This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions to your practice problems: Question 3: Use Gauss Elimination method to solve the linear systems of equations. The system of equations is: x_1 - x_2 + x_3 = 0 -x_1 + x_2 - x_3 = 0 10x_1 + 25x_3 = 90 20x_1 + 10x_2 = 80 Step 1: Write the augmented matrix for the system. 1 & -1 & 1 & | & 0 \\ -1 & 1 & -1 & | & 0 \\ 10 & 0 & 25 & | & 90 \\ 20 & 10 & 0 & | & 80 Step 2: Perform row operations to achieve row echelon form. R_2 R_2 + R_1 R_3 R_3 - 10R_1 R_4 R_4 - 20R_1 1 & -1 & 1 & | & 0 \\ 0 & 0 & 0 & | & 0 \\ 0 & 10 & 15 & | & 90 \\ 0 & 30 & -20 & | & 80 Step 3: Swap rows to bring non-zero rows to the top and simplify R_3. Swap R_2 with R_3. 1 & -1 & 1 & | & 0 \\ 0 & 10 & 15 & | & 90 \\ 0 & 0 & 0 & | & 0 \\ 0 & 30 & -20 & | & 80 Swap R_3 with R_4. 1 & -1 & 1 & | & 0 \\ 0 & 10 & 15 & | & 90 \\ 0 & 30 & -20 & | & 80 \\ 0 & 0 & 0 & | & 0 R_2 (1)/(10)R_2 1 & -1 & 1 & | & 0 \\ 0 & 1 & 1.5 & | & 9 \\ 0 & 30 & -20 & | & 80 \\ 0 & 0 & 0 & | & 0 Step 4: Continue row operations. R_3 R_3 - 30R_2 1 & -1 & 1 & | & 0 \\ 0 & 1 & 1.5 & | & 9 \\ 0 & 0 & -20 - 30(1.5) & | & 80 - 30(9) \\ 0 & 0 & 0 & | & 0 1 & -1 & 1 & | & 0 \\ 0 & 1 & 1.5 & | & 9 \\ 0 & 0 & -65 & | & -190 \\ 0 & 0 & 0 & | & 0 Step 5: Normalize the third row. R_3 -(1)/(65)R_3 1 & -1 & 1 & | & 0 \\ 0 & 1 & 1.5 & | & 9 \\ 0 & 0 & 1 & | & (190)/(65) \\ 0 & 0 & 0 & | & 0 Simplify the fraction: (190)/(65) = (38)/(13). 1 & -1 & 1 & | & 0 \\ 0 & 1 & 1.5 & | & 9 \\ 0 & 0 & 1 & | & (38)/(13) \\ 0 & 0 & 0 & | & 0 Step 6: Use back substitution to solve for x_1, x_2, x_3. From the third row: x_3 = (38)/(13). From the second row: x_2 + 1.5x_3 = 9 x_2 + (3)/(2)((38)/(13)) = 9 x_2 + (57)/(13) = 9 x_2 = 9 - (57)/(13) = (117 - 57)/(13) = (60)/(13). From the first row: x_1 - x_2 + x_3 = 0 x_1 - (60)/(13) + (38)/(13) = 0 x_1 - (22)/(13) = 0 x_1 = (22)/(13). The solution is x_1 = (22)/(13), x_2 = (60)/(13), x_3 = (38)/(13). Question 4: Find the current I(t) in an RLC circuit with R=11, L=0.1H, C=10^-2F which is connected to the source of emf E(t) = 110(120 t) at a frequency of 60Hz, assuming the current and capacitor charge are 0 when t=0 seconds. Given: R=11, L=0.1H, C=10^-2F, E(t) = 110(120 t). Angular frequency = 120 rad/s. Initial conditions: I(0)=0 and Q(0)=0. Step 1: Set up the differential equation for the charge Q(t) in a series RLC circuit. L (d^2Q)/(dt^2) + R (dQ)/(dt) + (1)/(C)Q = E(t) Substitute the given values: 0.1 (d^2Q)/(dt^2) + 11 (dQ)/(dt) + (1)/(10^-2)Q = 110(120 t) 0.1 (d^2Q)/(dt^2) + 11 (dQ)/(dt) + 100Q = 110(120 t) Step 2: Find the homogeneous solution Q_h(t) and the particular solution Q_p(t). The characteristic equation for the homogeneous part is 0.1r^2 + 11r + 100 = 0, or r^2 + 110r + 1000 = 0. Using the quadratic formula, r = -110 ± sqrt(110^2 - 4(1000))2 = -110 ± sqrt(12100 - 4000)2 = -110 ± sqrt(8100)2 = (-110 ± 90)/(2). The roots are r_1 = -10 and r_2 = -100. So, Q_h(t) = A e^-10t + B e^-100t. For the particular solution Q_p(t), we use the phasor method for steady-state current. The impedance Z = R + j( L - (1)/( C)). L = 120 × 0.1 = 12 ≈ 37.6991. (1)/( C) = (1)/(120 × 10^-2) = (1)/(1.2) ≈ 0.2653. Z = 11 + j(12 - (1)/(1.2)) ≈ 11 + j(37.6991 - 0.2653) = 11 + j37.4338. The magnitude of impedance |Z| = sqrt(11^2 + (37.4338)^2) = sqrt(121 + 1399.98) = sqrt(1520.98) ≈ 38.9997. The phase angle of impedance _Z = ((37.4338)/(11)) ≈ 73.61^ ≈ 1.2848 radians. The peak current I_peak = E_peak|Z| = (110)/(38.9997) ≈ 2.8205 A. The steady-state current is I_p(t) = I_peak ( t - _Z) = 2.8205 (120 t - 73.61^). The steady-state charge Q_p(t) leads the current by 90^ in a capacitive circuit, or lags the voltage by 90^. Q_p(t) = I_peak ( t - _Z - ()/(2)) = (2.8205)/(120) (120 t - 73.61^ - 90^) Q_p(t) = 0.00747 (120 t - 163.61^). Alternatively, Q_p(t) = C_1 (120 t) + C_2 (120 t), where C_1 ≈ -0.002108 and C_2 ≈ -0.007173 (from previous thought process). Step 3: Apply initial conditions to the total solution Q(t) = Q_h(t) + Q_p(t). Q(t) = A e^-10t + B e^-100t + 0.00747 (120 t - 163.61^). I(t) = (dQ)/(dt) = -10A e^-10t - 100B e^-100t + 2.8205 (120 t - 73.61^). Using Q(0)=0: 0 = A e^0 + B e^0 + 0.00747 (-163.61^) 0 = A + B + 0.00747(-0.2822) 0 = A + B - 0.002108 A + B = 0.002108 (Eq. 1) Using I(0)=0: 0 = -10A e^0 - 100B e^0 + 2.8205 (-73.61^) 0 = -10A - 100B + 2.8205(-0.9599) 0 = -10A - 100B - 2.707 -10A - 100B = 2.707 (Eq. 2) Step 4: Solve the system of equations for A and B. From (Eq. 1), A = 0.002108 - B. Substitute into (Eq. 2): -10(0.002108 - B) - 100B = 2.707 -0.02108 + 10B - 100B = 2.707 -90B = 2.707 + 0.02108 -90B = 2.72808 B = -(2.72808)/(90) ≈ -0.030312 A = 0.002108 - (-0.030312) = 0.002108 + 0.030312 = 0.03242. The current I(t) is: I(t) = -10(0.03242) e^-10t - 100(-0.030312) e^-100t + 2.8205 (120 t - 73.61^) I(t) = -0.3242 e^-10t + 3.0312 e^-100t + 2.8205 (120 t - 73.61^) A Drop the next question! 📸