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
Step 1: Analyze the circuit with V_S1 active and V_S2 turned off. To turn off V_S2, replace it with a short circuit. The circuit then consists of R_1 in series with the parallel combination of R_2 and R_3. The equivalent resistance of R_2 and R_3 in parallel is: R_23 = (R_2 × R_3)/(R_2 + R_3) = (470\, × 470\,)/(470\, + 470\,) = (220900\,^2)/(940\,) = 235\, The total resistance seen by V_S1 is: R_total1 = R_1 + R_23 = 470\, + 235\, = 705\, The total current flowing from V_S1 is: I_total1 = V_S1R_total1 = (12\,V)/(705\,) This current splits between R_2 and R_3. Since R_2 = R_3, the current divides equally. The current through R_2 due to V_S1 (I_R2,S1) is: I_R2,S1 = I_total12 = (12\,V / 705\,)/(2) = (6\,V)/(705\,) ≈ 0.008511\,A = 8.511\,mA The direction of this current is downwards through R_2. The voltage across R_2 due to V_S1 (V_R2,S1) is: V_R2,S1 = I_R2,S1 × R_2 = (6\,V)/(705\,) × 470\, = (2820)/(705)\,V = 4\,V The polarity is positive at the top of R_2. Step 2: Analyze the circuit with V_S2 active and V_S1 turned off. To turn off V_S1, replace it with a short circuit. The circuit then consists of R_3 in series with the parallel combination of R_1 and R_2. The equivalent resistance of R_1 and R_2 in parallel is: R_12 = (R_1 × R_2)/(R_1 + R_2) = (470\, × 470\,)/(470\, + 470\,) = (220900\,^2)/(940\,) = 235\, The total resistance seen by V_S2 is: R_total2 = R_3 + R_12 = 470\, + 235\, = 705\, The total current flowing from V_S2 is: I_total2 = V_S2R_total2 = (18\,V)/(705\,) This current splits between R_1 and R_2. Since R_1 = R_2, the current divides equally. The current through R_2 due to V_S2 (I_R2,S2) is: I_R2,S2 = I_total22 = (18\,V / 705\,)/(2) = (9\,V)/(705\,) ≈ 0.012766\,A = 12.766\,mA The direction of this current is downwards through R_2. The voltage across R_2 due to V_S2 (V_R2,S2) is: V_R2,S2 = I_R2,S2 × R_2 = (9\,V)/(705\,) × 470\, = (4230)/(705)\,V = 6\,V The polarity is positive at the top of R_2. Step 3: Combine the results to find the total current and voltage for R_2. Since both I_R2,S1 and I_R2,S2 flow in the same direction (downwards through R_2), they add up. I_R2 = I_R2,S1 + I_R2,S2 = (6\,V)/(705\,) + (9\,V)/(705\,) = (15\,V)/(705\,)