
(a)
Interpretation:
The amount by which
Concept introduction:
Limiting condition of operational amplifier as a comparator implies that the comparator output voltage reaches the limiting value of output which is given here as +13 V on the positive side and
We have,
Corresponding to the positive saturation value and negative saturation value respectively. We need to find the difference in input voltage in each case.

Answer to Problem 3.1QAP
Explanation of Solution
Given the open loop gain A is 100,000
Operational amplifier as a comparator compares two voltages at its respective input terminals
We have to find the amount by which
Case 1: For
Here,
Hence,
This implies
Case 2: For
Here,
Hence,
This implies
(b)
Interpretation:
The amount by which
Concept introduction:
Limiting condition of operational amplifier as a comparator implies that the comparator output voltage reaches the limiting value of output which is given here as +13 V on the positive side and
We have,
Corresponding to the positive saturation value and negative saturation value respectively. We need to find the difference in input voltage in each case.

Answer to Problem 3.1QAP
Explanation of Solution
Given the open loop gain A is 600,000.
Operational amplifier as a comparator compares two voltages at its respective input terminals
We have to find the amount by which
Case 1: For
Here,
Hence,
This implies
Case 2: For
Here,
Hence,
This implies
(c)
Interpretation:
The amount by which
Concept introduction:
Limiting condition of operational amplifier as a comparator implies that the comparator output voltage reaches the limiting value of output which is given here as +13 V on the positive side and
We have,
Corresponding to the positive saturation value and negative saturation value respectively. We need to find the difference in input voltage in each case.

Answer to Problem 3.1QAP
Explanation of Solution
Given the open loop gain A is
Operational amplifier as a comparator compares two voltages at its respective input terminals
We have to find the amount by which
Case 1: For
Here,
Hence,
This implies
Case 2 : For
Here,
Hence,
This implies
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Chapter 3 Solutions
Principles of Instrumental Analysis
- Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely: (a) 0.000259 M HClO4arrow_forwardWhat is the pH of a 1.0 L buffer made with 0.300 mol of HF (Ka = 6.8 × 10⁻⁴) and 0.200 mol of NaF to which 0.160 mol of NaOH were added?arrow_forwardDetermine if the following salt is neutral, acidic or basic. If acidic or basic, write the appropriate equilibrium equation for the acid or base that exists when the salt is dissolved in aqueous solution. If neutral, simply write only NR. Be sure to include the proper phases for all species within the reaction. NaN₃arrow_forward
- A. Draw the structure of each of the following alcohols. Then draw and name the product you would expect to produce by the oxidation of each. a. 4-Methyl-2-heptanol b. 3,4-Dimethyl-1-pentanol c. 4-Ethyl-2-heptanol d. 5,7-Dichloro-3-heptanolarrow_forwardWhat is the pH of a 1.0 L buffer made with 0.300 mol of HF (Ka = 6.8 × 10⁻⁴) and 0.200 mol of NaF to which 0.160 mol of NaOH were added?arrow_forwardCan I please get help with this.arrow_forward
- Determine if the following salt is neutral, acidic or basic. If acidic or basic, write the appropriate equilibrium equation for the acid or base that exists when the salt is dissolved in aqueous solution. If neutral, simply write only NR. Be sure to include the proper phases for all species within the reaction. N₂H₅ClO₄arrow_forwardPlease help me with identifying these.arrow_forwardCan I please get help with this?arrow_forward
- Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
