(a)
Interpretation:
The design of an inverting amplifier with gain of
Concept introduction:
An inverting amplifier is a type of operational amplifier in which the output of the amplifier is given to the inverted terminal as an input through the feedback resistor. In inverting amplifier, the phase difference between the input and the output voltage is
The gain in the inverting amplifier is defined as
(b)
Interpretation:
The range of the usable input voltage of inverting amplifier of part (a) is to be stated.
Concept introduction:
An inverting amplifier is a type of operational amplifier in which the output of the amplifier is given to the inverted terminal as an input through the feedback resistor. In inverting amplifier, the phase difference between the input and the output voltage is equal to
The output voltage of the inverting amplifier is defined as
(c)
Interpretation:
The input resistance to the inverting amplifier of part (a) is to be stated.
Concept introduction:
Inverting amplifier is the type of operational amplifier in which the output of the amplifier is given to the inverted terminal as an input through the feedback resistor. In inverting amplifier the phase difference between the input and the output voltage is equal to
The gain in the inverting amplifier is defined as
(d)
Interpretation:
The ways to avoid the loading error if the voltage source were loaded by an input resistance as a result in the part (c).
Concept introduction:
Inverting amplifier is the type of operational amplifier in which the output of the amplifier is given to the inverted terminal as an input through the feedback resister. In inverting amplifier the phase difference between the input and the output voltage is
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Chapter 3 Solutions
Principles of Instrumental Analysis
- The Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials subjected to aging and stress. Use this distribution as a model for time (in hours) to failure of solid insulating specimens subjected to AC voltage. The values of the parameters depend on the voltage and temperature; suppose ? = 2.2 and ? = 220. (a) What is the probability that a specimen's lifetime is at most 250? Less than 250? More than 300? (Round your answers to five decimal places.) at most 250 less than 250more than 300 (b) What is the probability that a specimen's lifetime is between 100 and 250? (Round your answer to four decimal places.) (c) What value (in hr) is such that exactly 50% of all specimens have lifetimes exceeding that value? (Round your answer to three decimal places.) hrarrow_forwardIf you wish to decrease the amount of current in a resistor from 240mA to 180mA by changing the 24V source, what should be the new voltage setting be?arrow_forwardDescribe the difference between power-compensated, heat-flux, and modulated DSC instruments.arrow_forward
- What is the purpose of waiting 1s after a voltage pulse before measuring current in sampled current voltammetry?arrow_forwardA Dubosq colorimeter consists of a cell of fixed path length and a cell of variable path length. By adjusting the length of the latter until the transmission through the two cells is the same. the concentration of the second solution can be inferred from that of the former. Suppose that a plantdye of concentration 25 μg dm-3 is added to the fixed cell. the length of which is 1.55 em. Then a solution of the same dye. but of unknown concentration. is added to the second cel l. It is found that the same t ransmittance is obtained when the length of the second cel l is adjusted to 1.18 em. What is the concentration of the second solution?arrow_forwardAt 520 nm with a 1.00 cm cell a 8.24 ppm solution of KMnO4 has a transmittance of 0.126. What is the molar absorptivity of KMnO4 at 520 nm?arrow_forward
- Determine the unknown (Abs=0.812) if the absorption of the standard are the following: Standard 1, Abs = 0.354 Standard 2, Abs = 0.453 Standard 3, Abs = 0.563 Standard 4, Abs = 0.683 Blank, Abs = 0.011arrow_forwardA standard solution was prepared containing 10.0 ppm of an analyte and 15.0 ppm of an internal standard. Analysis of the sample gave signals for the analyte and internal standard of 0.155 and 0.233 (arbitrary units), respectively. Sufficient internal standard was added to an unknown sample to make it 15.0 ppm in the internal standard. Analysis of the unknown sample yielded signals for the analyte and internal standard of 0.274 and 0.198, respectively. Calculate the concentration of analyte in the unknown sample.arrow_forwardThe Ksp of Zn(OH)2 is 3.00 x 10^-17arrow_forward
- Calculate the specific absorbance ratio (SAR), where A620 = 0.136 and A280 = 1.392. Enter your result with 3 decimal places.arrow_forwardChoose all correct answers for the following.Percent Transmittance is defined as: Question 6 options: T = Po/P %T = (P/Po) x 100% T = P/Po %T = (Po/P) x 100%arrow_forward
Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning