5 + j0.866) = 200 + j346.4$$ $$200\angle 60^\circ = 200(0.5 + j0.866) = 100 + j173.2$$ 11. **Substitute values:** First equation right side: $$225j - (200 + j346.4) = -200 + j(225 - 346.4) = -200 - j121.4$$ System becomes: $$\begin{cases} (4 + 3j) V_1 - 4 V_2 = -200 - j121.4 \\ -2 V_1 + (2 + 3j) V_2 = 100 + j173.2 \end{cases}$$ 12. **Solve system using substitution or matrix method:** Matrix form: $$\begin{bmatrix}4+3j & -4 \\ -2 & 2+3j\end{bmatrix} \begin{bmatrix}V_1 \\ V_2\end{bmatrix} = \begin{bmatrix}-200 - j121.4 \\ 100 + j173.2\end{bmatrix}$$ Calculate determinant: $$D = (4+3j)(2+3j) - (-4)(-2) = (8 + 12j + 6j + 9j^2) - 8 = (8 + 18j - 9) - 8 = (-1 + 18j) - 8 = -9 + 18j$$ 13. **Calculate $V_1$ numerator determinant:** $$D_{V_1} = \begin{vmatrix}-200 - j121.4 & -4 \\ 100 + j173.2 & 2+3j\end{vmatrix} = (-200 - j121.4)(2+3j) - (-4)(100 + j173.2)$$ Calculate each term: $$(-200)(2) = -400$$ $$(-200)(3j) = -600j$$ $$(-j121.4)(2) = -j242.8$$ $$(-j121.4)(3j) = -3j^2 121.4 = 3 \times 121.4 = 364.2$$ Sum: $$-400 - 600j - j242.8 + 364.2 = (-400 + 364.2) + (-600j - 242.8j) = -35.8 - 842.8j$$ Second term: $$-(-4)(100 + j173.2) = 4(100 + j173.2) = 400 + j692.8$$ Sum total: $$-35.8 - 842.8j + 400 + 692.8j = 364.2 - 150j$$ 14. **Calculate $V_2$ numerator determinant:** $$D_{V_2} = \begin{vmatrix}4+3j & -200 - j121.4 \\ -2 & 100 + j173.2\end{vmatrix} = (4+3j)(100 + j173.2) - (-2)(-200 - j121.4)$$ Calculate first term: $$4 \times 100 = 400$$ $$4 \times j173.2 = j692.8$$ $$3j \times 100 = j300$$ $$3j \times j173.2 = 3j^2 173.2 = -3 \times 173.2 = -519.6$$ Sum: $$400 + j692.8 + j300 - 519.6 = (400 - 519.6) + j(692.8 + 300) = -119.6 + j992.8$$ Second term: $$-(-2)(-200 - j121.4) = -2(-200 - j121.4) = 400 + j242.8$$ Sum total: $$-119.6 + j992.8 - 400 - j242.8 = -519.6 + j750$$ 15. **Calculate $V_1$ and $V_2$:** $$V_1 = \frac{D_{V_1}}{D} = \frac{364.2 - 150j}{-9 + 18j}$$ $$V_2 = \frac{D_{V_2}}{D} = \frac{-519.6 + 750j}{-9 + 18j}$$ Multiply numerator and denominator by conjugate of denominator $-9 - 18j$: For $V_1$: $$V_1 = \frac{(364.2 - 150j)(-9 - 18j)}{(-9 + 18j)(-9 - 18j)}$$ Denominator: $$(-9)^2 + (18)^2 = 81 + 324 = 405$$ Numerator: $$364.2 \times -9 = -3277.8$$ $$364.2 \times -18j = -6555.6j$$ $$-150j \times -9 = 1350j$$ $$-150j \times -18j = 2700 j^2 = -2700$$ Sum numerator: $$(-3277.8 - 2700) + (-6555.6j + 1350j) = -5977.8 - 5205.6j$$ So: $$V_1 = \frac{-5977.8 - 5205.6j}{405} = -14.76 - 12.86j$$ For $V_2$: $$V_2 = \frac{(-519.6 + 750j)(-9 - 18j)}{405}$$ Numerator: $$-519.6 \times -9 = 4676.4$$ $$-519.6 \times -18j = 9352.8j$$ $$750j \times -9 = -6750j$$ $$750j \times -18j = -13500 j^2 = 13500$$ Sum numerator: $$(4676.4 + 13500) + (9352.8j - 6750j) = 18176.4 + 2602.8j$$ So: $$V_2 = \frac{18176.4 + 2602.8j}{405} = 44.9 + 6.43j$$ 16. **Final answers:** $$V_1 = -14.76 - 12.86j = 19.8 \angle -137.5^\circ \text{ V}$$ $$V_2 = 44.9 + 6.43j = 45.3 \angle 8.2^\circ \text{ V}$$ --- "slug": "voltages v1 v2", "subject": "electrical engineering", "desmos": {"latex": "V_1 = -14.76 - 12.86j, V_2 = 44.9 + 6.43j", "features": {"intercepts": false, "extrema": false}}, "q_count": 1