Understanding Current Through a Resistor Using Ohms Law and Circuit Analysis
Understanding Current Through a Resistor Using Ohm's Law and Circuit Analysis
Introduction
Determining the current through a resistor is a common problem in electrical engineering and physics. To calculate this, we need to apply principles such as Ohm's Law, Kirchhoff's Current Law (KCL), and basic power calculations. In this article, we will walk through a detailed example to understand how to find the current through a 3.0Ω resistor. We will also cover how to analyze circuits using KCL and calculate power consumed by resistors.Understanding Ohm's Law
Ohm's Law is a fundamental principle in electrical engineering that relates the voltage, current, and resistance in an electrical circuit. It is defined as:I V/R
Where: I is the current (in Amperes, A) V is the voltage (in Volts, V) R is the resistance (in Ohms, Ω) When determining the current through a resistor, we need to know the voltage across it and its resistance. If this information is not provided, we need to use the circuit configuration and other given details to find the necessary values.For instance, if we have a 3.0Ω resistor in a circuit, and the voltage across it is given or can be calculated, we can use Ohm's Law to find the current. However, without the voltage, we cannot calculate the current.
Circuit Analysis Using KCL
KCL is a law in circuit analysis that states the sum of the currents entering a node must equal the sum of the currents leaving the node. This principle can be applied to analyze complex circuits. Consider the following scenario: we have a circuit with a 18Ω resistor, a 3.0Ω resistor, and four 1.5V voltage sources arranged such that the right side of the circuit is grounded. Using KCL, we can find the voltage at a node.Let's denote the voltage at the node left of the 18Ω resistor as V. The KCL equation at this node would be:
V/1.5 - V/1.5 - V/1.5 V/18 0
Multiplying the equation by 18, we get:
9V - 1.5V 0
Thus, 10V 27/2, and V 27/20 1.35 volts.
The current into each 1.5V source can be calculated as:
(1.35 - 1.5) / 6 -1/40 -25mA
This means 25mA flows out of each battery and 75mA into the 18Ω resistor.
To verify this, let's calculate the power from the sources and the power consumed by the resistance:
Power from the sources 3 * 1.5V * 25mA 9/80 W 112.5mW.
Power consumed by the 3Ω resistor 3Ω * (25mA)^2 0.075W 75mW.
Power consumed by the 18Ω resistor 18Ω * (75mA)^2 1.125W 112.5mW.
The power consumed by the 3Ω resistor is 75mW, and the power consumed by the 18Ω resistor is 112.5mW. The total power consumed is 187.5mW. This power is supplied by the 3 sources, each delivering 112.5mW.