why resistors used in testing "the forward and reverse characteristics of diode" gets burned when high current is passed?
I'm sorry but I don't understand the question, can you clarify what exactly you are asking? You may benefit from reading this. Dave
High current in a resistor causes heat which must be dissipated by conduction, convection, or radiation. To compute the power in a resistor you square the current and multiply by the resistor value. If the current is expressed in Amperes and the resistance is expressed in Ohms then the power will be expressed in Watts. Example: An IR-LED has the desired brightness at a forward current of 150 mA at a forward voltage drop of 2.2 Volts. The supply is +5 VDC. The resistor for this application is 18 Ohms. Compute the required power dissipation rating of the resistor Code ( (Unknown Language)): (0.150)*(0.150)*18 = 405 mWatts If you use a quarter watt resistor in this application the magic smoke will very quickly exit the resistor body with the usual fetid acrid smell. Remember the following piece of doggerel: "Twinkle-twinkle little star Power's equal I squared R"