
Concept explainers
State which ungrounded conductors should be disconnected by the high-temperature limit control as per NEC 422.47 and provide the maximum water temperature that is limited by the high-temperature control.

Answer to Problem 1R
The high-temperature limit control should disconnect all the ungrounded conductors and the maximum water temperature limits to
Explanation of Solution
Discussion:
According to National Electric code NEC 422.47, the high-temperature limit control must disconnect all the ungrounded conductors.
Electric water heaters, which are utilized in residential, are listed under UL safety Standard 174, “Household Electric Storage Tank Water Heaters”.
The NEC and UL standard suggest that the electric water heater should be equipped with high temperature cut-off for safety purpose. The high-temperature control limits the water temperature into the maximum limit of
Conclusion:
Thus, the high-temperature limit control should disconnect all the ungrounded conductors and the maximum water temperature limits to
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Chapter 19 Solutions
Electrical Wiring Residential
- Consider the following circuit, assuming the switch has been in the same position for a long period of time before t = 0: Vx L iL R3 R2 R₁ Is + Vo - コロ >Where Is = 100 mA,R=2202, R2 = 4702,R3 =4702,L= 1 mH. As indicated on the diagram, before t = 0, the switch is closed, after t = 0 the switch is open. 1. What are Ve and Vo before the switch shown opens (answer to within 1% accuracy)? Vx = V, Vo = V 2. What is the T of the RL circuit after the switch operates (answer to within 1% accuracy)? T= μs 3. Complete the derivation for the inductor current in (t) differential equation below by filling in the blank coefficients (answer to within 1% accuracy): diy(1) dt di (0) + iz (t)+ = 0 4. Hence or otherwise, find the time domain expression for Vo(t) (answer to within 1% accuracy): Vo(t)= exp(arrow_forward確定圖5-38中的I1、I2和IB。 Vcc=-12 V;R1=33 kN;RC=1.8 kN;βDC = 150;RE=560;R2=5.6 kN 圖 5-38arrow_forwardConsider the following circuit: Vs R₁ Vx ww C'₁ R2 Where Vs = 3.3 cos(2000t-10) VR:=2202 R2 = 1002,L=1mH, Ci = 22 nF, C2 = 47 nF N.B. We have been using cosine as the basis for our phasors. 1. What is the impedance of each of R, Ci, L (answer to within 1%): Z RI = +j Q Zci= +j QZL= +jQ 心 2. Complete the following KCL for node Vo, assuming current flowing out of the node is positive (answer to within 1%): 0= +j )+Vo/ +j0)+Vo/(0+j 回回回 3. Hence or otherwise solve for Vo in phasor form (answer to within 1% amplitude and 5 degrees of phase): Vo = ° V 回 4. Convert this phasor to a time domain expression for Vo(t) (answer to within 1% amplitude and 5 degrees of phase): Vo(t) = cos( t+ Vo 1arrow_forward
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- EBK ELECTRICAL WIRING RESIDENTIALElectrical EngineeringISBN:9781337516549Author:SimmonsPublisher:CENGAGE LEARNING - CONSIGNMENT
