This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
here are the solutions to the circuit analysis problems. QUESTION 1 1.1 Use the superposition method to calculate the current flowing through the 25 resistor shown in FIGURE 1. The superposition method involves calculating the current due to each independent source separately and then summing them up. Step 1: Consider V_1 = 11 V acting alone (short-circuit V_2). The 25 and 21 resistors are in parallel: R_p1 = (25 × 21 )/(25 + 21 ) = (525)/(46) ≈ 11.413 This R_p1 is in series with the 22 resistor: R_s1 = 22 + R_p1 = 22 + (525)/(46) = (1012 + 525)/(46) = (1537)/(46) ≈ 33.413 This R_s1 is in parallel with the 30 resistor: R_p2 = 30 × R_s130 + R_s1 = (30 × 1537)/(46) 30 + (1537)/(46) = (46110)/(1380 + 1537) = (46110)/(2917) ≈ 15.807 The total resistance seen by V_1 is 60 in series with R_p2: R_total1 = 60 + R_p2 = 60 + (46110)/(2917) = (175020 + 46110)/(2917) = (221130)/(2917) ≈ 75.807 The total current from V_1 is: I_total1 = (V_1)/(R_total1) = 11 V(221130)/(2917) = (32087)/(221130) A ≈ 0.1451 A Using current division, the current flowing through the branch containing R_s1 (which includes the 25 resistor) is: I_branch1 = I_total1 × (30 )/(30 + R_s1) = (32087)/(221130) A × (30 )/(30 + 1537)46 = (32087)/(221130) × (1380)/(2917) A ≈ 0.0686 A This current I_branch1 then splits between 25 and 21 . The current through the 25 resistor is: I_25,V1 = I_branch1 × (21 )/(25 + 21 ) = (44279460)/(645065010) A × (21)/(46) ≈ 0.03133 A The direction of this current is downwards. Step 2: Consider V_2 = 9 V acting alone (short-circuit V_1). The 60 and 30 resistors are in parallel: R_p3 = (60 × 30 )/(60 + 30 ) = (1800)/(90) = 20 This R_p3 is in series with the 22 resistor: R_s2 = 22 + R_p3 = 22 + 20 = 42 This R_s2 is in parallel with the 25 resistor: R_p4 = 25 × R_s225 + R_s2 = (25 × 42 )/(25 + 42 ) = (1050)/(67) ≈ 15.672 The total resistance seen by V_2 is 21 in series with R_p4: R_total2 = 21 + R_p4 = 21 + (1050)/(67) = (1407 + 1050)/(67) = (2457)/(67) ≈ 36.672 The total current from V_2 is: I_total2 = (V_2)/(R_total2) = 9 V(2457)/(67) = (603)/(2457) A ≈ 0.2454 A Using current division, the current flowing through the 25 resistor is: I_25,V2 = I_total2 × R_s225 + R_s2 = (603)/(2457) A × (42 )/(25 + 42 ) = (603)/(2457) × (42)/(67) A ≈ 0.15385 A The direction of this current is downwards. Step 3: Sum the currents. Both currents I_25,V1 and I_25,V2 flow in the same direction (downwards) through the 25 resistor. I_25 = I_25,V1 + I_25,V2 = 0.03133 A + 0.15385 A = 0.18518 A The current flowing through the 25 resistor is 0.1852 A. 1.2 Study FIGURE 2 below. Using Thevenin's method, calculate the current flowing through the 15 resistor. Step 1: Remove the 15 resistor (R_5) and find V_Th (Thevenin voltage across terminals B and D). Assume the bottom wire (connected to C and D) is ground (0 V). Then V_A = 5 V. For the path A-B-C, the voltage at B is found using the voltage divider rule: V_B = V_A × (R_3)/(R_2 + R_3) = 5 V × (5 )/(35 + 5 ) = 5 V × (5)/(40) = 0.625 V For the path A-D-C, the voltage at D is found using the voltage divider rule: V_D = V_A × (R_4)/(R_1 + R_4) = 5 V × (10 )/(30 + 10 ) = 5 V × (10)/(40) = 1.25 V The Thevenin voltage V_Th is the potential difference between B and D. Let's define it as V_DB (since V_D > V_B). V_Th = V_DB = V_D - V_B = 1.25 V - 0.625 V = 0.625 V Step 2: Find R_Th (Thevenin resistance looking into terminals B and D). Short-circuit the voltage source V_T. This means points A and C are now electrically connected. Looking from terminal B, R_2 is connected to A and R_3 is connected to C. Since A and C are shorted, R_2 and R_3 are in parallel: R_BC' = R_2 || R_3 = (35 × 5 )/(35 + 5 ) = (175)/(40) = 4.375 Looking from terminal D, R_1 is connected to A and R_4 is connected to C. Since A and C are shorted, R_1 and R_4 are in parallel: R_DC' = R_1 || R_4 = (30 × 10 )/(30 + 10 ) = (300)/(40) = 7.5 The Thevenin resistance R_Th is the sum of these two parallel combinations, as they are effectively in series between B and D: R_Th = R_BC' + R_DC' = 4.375 + 7.5 = 11.875 Step 3: Calculate the current flowing through the 15 resistor (R_5). Now, connect the 15 resistor (R_5) to the Thevenin equivalent circuit. The current I_R5 is given by: I_R5 = V_ThR_Th + R_5 = 0.625 V11.875 + 15 = 0.625 V26.875 ≈ 0.0232558 A The current flowing through the 15 resistor is 0.0233 A. What's next?