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.
ScanSolve AI Answer
My apologies if you missed them in the previous message. Here are the solutions for questions 17-20 again: Question 17: Step 1: Identify the given current equation and resistance, and the formulas for mean power. The alternating current is I = 3 (50t). From this, the peak current is I_peak = 3 A. The resistive load is R = 10 Ω. The mean power dissipated in a resistive load is P_mean = I_rms^2 R. Step 2: Calculate the RMS current from the peak current. The relationship between RMS current (I_rms) and peak current (I_peak) is I_rms = I_peaksqrt(2). I_rms = 3 Asqrt(2) Step 3: Substitute the RMS current and resistance into the mean power formula. P_mean = ((3)/(sqrt(2)))^2 × 10 Ω P_mean = (9)/(2) × 10 W P_mean = 4.5 × 10 W P_mean = 45 W The mean power dissipated is 45 W. Question 18: Step 1: Identify the given power and RMS voltage, and the formulas for power and peak current. Power of the heater P = 4.0 kW = 4000 W. RMS voltage of the supply V_rms = 240 V. The power in an AC circuit is P = V_rms I_rms. The relationship between RMS current and peak current is I_peak = I_rms sqrt(2). Step 2: Calculate the RMS current. I_rms = (P)/(V_rms) I_rms = 4000 W240 V I_rms = (400)/(24) A = (50)/(3) A ≈ 16.67 A Step 3: Calculate the peak current. I_peak = I_rms sqrt(2) I_peak = (50)/(3) A × sqrt(2) I_peak ≈ 16.67 A × 1.414 I_peak ≈ 23.57 A The peak current drawn is 23.6 A. Question 19: Step 1: Identify the given voltages and secondary turns, and the transformer turns ratio formula. Primary voltage V_p = 220 V. Secondary voltage V_s = 7.5 V. Number of turns on the secondary coil N_s = 140 turns. For an ideal transformer, the ratio of voltages is equal to the ratio of turns: (V_p)/(V_s) = (N_p)/(N_s). Step 2: Rearrange the formula to solve for the number of turns on the primary coil (N_p) and substitute the values. N_p = N_s × (V_p)/(V_s) N_p = 140 turns × 220 V7.5 V N_p = 140 × 29.333... N_p ≈ 4106.67 Since the number of turns must be an integer, we round to the nearest whole number. The number of turns on the primary coil is 4107 turns. Question 20: Step 1: Identify the given voltages and secondary current, and the power conservation formula for an ideal transformer. Primary voltage V_p = 240 V. Secondary voltage V_s = 15 V. Secondary current I_s = 2.0 A. For an ideal transformer, the power in the primary coil equals the power in the secondary coil: P_p = P_s. This can be written as V_p I_p = V_s I_s. Step 2: Rearrange the formula to solve for the primary current (I_p) and substitute the values. I_p = (V_s I_s)/(V_p) I_p = (15 V) × (2.0 A)240 V I_p = 30 W240 V I_p = (1)/(8) A I_p = 0.125 A The current drawn from the primary coil is 0.125 A. That's 2 down. 3 left today — send the next one.