LED Resistor worksheet

1. Resistance needed: 125 Ω

1. The resistor takes whatever voltage the LED doesn't: subtract the forward voltage from the supply

   $\mathrm{V\_R}$$=5\text{\hspace{0.17em}V}-2\text{\hspace{0.17em}V}$$=3\text{\hspace{0.17em}V}$

2. Convert the current to amperes (1 mA = 0.001 A)

   $\frac{24\text{\hspace{0.17em}mA}}{\mathrm{1,000}}$$=0.024\text{\hspace{0.17em}A}$

3. Ohm's law (resistance = voltage ÷ current) gives the resistance that lets exactly this current through: R = V_R ÷ I

   $R$$=\frac{3\text{\hspace{0.17em}V}}{0.024\text{\hspace{0.17em}A}}$$=125\text{\hspace{0.17em}Ω}$

4. Work out the heat the resistor must handle, so you can pick a part that won't overheat: P = V_R × I

   $P$$=3\times 0.024$$=0.072\text{\hspace{0.17em}W}$

5. Round up to the next E24 value you can buy, so the current stays at or below the target. Then recheck the current

   $\text{130 Ω: I =}\frac{3}{130}\text{=}23.0769\text{\hspace{0.17em}mA}$

2. Resistance needed: 166.67 Ω

1. The resistor takes whatever voltage the LED doesn't: subtract the forward voltage from the supply

   $\mathrm{V\_R}$$=5\text{\hspace{0.17em}V}-2\text{\hspace{0.17em}V}$$=3\text{\hspace{0.17em}V}$

2. Convert the current to amperes (1 mA = 0.001 A)

   $\frac{18\text{\hspace{0.17em}mA}}{\mathrm{1,000}}$$=0.018\text{\hspace{0.17em}A}$

3. Ohm's law (resistance = voltage ÷ current) gives the resistance that lets exactly this current through: R = V_R ÷ I

   $R$$=\frac{3\text{\hspace{0.17em}V}}{0.018\text{\hspace{0.17em}A}}$$=166.6666666667\text{\hspace{0.17em}Ω}$

4. Work out the heat the resistor must handle, so you can pick a part that won't overheat: P = V_R × I

   $P$$=3\times 0.018$$=0.054\text{\hspace{0.17em}W}$

5. Round up to the next E12 value you can buy, so the current stays at or below the target. Then recheck the current

   $\text{180 Ω: I =}\frac{3}{180}\text{=}16.6667\text{\hspace{0.17em}mA}$

3. Resistance needed: 200 Ω

1. The resistor takes whatever voltage the LED doesn't: subtract the forward voltage from the supply

   $\mathrm{V\_R}$$=6\text{\hspace{0.17em}V}-2\text{\hspace{0.17em}V}$$=4\text{\hspace{0.17em}V}$

2. Convert the current to amperes (1 mA = 0.001 A)

   $\frac{20\text{\hspace{0.17em}mA}}{\mathrm{1,000}}$$=0.02\text{\hspace{0.17em}A}$

3. Ohm's law (resistance = voltage ÷ current) gives the resistance that lets exactly this current through: R = V_R ÷ I

   $R$$=\frac{4\text{\hspace{0.17em}V}}{0.02\text{\hspace{0.17em}A}}$$=200\text{\hspace{0.17em}Ω}$

4. Work out the heat the resistor must handle, so you can pick a part that won't overheat: P = V_R × I

   $P$$=4\times 0.02$$=0.08\text{\hspace{0.17em}W}$

5. Round up to the next E24 value you can buy, so the current stays at or below the target. Then recheck the current

   $\text{200 Ω: I =}\frac{4}{200}\text{=}20\text{\hspace{0.17em}mA}$

4. Resistance needed: 187.5 Ω

1. The resistor takes whatever voltage the LED doesn't: subtract the forward voltage from the supply

   $\mathrm{V\_R}$$=5\text{\hspace{0.17em}V}-2\text{\hspace{0.17em}V}$$=3\text{\hspace{0.17em}V}$

2. Convert the current to amperes (1 mA = 0.001 A)

   $\frac{16\text{\hspace{0.17em}mA}}{\mathrm{1,000}}$$=0.016\text{\hspace{0.17em}A}$

3. Ohm's law (resistance = voltage ÷ current) gives the resistance that lets exactly this current through: R = V_R ÷ I

   $R$$=\frac{3\text{\hspace{0.17em}V}}{0.016\text{\hspace{0.17em}A}}$$=187.5\text{\hspace{0.17em}Ω}$

4. Work out the heat the resistor must handle, so you can pick a part that won't overheat: P = V_R × I

   $P$$=3\times 0.016$$=0.048\text{\hspace{0.17em}W}$

5. Round up to the next E24 value you can buy, so the current stays at or below the target. Then recheck the current

   $\text{200 Ω: I =}\frac{3}{200}\text{=}15\text{\hspace{0.17em}mA}$

5. Resistance needed: 142.86 Ω

1. The resistor takes whatever voltage the LED doesn't: subtract the forward voltage from the supply

   $\mathrm{V\_R}$$=4\text{\hspace{0.17em}V}-2\text{\hspace{0.17em}V}$$=2\text{\hspace{0.17em}V}$

2. Convert the current to amperes (1 mA = 0.001 A)

   $\frac{14\text{\hspace{0.17em}mA}}{\mathrm{1,000}}$$=0.014\text{\hspace{0.17em}A}$

3. Ohm's law (resistance = voltage ÷ current) gives the resistance that lets exactly this current through: R = V_R ÷ I

   $R$$=\frac{2\text{\hspace{0.17em}V}}{0.014\text{\hspace{0.17em}A}}$$=142.8571428571\text{\hspace{0.17em}Ω}$

4. Work out the heat the resistor must handle, so you can pick a part that won't overheat: P = V_R × I

   $P$$=2\times 0.014$$=0.028\text{\hspace{0.17em}W}$

5. Round up to the next E24 value you can buy, so the current stays at or below the target. Then recheck the current

   $\text{150 Ω: I =}\frac{2}{150}\text{=}13.3333\text{\hspace{0.17em}mA}$

6. Resistance needed: 55.56 Ω

1. The resistor takes whatever voltage the LED doesn't: subtract the forward voltage from the supply

   $\mathrm{V\_R}$$=5\text{\hspace{0.17em}V}-4\text{\hspace{0.17em}V}$$=1\text{\hspace{0.17em}V}$

2. Convert the current to amperes (1 mA = 0.001 A)

   $\frac{18\text{\hspace{0.17em}mA}}{\mathrm{1,000}}$$=0.018\text{\hspace{0.17em}A}$

3. Ohm's law (resistance = voltage ÷ current) gives the resistance that lets exactly this current through: R = V_R ÷ I

   $R$$=\frac{1\text{\hspace{0.17em}V}}{0.018\text{\hspace{0.17em}A}}$$=55.5555555556\text{\hspace{0.17em}Ω}$

4. Work out the heat the resistor must handle, so you can pick a part that won't overheat: P = V_R × I

   $P$$=1\times 0.018$$=0.018\text{\hspace{0.17em}W}$

5. Round up to the next E24 value you can buy, so the current stays at or below the target. Then recheck the current

   $\text{56 Ω: I =}\frac{1}{56}\text{=}17.8571\text{\hspace{0.17em}mA}$

7. Resistance needed: 166.67 Ω

1. The resistor takes whatever voltage the LED doesn't: subtract the forward voltage from the supply

   $\mathrm{V\_R}$$=6\text{\hspace{0.17em}V}-2\text{\hspace{0.17em}V}$$=4\text{\hspace{0.17em}V}$

2. Convert the current to amperes (1 mA = 0.001 A)

   $\frac{24\text{\hspace{0.17em}mA}}{\mathrm{1,000}}$$=0.024\text{\hspace{0.17em}A}$

3. Ohm's law (resistance = voltage ÷ current) gives the resistance that lets exactly this current through: R = V_R ÷ I

   $R$$=\frac{4\text{\hspace{0.17em}V}}{0.024\text{\hspace{0.17em}A}}$$=166.6666666667\text{\hspace{0.17em}Ω}$

