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 five questions from your exam paper: Question 1: 1a) With reference to instrumentation, here are the definitions: • Tolerance: The maximum permissible error or deviation from the true or specified value. It defines the acceptable range within which a measurement or component value can vary. • Precision: The degree to which repeated measurements under unchanged conditions show the same results. It indicates the closeness of agreement among independent test results obtained under stipulated conditions. • Repeatability: The variation in measurements taken by a single person or instrument on the same item under the same conditions over a short period of time. It's a measure of the consistency of an instrument. • Span: The range of an instrument, defined as the algebraic difference between the upper and lower limits of the measurement range. For example, if an instrument measures from 0 to 100V, its span is 100V. • Linearity: The extent to which the output of an instrument is directly proportional to the input over its entire measurement range. A perfectly linear instrument would have a straight-line relationship between input and output. 1b) Here are the advantages and disadvantages of null and deflection types of measuring instruments: • Null-type instruments: • Advantages: High accuracy, independent of indicating device calibration, less susceptible to parameter changes, often more sensitive. • Disadvantages: Slower response time, more complex to operate, cannot be used for rapidly changing quantities, often more expensive. • Deflection-type instruments: • Advantages: Faster response time (direct readings), simpler to operate, can be used for rapidly changing quantities, generally less expensive. • Disadvantages: Lower accuracy (depends on scale calibration), more susceptible to errors from friction, temperature, and aging, can have a loading effect on the circuit. 1c) Null-type instruments are mainly used for high-precision measurements where accuracy is paramount, such as in laboratories, calibration standards, and industrial process control systems. They are preferred because they operate by balancing the unknown quantity against a known standard, meaning the measurement is taken when the detector indicates zero (null). This eliminates errors associated with the detector's calibration or non-linearity, making them inherently more accurate and reliable for precise work. Question 2: 2a) Here is the difference between systematic and random errors: • Systematic errors are consistent, repeatable errors inherent in the measurement system or method, causing measurements to consistently deviate from the true value in the same direction. They can often be identified and corrected. • Random errors are unpredictable, variable errors due to uncontrollable fluctuations in the measurement process, causing measurements to vary randomly around the true value. They cannot be eliminated but can be minimized by taking multiple measurements and averaging. 2b) Here are the typical sources of these two types of error: • Sources of Systematic Errors: Instrumental errors (faulty calibration, zero error), environmental errors (consistent external influences), observational errors (parallax), theoretical errors (model approximations). • Sources of Random Errors: Environmental fluctuations (unpredictable changes), human limitations (estimation, reaction time), instrument limitations (noise, resolution), variations in the measured quantity itself. 2c) To calculate the value of density and its possible error: Given: a = 100mm ± 1\%, b = 200mm ± 1\%, c = 300mm ± 1\%, m = 20kg ± 0.5\%. Density = (m)/(V), where V = a × b × c. Step 1: Calculate the nominal values. Convert dimensions to meters: a = 0.1 m, b = 0.2 m, c = 0.3 m. Nominal Volume V = 0.1 m × 0.2 m × 0.3 m = 0.006 m^3. Nominal Density = 20 kg0.006 m^3 = 3333.33 kg/m^3. Step 2: Calculate the maximum possible percentage error in density. For a product/quotient, the maximum percentage error is the sum of individual percentage errors. \% error_ = \% error_m + \% error_a + \% error_b + \% error_c \% error_ = 0.5\% + 1\% + 1\% + 1\% = 3.5\% Step 3: Calculate the absolute error in density. = × \% error_100 = 3333.33 kg/m^3 × (3.5)/(100) = 116.66655 kg/m^3 Step 4: State the final value of density with its possible error. Rounding to appropriate significant figures: = 3333 ± 117 kg/m^3 Question 3: 3a) All analogue electrical indicating instruments require three essential devices: 1. Deflecting System: This system produces a deflecting torque that causes the pointer to move away from its zero position, proportional to the measured quantity. 2. Controlling System: This system provides a controlling torque that opposes the deflecting torque, ensuring the pointer comes to rest at a specific position corresponding to the measured quantity. 3. Damping System: This system prevents the pointer from oscillating excessively around its final steady position, bringing it to rest quickly to allow for easy reading. 3b) To calculate the value of the shunt: Given: I_m = 40 mA = 0.04 A, R_m = 25, I = 50 A. Step 1: Calculate the voltage across the meter at full-scale deflection. V_m = I_m × R_m = 0.04 A × 25 = 1 V Step 2: Determine the current that must flow through the shunt. I_sh = I - I_m = 50 A - 0.04 A = 49.96 A Step 3: Calculate the resistance of the shunt (R_sh). Since the shunt is in parallel with the meter, V_sh = V_m. R_sh = V_shI_sh = 1 V49.96 A ≈ 0.020016 Rounding to four significant figures: R_sh = 0.0200 Question 4: Given: f_0 = 1.2 kHz = 1200 Hz, Q = 30, Z_0 = 50. At resonance, Z_0 = R, so R = 50. Step 1: Determine the angular resonant frequency (_0). _0 = 2 f_0 = 2 × 1200 Hz = 2400 rad/s ≈ 7539.82 rad/s Step 2: Determine the inductance (L). Using Q = (_0 L)/(R): L = (Q × R)/(_0) = (30 × 50)/(2400 rad/s) = (1500)/(2400) H = (5)/(8) H L ≈ 0.199 H Step 3: Determine the capacitance (C). Using _0 = (1)/(sqrt(LC)): C = (1)/(_0^2 L) = (1)/((2400 rad/s))^2 × (5)/(8) H = (1)/(5760000^2 × 5)8 F C = (8)/(28800000) F = (1)/(3600000) F ≈ 8.84 × 10^-8 F C ≈ 88.4 nF Step 4: Determine the bandwidth (BW). BW = (f_0)/(Q) = 1200 Hz30 = 40 Hz Step 5: Determine the lower and upper half-power frequencies (f_1 and f_2). f_1 = f_0 - BW2 = 1200 Hz - 40 Hz2 = 1180 Hz f_2 = f_0 + BW2 = 1200 Hz + 40 Hz2 = 1220 Hz f_1 = 1180 Hz f_2 = 1220 Hz Step 6: Determine the value of the circuit impedance at the half-power frequencies. At half-power frequencies, the impedance magnitude is sqrt(2) times the resistance. Z_half-power = sqrt(2) × R = sqrt(2) × 50 ≈ 70.71 Z_half-power = 70.7 Question 5: 5a) The circuit diagram to measure 3-phase power using the two-wattmeter method involves connecting two wattmeters to the three-phase supply and load. • Wattmeter 1 (W1): Its current coil is connected in series with line L1. Its voltage coil is connected between line L1 and line L2. • Wattmeter 2 (W2): Its current coil is connected in series with line L3. Its voltage coil is connected between line L3 and line L2. Line L2 serves as the common potential point for the voltage coils of both wattmeters. 5b) To derive the equation for the power output and power angle from the two-wattmeter readings: Let W_1 and W_2 be the readings of the two wattmeters. For a balanced 3-phase load, the readings are given by: W_1 = V_L I_L (30^ - ) W_2 = V_L I_L (30^ + ) where V_L is the line voltage, I_L is the line current, and is the phase angle of the load. Step 1: Total Power Output (P_T). The total power is the sum of the two readings: P_T = W_1 + W_2 = V_L I_L [(30^ - ) + (30^ + )] Using the identity (A-B) + (A+B) = 2 A B: P_T = V_L I_L [2 (30^) ()] = V_L I_L (2 × sqrt(3)2 × ()) P_T = sqrt(3) V_L I_L () Step 2: Power Angle (). Consider the difference between the two readings: W_1 - W_2 = V_L I_L [(30^ - ) - (30^ + )] Using the identity (A-B) - (A+B) = 2 A B: W_1 - W_2 = V_L I_L [2 (30^) ()] = V_L I_L (2 × (1)/(2) × ()) W_1 - W_2 = V_L I_L () Now, divide (W_1 - W_2) by (W_1 + W_2): (W_1 - W_2)/(W_1 + W_2) = (V_L I_L ())/(sqrt(3) V_L I_L ()) = (1)/(sqrt(3)) () Rearranging for : = (sqrt(3) (W_1 - W_2)/(W_1 + W_2)) 5c) Given: f = 50 Hz, I_L = 25 A. Assumption: The question states "a single wattmeter connected to measure the input power indicates 5.185KW and 10.37KW respectively." This is contradictory. Assuming these are the readings from two wattmeters, W_1 = 10.37 kW and W_2 = 5.185 kW. The coils are star-connected. i) Draw the circuit diagram of the setup This refers to the two-wattmeter method diagram as described in 5a. ii) Calculate the line and phase voltages Step 1: Calculate total power (P_T). P_T = W_1 + W_2 = 10.37 kW + 5.185 kW = 15.555 kW Step 2: Calculate the power factor angle (). () = sqrt(3) (W_1 - W_2)/(W_1 + W_2) = sqrt(3) 10.37 kW - 5.185 kW10.37 kW + 5.185 kW () = sqrt(3) 5.185 kW15.555 kW = sqrt(3) × (1)/(3) = (1)/(sqrt(3)) = ((1)/(sqrt(3))) = 30^ Step 3: Calculate the power factor (). = (30^) = sqrt(3)2 ≈ 0.866 Step 4: Calculate the line voltage (V_L). Using P_T = sqrt(3) V_L I_L : V_L = (P_T)/(sqrt(3) I_L ) = 15555 Wsqrt(3) × 25 A × 0.866 = (15555)/(37.5) ≈ 414.8 V Line Voltage V_L = 414.8 V Step 5: Calculate the phase voltage (V_ph). For a star-connected system, V_L = sqrt(3) V_ph: V_ph = (V_L)/(sqrt(3)) = 414.8 V1.732 ≈ 239.5 V Phase Voltage V_ph = 239.5 V iii) Calculate the resistance and reactance of each coil. For a star-connected system, I_ph = I_L = 25 A. Step 1: Calculate the impedance per phase (Z_ph). Z_ph = V_phI_ph = 239.5 V25 A = 9.58 Step 2: Calculate the resistance per phase (R_ph). R_ph = Z_ph = 9.58 × 0.866 ≈ 8.29 Resistance of each coil R = 8.29 Step 3: Calculate the reactance per phase (X_ph). X_ph = Z_ph Since = 30^, = (30^) = 0.5. X_ph = 9.58 × 0.5 = 4.79 Reactance of each coil X = 4.79 What's next?