Show that the output voltage vo of the circuit shown below is proportional to the derivative of the input voltage vi, If C = 10-5 F, what value of R is required if vo = -dvi/dt, assume the operational amplifier is ideal.

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Asked on June 26, 2026Physics

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.

Accepted Answer · 2026-06-26T23:59:41.341Z

Here are the solutions to the questions. 4a. Two types of inductive transducers are: Self-inductive transducers*: These transducers operate by changing their self-inductance due to the displacement of a ferromagnetic core or a change in the magnetic circuit. Examples: Variable reluctance transducer Eddy current transducer Proximity sensor Mutual-inductive transducers*: These transducers operate by changing the mutual inductance between two or more coils, typically due to the displacement of a core. Examples: Linear Variable Differential Transformer (LVDT) Rotary Variable Differential Transformer (RVDT) Synchros 4b. Advantage of a.c. bridges over d.c. bridges used in strain gauges*: A.C. bridges are less susceptible to thermoelectric EMFs (stray DC voltages) that can cause errors in D.C. bridge measurements. They can be used to measure reactive components (inductance and capacitance) in addition to resistance, offering more versatility. They can offer higher sensitivity for detecting small changes in resistance, especially when coupled with phase-sensitive detectors. Disadvantage of a.c. bridges over d.c. bridges used in strain gauges*: A.C. bridges are generally more complex to design and operate, requiring A.C. excitation sources and phase-sensitive null detectors. They are susceptible to errors caused by stray capacitance and inductance, which can be significant at higher frequencies and require careful shielding. Balancing an A.C. bridge requires adjusting both magnitude and phase, making the balancing procedure more intricate than for D.C. bridges. 4c. Given: Capacitance of crystal C_crystal = 10^9 \, F Capacitance of cable C_cable = 3 × 10^-10 \, F Charge constant of crystal d = 4 × 10^-6 \, C/cm Oscilloscope resistance R_scope = 1 \, M = 1 × 10^6 \, Oscilloscope capacitance C_scope = 10^-10 \, F Harmonic deformation amplitude x_0 = 10^-3 \, mm = 10^-4 \, cm Frequency f = 200 \, Hz Step 1: Calculate the total capacitance (C_total). The crystal, cable, and oscilloscope capacitances are in parallel. C_total = C_crystal + C_cable + C_scope C_total = 10^9 \, F + 3 × 10^-10 \, F + 10^-10 \, F C_total = 10^9 \, F + 4 × 10^-10 \, F ≈ 10^9 \, F (Note: The given crystal capacitance of 10^9 \, F is an extremely large value for a piezoelectric crystal, typically they are in pF or nF. We proceed with the given value.) Step 2: Calculate the peak charge generated by the crystal (Q_0). Q_0 = d × x_0 = (4 × 10^-6 \, C/cm) × (10^-4 \, cm) Q_0 = 4 × 10^-10 \, C Step 3: Calculate the output voltage amplitude (V_o). For a piezoelectric transducer connected to a parallel load (resistance and capacitance), the output voltage amplitude is approximately V_o = (Q_0)/(C_total) when the capacitive impedance dominates the resistive impedance, which is the case here due to the extremely large C_total. V_o = 4 × 10^-10 \, C10^9 \, F V_o = 4 × 10^-19 \, V The amplitude of the output voltage is 4 × 10^-19 \, V. 5a. Devices used for measuring power in rotating systems are referred to as dynamometers. 5b. Three (3) categories of dynamometers are: Absorption dynamometers* (e.g., Prony brake, hydraulic dynamometer) Transmission dynamometers* (e.g., torsion dynamometer, strain gauge dynamometer) Driving dynamometers* (or Engine dynamometers, which can both absorb and drive) 5c. Given: Strain gauge resistance R = 120 \, Bridge supply voltage V_s = 9 \, V Power transmitted P = 5 \, hp Speed of shaft N = 900 \, rpm Diameter of shaft d = 2 \, cm = 0.02 \, m Modulus of rigidity G = 8 × 10^10 \, N/m^2 Gauge factor GF = 2.5 Step 1: Convert power to Watts and speed to rad/s. P = 5 \, hp × 745.7 \, W/hp = 3728.5 \, W = N × (2)/(60) = 900 \, rpm × (2)/(60) \, rad/(s·rpm) = 30 \, rad/s ≈ 94.248 \, rad/s Step 2: Calculate the torque (T) transmitted by the shaft. T = (P)/() = 3728.5 \, W94.248 \, rad/s ≈ 39.56 \, N·m Step 3: Calculate the polar moment of inertia (J) of the shaft. For a solid circular shaft, J = ( d^4)/(32). J = (0.02 \, m)^432 = × 16 × 10^-832 = ()/(2) × 10^-8 \, m^4 ≈ 1.5708 × 10^-8 \, m^4 Step 4: Calculate the maximum shear stress (_max) in the shaft. _max = (T r)/(J) = (39.56 \, N·m) × (0.01 \, m)1.5708 × 10^-8 \, m^4 ≈ 2.518 × 10^7 \, N/m^2 Step 5: Calculate the shear strain () and longitudinal strain (). = _maxG = 2.518 × 10^7 \, N/m^28 × 10^10 \, N/m^2 ≈ 3.1475 × 10^-4 For strain gauges placed at 45^ to the shaft axis, the longitudinal strain is = ()/(2). = 3.1475 × 10^-42 = 1.57375 × 10^-4 Step 6: Calculate the output voltage of the Wheatstone bridge (V_o). For a full Wheatstone bridge with four active strain gauges (two in tension, two in compression), the output voltage is given by: V_o = V_s × GF × V_o = 9 \, V × 2.5 × 1.57375 × 10^-4 V_o ≈ 0.003541 \, V = 3.541 \, mV The output voltage is 3.541 mV. 6a. A thermistor is a type of resistor whose electrical resistance changes significantly with temperature. It is typically made from semiconductor materials and is highly sensitive, exhibiting a large change in resistance for a small change in temperature, making it suitable for precise temperature measurement and control. 6b. The function of a thermostat as a controlling device is to sense the temperature of a system and automatically maintain it within a desired range. It does this by comparing the measured temperature to a set point and then activating or deactivating a heating or cooling mechanism to correct any deviation from the set point. 6c. Given: Copper-constantan thermocouple, linear calibration. EMF at 400 \, ^ C (with 0 \, ^ C reference) = 20.68 \, mV. Step 1: Determine the Seebeck coefficient (S) of the thermocouple. S = EMFTemperature difference = 20.68 \, mV400 \, ^ C - 0 \, ^ C = 0.0517 \, mV/^ C i) Determine the correction for cold junction at 25 \, ^ C. The correction is the EMF generated by the cold junction at 25 \, ^ C. E_correction = S × T_cold = (0.0517 \, mV/^ C) × (25 \, ^ C) E_correction = 1.2925 \, mV The correction to be made to the indicated e.m.f. is 1.2925 mV. ii) If the indicated e.m.f. is 8.92 \, mV, determine the temperature of the hot Junction. Indicated EMF E_ind = 8.92 \, mV. The actual EMF generated by the hot junction (relative to 0 \, ^ C) is the indicated EMF plus the cold junction correction. E_actual = E_ind + E_correction = 8.92 \, mV + 1.2925 \, mV E_actual = 10.2125 \, mV Now, calculate the hot junction temperature T_hot: T_hot = E_actualS = 10.2125 \, mV0.0517 \, mV/^ C T_hot ≈ 197.53 \, ^ C The temperature of the hot junction is 197.53^. 7a. Three (3) prominent amplifiers used in measurement and instrumentation are: Operational Amplifier (Op-Amp)* Instrumentation Amplifier* Isolation Amplifier* 7b. Three (3) characteristics of an ideal amplifier (operational amplifier) are: Infinite input impedance*: No current flows into the input terminals. Zero output impedance*: The amplifier can supply any current to the load without any voltage drop at its output. Infinite open-loop gain*: Even a tiny difference between the input terminals results in a very large output voltage. 7c. The circuit shown is an inverting differentiator. Step 1: Apply ideal op-amp rules. For an ideal op-amp, there is a virtual short between the inverting (V_-) and non-inverting (V_+) inputs. Since the non-inverting input is grounded, V_+ = 0 \, V, therefore V_- = 0 \, V. Also, no current flows into the input terminals of the op-amp. Step 2: Apply Kirchhoff's Current Law (KCL) at the inverting input node (V_-). The current through the capacitor (I_C) flows into the node, and the current through the feedback resistor (I_R) flows out of the node towards the output. I_C = I_R The current through the capacitor is given by: I_C = C (d(V_i - V_-))/(dt) Since V_- = 0 \, V: I_C = C (dV_i)/(dt) The current through the feedback resistor is given by: I_R = (V_- - V_o)/(R) Since V_- = 0 \, V: I_R = (-V_o)/(R) Equating I_C and I_R: C (dV_i)/(dt) = (-V_o)/(R) Rearranging to solve for V_o: V_o = -RC (dV_i)/(dt) This equation shows that the output voltage V_o is proportional to the derivative of the input voltage V_i. Step 3: Determine the value of R. Given C = 10^-5 \, F and the desired output V_o = (-dV_i)/(dt). Comparing this with our derived equation V_o = -RC (dV_i)/(dt): -RC (dV_i)/(dt) = -1 × (dV_i)/(dt) Therefore, RC = 1. R = (1)/(C) = (1)/(10^-5) \, F R = 10^5 \, = 100 \, k The value of R required is 100 k. What's next?

Accepted Answer · 2026-06-26T23:59:41.341Z

Here are the solutions to the questions. 4a. Two types of inductive transducers are: Self-inductive transducers*: These transducers operate by changing their self-inductance due to the displacement of a ferromagnetic core or a change in the magnetic circuit. Examples: Variable reluctance transducer Eddy current transducer Proximity sensor Mutual-inductive transducers*: These transducers operate by changing the mutual inductance between two or more coils, typically due to the displacement of a core. Examples: Linear Variable Differential Transformer (LVDT) Rotary Variable Differential Transformer (RVDT) Synchros 4b. Advantage of a.c. bridges over d.c. bridges used in strain gauges*: A.C. bridges are less susceptible to thermoelectric EMFs (stray DC voltages) that can cause errors in D.C. bridge measurements. They can be used to measure reactive components (inductance and capacitance) in addition to resistance, offering more versatility. They can offer higher sensitivity for detecting small changes in resistance, especially when coupled with phase-sensitive detectors. Disadvantage of a.c. bridges over d.c. bridges used in strain gauges*: A.C. bridges are generally more complex to design and operate, requiring A.C. excitation sources and phase-sensitive null detectors. They are susceptible to errors caused by stray capacitance and inductance, which can be significant at higher frequencies and require careful shielding. Balancing an A.C. bridge requires adjusting both magnitude and phase, making the balancing procedure more intricate than for D.C. bridges. 4c. Given: Capacitance of crystal C_crystal = 10^9 \, F Capacitance of cable C_cable = 3 × 10^-10 \, F Charge constant of crystal d = 4 × 10^-6 \, C/cm Oscilloscope resistance R_scope = 1 \, M = 1 × 10^6 \, Oscilloscope capacitance C_scope = 10^-10 \, F Harmonic deformation amplitude x_0 = 10^-3 \, mm = 10^-4 \, cm Frequency f = 200 \, Hz Step 1: Calculate the total capacitance (C_total). The crystal, cable, and oscilloscope capacitances are in parallel. C_total = C_crystal + C_cable + C_scope C_total = 10^9 \, F + 3 × 10^-10 \, F + 10^-10 \, F C_total = 10^9 \, F + 4 × 10^-10 \, F ≈ 10^9 \, F (Note: The given crystal capacitance of 10^9 \, F is an extremely large value for a piezoelectric crystal, typically they are in pF or nF. We proceed with the given value.) Step 2: Calculate the peak charge generated by the crystal (Q_0). Q_0 = d × x_0 = (4 × 10^-6 \, C/cm) × (10^-4 \, cm) Q_0 = 4 × 10^-10 \, C Step 3: Calculate the output voltage amplitude (V_o). For a piezoelectric transducer connected to a parallel load (resistance and capacitance), the output voltage amplitude is approximately V_o = (Q_0)/(C_total) when the capacitive impedance dominates the resistive impedance, which is the case here due to the extremely large C_total. V_o = 4 × 10^-10 \, C10^9 \, F V_o = 4 × 10^-19 \, V The amplitude of the output voltage is 4 × 10^-19 \, V. 5a. Devices used for measuring power in rotating systems are referred to as dynamometers. 5b. Three (3) categories of dynamometers are: Absorption dynamometers* (e.g., Prony brake, hydraulic dynamometer) Transmission dynamometers* (e.g., torsion dynamometer, strain gauge dynamometer) Driving dynamometers* (or Engine dynamometers, which can both absorb and drive) 5c. Given: Strain gauge resistance R = 120 \, Bridge supply voltage V_s = 9 \, V Power transmitted P = 5 \, hp Speed of shaft N = 900 \, rpm Diameter of shaft d = 2 \, cm = 0.02 \, m Modulus of rigidity G = 8 × 10^10 \, N/m^2 Gauge factor GF = 2.5 Step 1: Convert power to Watts and speed to rad/s. P = 5 \, hp × 745.7 \, W/hp = 3728.5 \, W = N × (2)/(60) = 900 \, rpm × (2)/(60) \, rad/(s·rpm) = 30 \, rad/s ≈ 94.248 \, rad/s Step 2: Calculate the torque (T) transmitted by the shaft. T = (P)/() = 3728.5 \, W94.248 \, rad/s ≈ 39.56 \, N·m Step 3: Calculate the polar moment of inertia (J) of the shaft. For a solid circular shaft, J = ( d^4)/(32). J = (0.02 \, m)^432 = × 16 × 10^-832 = ()/(2) × 10^-8 \, m^4 ≈ 1.5708 × 10^-8 \, m^4 Step 4: Calculate the maximum shear stress (_max) in the shaft. _max = (T r)/(J) = (39.56 \, N·m) × (0.01 \, m)1.5708 × 10^-8 \, m^4 ≈ 2.518 × 10^7 \, N/m^2 Step 5: Calculate the shear strain () and longitudinal strain (). = _maxG = 2.518 × 10^7 \, N/m^28 × 10^10 \, N/m^2 ≈ 3.1475 × 10^-4 For strain gauges placed at 45^ to the shaft axis, the longitudinal strain is = ()/(2). = 3.1475 × 10^-42 = 1.57375 × 10^-4 Step 6: Calculate the output voltage of the Wheatstone bridge (V_o). For a full Wheatstone bridge with four active strain gauges (two in tension, two in compression), the output voltage is given by: V_o = V_s × GF × V_o = 9 \, V × 2.5 × 1.57375 × 10^-4 V_o ≈ 0.003541 \, V = 3.541 \, mV The output voltage is 3.541 mV. 6a. A thermistor is a type of resistor whose electrical resistance changes significantly with temperature. It is typically made from semiconductor materials and is highly sensitive, exhibiting a large change in resistance for a small change in temperature, making it suitable for precise temperature measurement and control. 6b. The function of a thermostat as a controlling device is to sense the temperature of a system and automatically maintain it within a desired range. It does this by comparing the measured temperature to a set point and then activating or deactivating a heating or cooling mechanism to correct any deviation from the set point. 6c. Given: Copper-constantan thermocouple, linear calibration. EMF at 400 \, ^ C (with 0 \, ^ C reference) = 20.68 \, mV. Step 1: Determine the Seebeck coefficient (S) of the thermocouple. S = EMFTemperature difference = 20.68 \, mV400 \, ^ C - 0 \, ^ C = 0.0517 \, mV/^ C i) Determine the correction for cold junction at 25 \, ^ C. The correction is the EMF generated by the cold junction at 25 \, ^ C. E_correction = S × T_cold = (0.0517 \, mV/^ C) × (25 \, ^ C) E_correction = 1.2925 \, mV The correction