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I am beginner and trying to understand how to calculate the required resistance and required current in series circuits.

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In this first picture I would like to know if my calculations are correct or not correct. I have an LED and a resistor connected to a 5V 500mA. The LED requires 5V and .30mA current. I need to know how much resistance I need. I did the following calculations.

Volt (V) = 5V
Current (I) = .03 Amps
R = V/I = 5V/.03 Amps = 166 OHMS.

In the calculations above I calculated a 166 OHM resistor would output 5 Volts with a current of .03 Amps if my calculations correct or not? If my calculations are not correct please correct them and explain your correction.

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In the second circuit I would like to know how to calculate the total Amps required by all of the LEDs in the circuit. One LED requires 30mA, so I multiplied 30mA by 3 which is 90mA. Is this the correct why to know the Amps required by all the LEDs? If not please correct me and explain the correction.

Third question. I have 4 AA batteries. One AA battery outputs 1.5V and 2Amps. If I have 4 batteries does that mean I have 6V and 2Amps output?

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  • \$\begingroup\$ Please double check your values for the first picture. You write "connected to a 5V 500ma" when the voltage source seems to be labeled as 6V, not 5V. You also wrote the LED requires 5V and 0.30mA when it is labeled as 30mA, but then your calculation uses 0.3A which is actually 300mA. \$\endgroup\$
    – InBedded16
    Commented Jun 13, 2023 at 17:59

2 Answers 2

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For the first question, I think you misunderstood V=iR as it applies to a resistor. The V in this case is not 5V, but rather the voltage drop that you are expecting across the resistor. For this you also need to know the forward voltage of the LED, which I suspect is not actually 5V, and then find Vresistor=5V-VLED.That value you can plug into R=V/i.

For the second question, you are incorrect that you add the currents. An individual path does not lose or gain current at any point on the path. Voltage rises and drops over the course of a path. So you need to multiply the forward voltage of the LED by how many LEDs you want to connect in series, than provide that much voltage from your source plus a little extra for the voltage drop across the resistor. (In other words, a 5V power supply will not be able to support more than 1 of those LEDs in series. You can wire them in parallel instead, which keeps the voltage required at 3.2V but changes the current requirement.) The current stays fixed at 30mA no matter how many LEDs you put in series.

Similar to the previous concept, batteries will play together differently depending on if they are connected in series or in parallel. You are correct that if connected in series you will get 6V 2A, but in parallel you would get 1.5V 8A.

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    \$\begingroup\$ These aren't practice problems. I am trying to design a simple LED circuit and apply the correct calculations on it. I edited the circuit diagrams as you asked. Please check my first question to see if it is correct or not. Thanks \$\endgroup\$
    – Mahmoud Abdel-Rahman
    Commented Jun 13, 2023 at 18:12
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    \$\begingroup\$ @MahmoudAbdel-Rahman If you are trying to design a circuit, please provide specific parts that you want to use. Specifically a datasheet for your desired LED would be useful. \$\endgroup\$
    – InBedded16
    Commented Jun 13, 2023 at 18:15
  • \$\begingroup\$ The pdf for the LED I am using tme.eu/Document/e17ac5a5d91f1e843e5681cfa4a5ba4b/… \$\endgroup\$
    – Mahmoud Abdel-Rahman
    Commented Jun 13, 2023 at 18:20
  • \$\begingroup\$ @MahmoudAbdel-Rahman okay so 2 things I see in that datasheet. 1) The current through the LED will likely be more like 20mA, 30mA is the absolute maximum. 2) Use the Typ. Forward Voltage value on page 3 as the voltage drop across the LED. So for green/blue, 3.2V. \$\endgroup\$
    – InBedded16
    Commented Jun 13, 2023 at 18:24
  • \$\begingroup\$ In your answer you wrote "You can wire them in parallel instead, which keeps the voltage required at 3.2V but changes the current requirement.) ". Does this mean that I would need to up the current? \$\endgroup\$
    – Mahmoud Abdel-Rahman
    Commented Jun 13, 2023 at 19:25
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Here is a little mantra. First, lets define a "simple" series circuit as one that does not have other circuit connections, such as two resistors in parallel acting as a single resistor. Both of your circuits meet this requirement.

In a simple series circuit, 100% of the current goes through 100% of the components 100% of the time.

A similar statement can be made about parallel circuits.

In a simple parallel circuit, 100% of the voltage appears across 100% of the components 100% of the time.

Next, some guidance. An LED is not a resistor, and cannot be treated as one when analyzing or calculating a circuit. It does not obey or conform to Ohm's Law. I won't give away how that affects your problem unless you ask.

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