
(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.
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Chapter 21 Solutions
COLLEGE PHYSICS
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