Samiel Mehary

[Samiel Mehary]

can you email me #7 answer that you hid it.

[Samiel Mehary]

#9 or we dont have 240v ac

[Samiel Mehary]

#9 we have loose connection on L1 & L2

[Samiel Mehary]

So that mean, are booster and ignitor in a parallel circuit?

[Samiel Mehary]

the booster and main in a series circuit so they will share the voltage drop from the source voltage. Right.
#9 if the voltage is 0v that mean no complete circuit and no current flowing. Right.

[Samiel Mehary]

I hope I made you happy now. kkkkkkkkkkkkk

[Samiel Mehary]

#9
Meter 1: 0 vac across the heating element suppose to be 240V Ac but there is a problem
to find out the problem I have to measure as it shows on figure 2
disconnect L1 to make sure why not get 240V from the power source therefor on figure 2 = #1 & 3 is 0v that mean the unite didn’t get 240V from power source. So, I will check the outlet, breaker or cord.

[Samiel Mehary]

#8
oh ignitor will flow from L1 to ignitor through detector then N detector is not load it is a switch therefor 120v drop.

[Samiel Mehary]

#7
4. If you increase the resistance in a circuit, then the current will decrease
therefore the answer is decrease

[Samiel Mehary]

Big Sorry for #5 #5 = I/ R1+R2+R3
Therefore:
1/ (1/10ohm+1/20ohm) = .01 + 0.05 =
1/ 0.15 = 6.666Ohm

[Samiel Mehary]

#7
[answers hidden]
#8 In a parallel circuit connected independently to the supply power therefore,
1. the ignitor =120 v drop because in parallel circuit.
2. the booster coil and the main will share =120 v drop because they are in a Series and it dependent which has more resistance. The more resistance it has the more voltage drop will has.
3. the safety coil =120 V drop because in parallel circuit

#9
1.Suppose to be 240 ac V but as it shows L1 & L2 have no power supply both line somewhere disconnected.
2. I proved my fig 1 measurements that L1 & L2 have no power no connection if I get 12o ac V only L1 has lose connection and work to connect L1. In this case I have to connect L1 & L2.

[Samiel Mehary]

#4= I=V/R [answer hidden]
#5 = I/ R1+R2+R3
Therefore:
1/ (1/10ohm+1/20ohm+1/40ohm) 1/ (0.1+0.05+0.025)
1/0.175 = 5.71ohm

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