4. Work out the heat the resistor must handle, so you can pick a part that won't overheat: P = V_R × I

   $P$$=4\times 0.024$$=0.096\text{\hspace{0.17em}W}$

5. Round up to the next E24 value you can buy, so the current stays at or below the target. Then recheck the current

   $\text{180 Ω: I =}\frac{4}{180}\text{=}22.2222\text{\hspace{0.17em}mA}$

8. Resistance needed: 157.89 Ω

1. The resistor takes whatever voltage the 2 LEDs don't: subtract their combined forward voltage from the supply

   $\mathrm{V\_R}$$=5\text{\hspace{0.17em}V}-2\times 1\text{\hspace{0.17em}V}$$=3\text{\hspace{0.17em}V}$

2. Convert the current to amperes (1 mA = 0.001 A)

   $\frac{19\text{\hspace{0.17em}mA}}{\mathrm{1,000}}$$=0.019\text{\hspace{0.17em}A}$

3. Ohm's law (resistance = voltage ÷ current) gives the resistance that lets exactly this current through: R = V_R ÷ I

   $R$$=\frac{3\text{\hspace{0.17em}V}}{0.019\text{\hspace{0.17em}A}}$$=157.8947368421\text{\hspace{0.17em}Ω}$

4. Work out the heat the resistor must handle, so you can pick a part that won't overheat: P = V_R × I

   $P$$=3\times 0.019$$=0.057\text{\hspace{0.17em}W}$

5. Round up to the next E12 value you can buy, so the current stays at or below the target. Then recheck the current

   $\text{180 Ω: I =}\frac{3}{180}\text{=}16.6667\text{\hspace{0.17em}mA}$

9. Resistance needed: 55.56 Ω

1. The resistor takes whatever voltage the 2 LEDs don't: subtract their combined forward voltage from the supply

   $\mathrm{V\_R}$$=5\text{\hspace{0.17em}V}-2\times 2\text{\hspace{0.17em}V}$$=1\text{\hspace{0.17em}V}$

2. Convert the current to amperes (1 mA = 0.001 A)

   $\frac{18\text{\hspace{0.17em}mA}}{\mathrm{1,000}}$$=0.018\text{\hspace{0.17em}A}$

3. Ohm's law (resistance = voltage ÷ current) gives the resistance that lets exactly this current through: R = V_R ÷ I

   $R$$=\frac{1\text{\hspace{0.17em}V}}{0.018\text{\hspace{0.17em}A}}$$=55.5555555556\text{\hspace{0.17em}Ω}$

4. Work out the heat the resistor must handle, so you can pick a part that won't overheat: P = V_R × I

   $P$$=1\times 0.018$$=0.018\text{\hspace{0.17em}W}$

5. Round up to the next E12 value you can buy, so the current stays at or below the target. Then recheck the current

   $\text{56 Ω: I =}\frac{1}{56}\text{=}17.8571\text{\hspace{0.17em}mA}$

10. Resistance needed: 136.36 Ω

1. The resistor takes whatever voltage the LED doesn't: subtract the forward voltage from the supply

   $\mathrm{V\_R}$$=5\text{\hspace{0.17em}V}-2\text{\hspace{0.17em}V}$$=3\text{\hspace{0.17em}V}$

2. Convert the current to amperes (1 mA = 0.001 A)

   $\frac{22\text{\hspace{0.17em}mA}}{\mathrm{1,000}}$$=0.022\text{\hspace{0.17em}A}$

3. Ohm's law (resistance = voltage ÷ current) gives the resistance that lets exactly this current through: R = V_R ÷ I

   $R$$=\frac{3\text{\hspace{0.17em}V}}{0.022\text{\hspace{0.17em}A}}$$=136.3636363636\text{\hspace{0.17em}Ω}$

4. Work out the heat the resistor must handle, so you can pick a part that won't overheat: P = V_R × I

   $P$$=3\times 0.022$$=0.066\text{\hspace{0.17em}W}$

5. Round up to the next E12 value you can buy, so the current stays at or below the target. Then recheck the current

   $\text{150 Ω: I =}\frac{3}{150}\text{=}20\text{\hspace{0.17em}mA}$