should be the Intensity (minimum) of the BULB such that the LED doesn’t Burn out?If the LED can handle only 6 V across it and has an Internal Resistance of 9 Ω
Answers
Explanation:
Light emitting diode is a forward biased P-N junction which emits light.
The voltage across it = 2 v
Battery voltage = 6 v
Hence total voltage in the circuit = V = 6 - 2 = 4 v
since a LED has 0 resistance in the forward biased region, total resistance in the circuit
= 0 + r = r
current I = 10 mA = 10/1000 A = 0.01 A
we know that,
R = V/I
=> r = 4/0.01 = 400Ω
Hence the value of the limiting resistor is 400Ω.
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Answer:
Using the circuit above, you will need to know three values in order to determine the current limiting resistor value.
i = LED forward current in Amps (found in the LED datasheet)
Vf = LED forward voltage drop in Volts (found in the LED datasheet)
Vs = supply voltage
Once you have obtained these three values, plug them into this equation to determine the current limiting resistor:
Also, keep in mind these two concepts when referring to the circuit above.
The current, i, coming out of the power source, through the resistor and LED, and back to ground is the same. (KCL)
The voltage drop across the resistor, in addition to the forward voltage drop of the LED equals the supply voltage. (KVL)