Question

You want to power a green LED (ΔVOpenGΔVOpenG = 2.1 VV) and a red LED (ΔVOpenRΔVOpenR = 1.6 VV) using a 9.0-VV battery. You decide to connect the LEDs in series. In order to limit the current to the allowed value of 24 mAmA, you need to add a resistor in series with the LEDs.

Determine the resistance of the resistor that you need to connect to limit the current.

Express your answer with the appropriate units.

Answer 1

We know that In series circuit current in each element will be same and equal to total current.

Given that limit value of current in LED's = 24 mA = 0.024 A

dV_openG = 2.1 V

So Resistance of green LED = R = V/I (From ohm's law)

R1 = 2.1/0.024 = 87.5 ohm

dV_openR = 1.6 V

So Resistance of red LED = R = V/I (From ohm's law)

R2 = 1.6/0.024 = 66.7 ohm

EMF of battery = 9.0 V

Current through battery = 0.024 A

Resistance of battery = Req = V/I = 9.0/0.024 = 375 ohm

Now As given we need to add a resistor (R3 Resistance) in series with LED's to limit the current through them, So in series circuit:

Req = R1 + R2 + R3

R3 = Req - R1 - R2

R3 = 375 - 87.5 - 66.7

R3 = 220.8 ohm = resistance of resistor that needs to be added

Let me know if you've any query.