Diagrams showing wave fronts and definitions of interference and wave phenomena.

Hey bilgate, good to see you again. Here are the solutions for the questions from the image: 2. (a) (i) Copy each diagram and draw the wave fronts after the obstacle. Figure 1 (X - Refraction): When plane waves move from deep water to shallow water, their speed decreases, and their wavelength decreases. If the boundary is at an angle, the waves will refract (bend) towards the normal. The wave fronts in the shallow water will be closer together and bent. (Imagine drawing parallel lines for wave fronts in deep water, then drawing them closer together and angled differently in shallow water, bending towards the normal to the boundary.) Figure 1 (Y - Diffraction): When plane waves pass through a narrow slit (obstacle Y), they diffract. This means the waves spread out after passing through the opening, forming circular or semi-circular wave fronts. (Imagine drawing parallel lines for wave fronts approaching the slit, then drawing semi-circular wave fronts spreading out from the slit.) (ii) Name the wave phenomenon illustrated at X of figure 1. The wave phenomenon illustrated at X is refraction. 2. (b) (i) Define interference. Interference is the superposition of two or more waves that results in a new wave pattern, where the amplitudes of the waves either add up (constructive interference) or cancel out (destructive interference). (ii) Two similar waves, each of amplitude 4 cm undergo constructive interference. Determine the amplitude of the resultant wave. Step 1: Understand constructive interference. In constructive interference, the amplitudes of the individual waves add up. Step 2: Add the amplitudes. Resultant Amplitude = Amplitude_1 + Amplitude_2 Resultant Amplitude = 4 cm + 4 cm Resultant Amplitude = 8 cm The amplitude of the resultant wave is 8 cm. 3. Using the information on the diagram, calculate: (i) the output voltage, V₀. Step 1: Use the transformer equation relating voltage and turns. (V_p)/(V_s) = (N_p)/(N_s) Where V_p is primary voltage, V_s is secondary voltage (output voltage V_o), N_p is primary turns, and N_s is secondary turns. Step 2: Substitute the given values. V_p = 240 V N_p = 4000 turns N_s = 250 turns 240 VV_o = (4000)/(250) Step 3: Solve for V_o. V_o = 240 V × (250)/(4000) V_o = 240 V × (1)/(16) V_o = 15 V The output voltage, V₀, is 15 V. (ii) the current through the 6 Ω resistor. Step 1: Use Ohm's Law. The output voltage V_o is across the 6 Ω resistor. I = (V_o)/(R) Step 2: Substitute the values. V_o = 15 V (from part (i)) R = 6 \, I = 15 V6 \, Step 3: Calculate the current. I = 2.5 A The current through the 6 Ω resistor is 2.5 A. (iii) explain whether or not the transformer will function normally if the alternating current in the primary circuit is replaced with a direct current. No, the transformer will not function normally if the alternating current (AC) is replaced with a direct current (DC). Transformers work on the principle of mutual induction, which requires a changing magnetic flux in the core. An alternating current produces a continuously changing magnetic field, inducing a voltage in the secondary coil. A direct current, however, produces a constant magnetic field, which does not induce any voltage in the secondary coil after the initial switch-on transient. 4. Determine: (a) the potential difference across the resistor, R. Step 1: Understand voltage in a series circuit. In a series circuit, the total voltage supplied by the battery is divided among the components. V_total = V_R + V_15 Step 2: Substitute the known values. V_total = 12 V V_15 = 3 V (measured by the voltmeter) 12 V = V_R + 3 V Step 3: Solve for V_R. V_R = 12 V - 3 V V_R = 9 V The potential difference across the resistor, R, is 9 V. (b) the current through the 15 Ω resistor. Step 1: Use Ohm's Law for the 15 Ω resistor. I = V_15R_15 Step 2: Substitute the values. V_15 = 3 V R_15 = 15 \, I = 3 V15 \, Step 3: Calculate the current. I = 0.2 A The current through the 15 Ω resistor is 0.2 A. (c) the resistance of the resistor, R. Step 1: Understand current in a series circuit. In a series circuit, the current is the same through all components. Therefore, the current through resistor R is the same as the current through the 15 Ω resistor. I_R = I_15 = 0.2 A Step 2: Use Ohm's Law for resistor R. R = (V_R)/(I_R) Step 3: Substitute the values. V_R = 9 V (from part (a)) I_R = 0.2 A (from part (b)) R = 9 V0.2 A Step 4: Calculate the resistance. R = 45 \, The resistance of the resistor, R, is 45 \, . Drop the next question! 📸