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.
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Here are the solutions to the questions from the image. Question 1: Series RLC Circuit Given: Capacitance C = 58 \, = 58 × 10^-6 \, F Resistance R = 15 \, Inductance L = 128 \, mH = 128 × 10^-3 \, H Supply voltage V = 100 \, V (RMS, assumed) 1.1 Calculate the resonant frequency. Step 1: Use the formula for resonant frequency in a series RLC circuit. The resonant frequency f_0 is given by: f_0 = (1)/(2(LC)) Step 2: Substitute the given values and calculate f_0. f_0 = (1)/(2((128 × 10^-3) \, H))(58 × 10^-6 \, F) f_0 = (1)/(2(7.424 × 10^-6)) f_0 = (1)/(2 × 2.7247 × 10^-3) f_0 = (1)/(0.017119) f_0 ≈ 58.42 \, Hz 1.2 By means of calculations, show that the impedance is equal to the resistance of the coil. Step 1: Calculate the inductive reactance (X_L) and capacitive reactance (X_C) at the resonant frequency f_0. X_L = 2 f_0 L X_L = 2 (58.42 \, Hz)(128 × 10^-3 \, H) X_L ≈ 47.05 \, X_C = (1)/(2 f_0 C) X_C = (1)/(2 (58.42 \, Hz))(58 × 10^-6 \, F) X_C ≈ (1)/(0.02133) X_C ≈ 46.88 \, (Note: The slight difference between X_L and X_C is due to rounding f_0. If f_0 is used with full precision, they would be equal.) Step 2: Calculate the total impedance Z of the series RLC circuit. Z = sqrt(R^2 + (X_L - X_C)^2) Using the calculated values: Z = sqrt((15 \, )^2 + (47.05 \, - 46.88 \, )^2) Z = sqrt(225 + (0.17)^2) Z = sqrt(225 + 0.0289) Z = sqrt(225.0289) Z ≈ 15.00096 \, Since R = 15 \, , the impedance Z is approximately equal to the resistance R at resonance. Z ≈ 15 \, This shows that Z ≈ R at resonance. 1.3 Draw the phasor diagram. (Since I cannot draw an image, I will describe the key features of the phasor diagram at resonance). • The current phasor (I) is typically drawn along the positive x-axis (reference). • The voltage across the resistor (V_R = IR) is in phase with the current, so V_R is also along the positive x-axis. • The voltage across the inductor (V_L = IX_L) leads the current by 90^, so V_L points along the positive y-axis. • The voltage across the capacitor (V_C = IX_C) lags the current by 90^, so V_C points along the negative y-axis. • At resonance, V_L and V_C are equal in magnitude and opposite in direction, so they cancel each other out. • The total supply voltage (V) is equal to V_R and is in phase with the current. The diagram would show I and V_R on the positive x-axis, V_L on the positive y-axis, and V_C on the negative y-axis, with V_L and V_C having equal lengths. The resultant voltage V would coincide with V_R. Question 2: Parallel RLC Circuit (Implied) Given: Coil resistance R = 15 \, Coil inductance L = 200 \, mH = 200 × 10^-3 \, H 2.1 Determine the frequency at which the resulting current is in phase with the source voltage. Assumption: This question implies a parallel resonant circuit where the coil (R and L in series) is connected in parallel with a capacitor. However, the value of the capacitor is not provided in the question. Without the capacitor value, it is not possible to calculate a specific non-zero resonant frequency. If the circuit consists only of the coil (R and L in series) connected to the source, the current will always lag the voltage at any non-zero frequency. The current would only be in phase with the voltage if the circuit were purely resistive, which means the inductance L would have to be zero or the frequency would have to be zero (DC). Therefore, based on the information given, this question cannot be solved for a non-zero frequency. Missing information: Value of the parallel capacitor. 2.2 Calculate the related Q factor value. The Q factor (Quality factor) for a parallel resonant circuit where the coil (R and L in series) is in parallel with a capacitor is given by: Q = (_0 L)/(R) or Q = (1)/(_0 C R) Since the resonant frequency _0 (or f_0) cannot be determined without the capacitor value (as explained in 2.1), the Q factor also cannot be calculated. Missing information: Value of the parallel capacitor (and thus resonant frequency). Question 3: Series RLC Circuit with Variable Capacitor Given: Supply voltage V = 100 \, V (RMS, assumed) Supply frequency f = 50 \, Hz Coil resistance R = 10 \, Coil inductance L = 210 \, mH = 210 × 10^-3 \, H A variable capacitor is connected in series. 3.1 Capacitance required to produce resonance. Step 1: Use the formula for resonant frequency in a series RLC circuit. At resonance, f_0 = f = 50 \, Hz. f_0 = (1)/(2(LC)) Step 2: Rearrange the formula to solve for C. 2 f_0 = (1)/(sqrt(LC)) (2 f_0)^2 = (1)/(LC) C = (1)/(L(2 f_0)^2) Step 3: Substitute the given values and calculate C. C = (1)/((210 × 10^-3) \, H)(2 × 50 \, Hz)^2 C = (1)/((0.210)(314.159)^2) C = (1)/((0.210)(98696.04)) C = (1)/(20726.168) C ≈ 4.824 × 10^-5 \, F C ≈ 48.24 \, 3.2 Value of the capacitor. This is the same as Question 3.1. The value of the capacitor is 48.24 \, . 3.3 Voltage drop across the coil and capacitor. Step 1: Calculate the inductive reactance (X_L) and capacitive reactance (X_C) at resonance. At resonance, X_L = X_C. X_L = 2 f L X_L = 2 (50 \, Hz)(210 × 10^-3 \, H) X_L = 2 (50)(0.210) X_L = 65.97 \, So, X_C = 65.97 \, . Step 2: Calculate the total current (I) in the series circuit at resonance. At resonance, the impedance Z = R. I = (V)/(Z) = (V)/(R) I = 100 \, V10 \, I = 10 \, A Step 3: Calculate the voltage drop across the coil (V_coil) and the capacitor (V_C). The coil has resistance R and inductance L. The voltage drop across the coil is V_coil = I · Z_coil, where Z_coil = sqrt(R^2 + X_L^2). Z_coil = sqrt((10 \, )^2 + (65.97 \, )^2) Z_coil = sqrt(100 + 4351.94) Z_coil = sqrt(4451.94) Z_coil ≈ 66.72 \, V_coil = I · Z_coil = (10 \, A)(66.72 \, ) V_coil = 667.2 \, V The voltage drop across the capacitor is V_C = I · X_C. V_C = (10 \, A)(65.97 \, ) V_C = 659.7 \, V 3.4 Quality factor. Step 1: Use the formula for the Q factor of a series RLC circuit. Q = (X_L)/(R) or Q = (1)/(R)sqrt((L)/(C)) Step 2: Substitute the values for X_L and R. Q = (65.97 \, )/(10 \, ) Q = 6.60 Got more? Send 'em!