(a)
Interpretation:
Time constant for the circuit should be calculated.
Concept introduction:
The product of RC is referred to as time constant for the circuit and is a measure of the time required for a capacitor to charge or discharge.
(b)
Interpretation:
The current, voltage drops across the capacitor and the resistor during a charging cycle at given times should be calculated.
Concept introduction:
The product of RC is referred to as time constant for the circuit and is a measure of the time required for a capacitor to charge or discharge.
Ohm’s law:
Ohm’s law describes the relationship among voltage, resistance, and current in a resistive series circuit.
V = IR
Connection between initial current and current across the capacitor (i) at given time during the charging is given by
The value of the voltage across the capacitor (Vc) at given time during the charging period can be given like this
Vc = Voltage across the capacitor
Vs= Supply voltage
t = time
RC = time constant for RC circuit
(c)
Interpretation:
The current and voltage drops across the capacitor and the resistor during a discharging cycle at time 10 ms should be calculated.
Concept introduction:
The product of RC is referred to as time constant for the circuit and is a measure of the time required for a capacitor to charge or discharge.
Ohm’s law:
Ohm’s law describes the relationship among voltage, resistance, and current in a resistive series circuit.
V = IR
The value of the voltage across the capacitor (Vc) at given time during the charging period can be given like this;
Vc = Voltage across the capacitor
Vs= Supply voltage
t = time
RC = time constant for RC circuit
Connection between initial current and current across the capacitor (i) at given time during the discharging is given by;
Want to see the full answer?
Check out a sample textbook solutionChapter 2 Solutions
Principles of Instrumental Analysis
- Calculate the dielectric constant of the medium for the zone between a metal and the OHP layer from the Helmholtz model, when the differential capacitance of the electrode is 1.4 mF cm-2. Indicate, according to this model, what this differential capacity will be like when the potential is reduced by half. Data: distance between metal and OHP layer 4 A, vacuum dielectric constant epsilon0 = 8.85·10-12 F m-1arrow_forwardMehuuul.ou9arrow_forwardShow complete solution for the given aswers below.arrow_forward
- 4. What is the capacitance of an empty parallel-plate capacitor with metal plates that each have an area of 1.00m², separated by 1.00 mm?arrow_forwardA current of 21.0 mA is maintained in a single circular loop of 1.40 m circumference. A magnetic field of 0.520 T is directed parallel to the plane of the loop. (a) Calculate the magnetic moment of the loop. mA. m² (b) What is the magnitude of the torque exerted by the magnetic field on the loop? Submit Answer mN m .arrow_forward-1! The concentrations of three dilute solutions of sodium acetate were measured in a con- ductance cell in which the parallel electrodes were 1 cm in area and 0.25 cm apart. Resistances of 274,700, 91,000 and 18,320 ohms were determined for the three solu- tions. Calculate the normality of each solution. T = 25°C.arrow_forward
- An air-gap parallel plate capacitor whose plates have an area of A=0,38m² are separated by a distance d=0,61m and are connected to a battery of V₁ = 1,8 V. The space between the parallel plates is partially filled (x=0,26m) with some material of dielectric constant K=4,56. Find the electric potential energy stored in the capacitor in terms of permitivity of free-space €0. Express your answer using 2 decimal places.arrow_forwardA student measured the resistance of a 0.5 M NaCl solution to be 45 Ω. Calculate the experimental value of the molar conductivity of this solution if the value of Kcell is 0.0083 cm-1 .arrow_forwardSuggest possible reasons for the difference between the experimental and calculated voltages.arrow_forward
- Direct currents are uniform and have uniform voltage. OTrue O Falsearrow_forwardAt 25 °C, 0.05 M of MgCl2 electrolyte solution showed molar conductivity of 194.2 S cm? mol-. A cell with electrodes that 1.50 cm² in surface area and 0.50 cm apart is filled with the above electrolyte. Determine the resistance of the solution.arrow_forward(d) Calculate the electrical conductivity of a 6.0 mm diameter cylindrical silicon specimen of 60 mm long in which a current of 0.5 A passes in an axial direction. A voltage of 12.5 V is measured across two probes that are separated by 40 mm. i. Compute the electrical conductivity of the specimen. C12334 ii. Compute the resistance over the entire 60 mm of the specimen using the data in (1).arrow_forward
- Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning