Case Study | Application of Ohm's Law in DC Circuit Analysis

Latest update time：2019-06-26

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Keywords : DC circuit, Ohm's law, application , series connection , parallel connection, series-parallel connection

When using an electronic component, you first need to know how to calculate the current, resistance, and voltage drop. Once you know two of these three parameters, you can calculate the third one using Ohm's law. Let's look at this calculation based on a few simple circuits.

Series Resistors Circuit

The circuit below is two resistors in series, powered by a 12VDC power supply. The first step is to calculate the total resistance of the circuit, and then calculate the current of the circuit. In a series circuit, the current of the entire circuit is the same.

The total resistance of the circuit is 11K ohms , and the total voltage drop across the load is 12VDC . The current can be calculated using Ohm's law: I = V / R = 0.00109A

Then use Ohm's law to calculate the voltage drop across each resistor: V = IR

The voltage drop across a 1K ohm resistor is: V = 0.00109 X 1000 = 1.09V
The voltage drop across a 10K ohm resistor is: V = 0.00109 X 10000 = 10.9V

Parallel resistors Circuit

The circuit in the figure below uses the same 12VDC power supply, but the load is connected in parallel. The two resistors in parallel have the same voltage across them, so the voltage drop across the two resistors in this circuit is 12V , but due to the different resistance values, the current flowing through the two resistors will be different.

The current flowing through the two resistors can then be calculated using Ohm's law: I = V/R

The current on a 1K ohm resistor is: I = 12/1000 = 0.012A
The current on the 10K ohm resistor is: I = 12/10000 = 0.0012A
The total current of the circuit is: 0.0012A + 0.012A = 0.0132 A

According to the above calculation results, the power consumption of the series circuit and the parallel circuit can be calculated respectively:

The power dissipated in the series circuit is: 12V x 0.00109A = 0.01308W
The power dissipation in parallel is: 12V x 0.0132A = 0.1584W

Series-parallel circuit

The following figure shows a simple series-parallel circuit. Many beginners will build this circuit on a breadboard. It is important to understand how to calculate.

The parallel circuit part is two LEDs with the same parameters connected in parallel. We know that in a parallel circuit, the total current is equal to the sum of the branch currents, so the total current is 40mA . Now there are enough parameters to calculate the resistance value of R1 . The voltage drop of the LED is 2.2V , so the voltage drop across the resistor is 9.8V . According to Ohm's law: R = V/I .