to be made to the indicated e.m.f. is 1.2925 mV. ii) If the indicated e.m.f. is 8.92 \, mV, determine the temperature of the hot Junction. Indicated EMF E_ind = 8.92 \, mV. The actual EMF generated by the hot junction (relative to 0 \, ^ C) is the indicated EMF plus the cold junction correction. E_actual = E_ind + E_correction = 8.92 \, mV + 1.2925 \, mV E_actual = 10.2125 \, mV Now, calculate the hot junction temperature T_hot: T_hot = E_actualS = 10.2125 \, mV0.0517 \, mV/^ C T_hot ≈ 197.53 \, ^ C The temperature of the hot junction is 197.53^. 7a. Three (3) prominent amplifiers used in measurement and instrumentation are: Operational Amplifier (Op-Amp)* Instrumentation Amplifier* Isolation Amplifier* 7b. Three (3) characteristics of an ideal amplifier (operational amplifier) are: Infinite input impedance*: No current flows into the input terminals. Zero output impedance*: The amplifier can supply any current to the load without any voltage drop at its output. Infinite open-loop gain*: Even a tiny difference between the input terminals results in a very large output voltage. 7c. The circuit shown is an inverting differentiator. Step 1: Apply ideal op-amp rules. For an ideal op-amp, there is a virtual short between the inverting (V_-) and non-inverting (V_+) inputs. Since the non-inverting input is grounded, V_+ = 0 \, V, therefore V_- = 0 \, V. Also, no current flows into the input terminals of the op-amp. Step 2: Apply Kirchhoff's Current Law (KCL) at the inverting input node (V_-). The current through the capacitor (I_C) flows into the node, and the current through the feedback resistor (I_R) flows out of the node towards the output. I_C = I_R The current through the capacitor is given by: I_C = C (d(V_i - V_-))/(dt) Since V_- = 0 \, V: I_C = C (dV_i)/(dt) The current through the feedback resistor is given by: I_R = (V_- - V_o)/(R) Since V_- = 0 \, V: I_R = (-V_o)/(R) Equating I_C and I_R: C (dV_i)/(dt) = (-V_o)/(R) Rearranging to solve for V_o: V_o = -RC (dV_i)/(dt) This equation shows that the output voltage V_o is proportional to the derivative of the input voltage V_i. Step 3: Determine the value of R. Given C = 10^-5 \, F and the desired output V_o = (-dV_i)/(dt). Comparing this with our derived equation V_o = -RC (dV_i)/(dt): -RC (dV_i)/(dt) = -1 × (dV_i)/(dt) Therefore, RC = 1. R = (1)/(C) = (1)/(10^-5) \, F R = 10^5 \, = 100 \, k The value of R required is 100 k. What's next?

Show that the output voltage vo of the circuit shown below is proportional to the derivative of the input voltage vi, If C = 10-5 F, what value of R is required if vo = -dvi/dt, assume the operational amplifier is ideal.

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Show that the output voltage vo of the circuit shown below is proportional to the derivative of the input voltage vi, If C = 10-5 F, what value of R is required if vo = -dvi/dt, assume the operational amplifier is ideal.

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