(a)
The impedance of the circuit containing a heating coil and an ultra-capacitor connected in series to a 60.0 Hz 120 V rms supply.
Answer to Problem 81QAP
The impedance of the circuit containing a heating coil and an ultra-capacitor connected in series to a 60.0 Hz 120 V rms supply is 2.18 O.
Explanation of Solution
Given:
Number of windings in the coil
Length of the coil
Diameter of the coil
Resistance of the coil
Capacitance of the capacitor
Frequency of the supply
The rms voltage of the supply
Formula used:
The heating coil behaves as an inductor. The inductance L of the coil is given by,
Here,
And r is the radius of the coil.
Therefore, the inductance is given by,
The impedance of the circuit is given by,
Calculation:
Determine the value of inductance of the heating coil by substituting the values of variables in equation (1).
Substitute the value of the known variables in equation (2) and calculate the value of impedance.
Conclusion:
Thus, the impedance of the circuit containing a heating coil and an ultra-capacitor connected in series to a 60.0 Hz 120 V rms supply is 2.18 O.
(b)
The rms current through of the circuit containing a heating coil and an ultra-capacitor connected in series to a 60.0 Hz 120 V rms supply.
Answer to Problem 81QAP
The rms current through of the circuit containing a heating coil and an ultra-capacitor connected in series to a 60.0 Hz 120 V rms supply is 55 A.
Explanation of Solution
Given:
The rms voltage of the supply
Impedance
Formula used:
The rms current flowing in the circuit is given by,
Calculation:
Calculate the rms value of the current by substituting the given values of variables in the formula.
Conclusion:
Thus the rms current through of the circuit containing a heating coil and an ultra-capacitor connected in series to a 60.0 Hz 120 V rms supply is 55 A.
(c)
The peak value of current in the circuit.
Answer to Problem 81QAP
The peak value of current in the circuit is 77.7 A.
Explanation of Solution
Given:
The rms value of current
Formula used:
The peak value of current is given by,
Calculation:
Calculate the value of the peak current by substituting the value of
Conclusion:
Thus the peak value of current in the circuit is 77.7 A.
Want to see more full solutions like this?
Chapter 21 Solutions
COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
- An ac source of voltage amplitude 10 V delivers electric energy at a rate of 0.80 W when its current output is 2.5 A. What is the phase angle between the emf and the current?arrow_forwardA 40-mH inductor is connected to a 60-Hz AC source whose voltage amplitude is 50 V. If an AC voltmeter is placed across the inductor, what does it read?arrow_forwardFor an RLC series circuit, the voltage amplitude and frequency of the source are 100 V and 500 Hz, respectively; R=500 ; and L = 0.20H . Find the average power dissipated in the resistor for the following values for the capacitance: (a) C=2.0F and (b) C=2.0F .arrow_forward
- Calculate the rms currents for an ac source is given by v(t)=v0sint , where V0=100V and =200rad/s when connected across (a) a 20F capacitor, (b) a 20-mH inductor, and (c) a 50 resistor.arrow_forwardA 1.5k resistor and 30-mH inductor are connected in series, as below, across a120-V(rms)ac power source oscillating at 60-Hz frequency. (a) Find the current in the circuit. (b) Find the voltage drops across the resistor and inductor. (C) Find the impedance of the circuit. (d) Find the power dissipated in the resistor. (e) Find the power dissipated in the inductor. (1) Find the power produced by the source.arrow_forwardThe emf of an ac source is given by v(t)=V0sint, where V0=100V and =200 . Find an expression that represents the output current of the source if it is connected across (a) a 20-pF capacitor, (b) a 20-mH inductor, and (c) a 50 resistor.arrow_forward
- In a purely inductive AC circuit as shown in Figure P21.15, Vmax = 100. V. (a) The maximum current is 7.50 A at 50.0 Hz. Calculate the inductance L. (b) At what angular frequency is the maximum current 2.50A? Figure p21.15arrow_forwardAn AC generator with an rms emf of 15.0 V is connected in series with a 0.54-H inductor. The frequency of the source emf is 70.0 Hz. Draw a phasor diagram for this circuit, including the current, the potential difference across the inductor, and the source emf. Draw your diagram with the current phasor pointing toward the right along the horizontal axis.arrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningGlencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-HillCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPhysics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning