Recent content by hisham.i

Lazyaditya, Your work was correct until the last line of the second picture you attached. You substituted 430 uA for Ic which is not correct. 430uA is the total current = Ic + I(4 Kohm resistor). The solution in the book is correct, you don't need to substitute each current value...

Abhiroop, Such a signal with this small amplitude is very sensitive to noise, so ofcourse aligator clips will introduce sufficient amount of unwanted signal especially from power line source. In all cases you will need for band pass filter to filter out the noise & get the usefull info...

Thanks berkman, You are absolutely correct, NEC can be very helpful. Actually i am working for offshore platforms installation design, so if you have any other document to share, it will be very helpful. Regards, Hisham

Hello, I am design electrical engieer, & as part of my job i am responsible to prepare fabrication & detail drawings. However this requires extensive experience in yard & fabrication work, since there is a lot of consumbale items that need previous experience to make the correct details &...

Hello guys, I was searching for what voltage does IEC standard call for Di electric test that should be done to low voltage panels? is it 2.5 KV for 1 minutes? If yes how how low voltage cable rated 600/1000 V can withstand 2.5KV without insulation failure?

Agreed that it will provide more strength on horizantal forces, but how it will affect the loading? as "the_emi_guy". Since return flange will not provide any strength for the loading capacity.

Thanks for your helpful info. But i don't understand how can a flange on the top of the cable tray make it stronger?

I am designing cable tray for installation, while i am looking on different types of cables trays i found that there is cable trays called "Cable trays with return flange". So i am wondering where such type of trays is used? Thanks, Hisham

No i want to use the resistor & optocoupler configuration, but the value of 20 mA is not the minimum current that can be drawn through the optocoupler to open the photo transistor. My question is what is the minimum current that can be drawn through the optocoupler to open the photo...

Hi davenn, But drawing droping the voltage through the resistor from 354V to 2 V at the terminals for the optocoupler will result is a high resistance required (This is not a problem i can use many resistors in series), but if you take 20 mA current the power for the resistor to be selected...

In your connection the resistor is useless, in order to limit the current the resistor should be in series with the battery. Place a variable resistor with correct power in series with ammeter and the battery, then change the resistance to adjust the charging current. Also remove the op amp the...

Hello, I want to design an AC voltage indicator. So i designed the circuit attached, but now i am trying to calculate the value of the resistor R1. I have the following problem: 1- The voltage should be given at the terminals of the optocoupler. 2- The minimum current that should flow...

USC_ME how much is your input voltage? And why you didn't use LM7812 voltage regulator instead of all this?

Both are the same, eddy current losses means the power loss due to eddy current.

It depends on the shape of the magnetic field you want to generate. When current passes through a wire, it creates magnetic field that is circular around the wire. While if you turn it in the shape of the coil as you mentioned you will get magnetic field in the shape of approximately straight...