Find the current in R1 in Fig. 17.80 .

Name: Problem 30, Chapter 17: Applied Physics
Uploaded: 2021-07-14T09:32:41-08:00
Duration: 7 min 28 s
Channel: Umar Sohail Qureshi
Description: Find the current in R1 in Fig. 17.80 .

Key Concepts

Circuit Simplification Techniques

Techniques such as Thevenin's and Norton's theorems enable the reduction of a complex circuit to a simpler equivalent circuit. These methods are valuable tools for isolating parts of a circuit to simplify the process of analyzing the current in specific components.

Series and Parallel Resistors

Understanding how resistors combine in series and parallel configurations is essential for simplifying circuits. Calculating the equivalent resistance helps in reducing complex networks into simpler forms, facilitating easier analysis of current flow through individual components.

Kirchhoff's Current Law (KCL)

KCL asserts that the total current entering a junction must equal the total current leaving it. This principle is crucial for analyzing and calculating the distribution of current among different branches in a circuit.

Ohm's Law

Ohm's Law establishes the fundamental relationship between voltage, current, and resistance. It is used to calculate one of these parameters when the other two are known, making it essential for analyzing current flow in electrical circuits.

Kirchhoff's Voltage Law (KVL)

This law states that the sum of the electrical potential differences (voltages) around any closed loop in a circuit is zero. It is a fundamental tool for setting up equations to solve for unknown voltages, which indirectly assists in determining currents in various circuit components.

Transcript

00:01 In this problem, we have to find the current across the resistor r1, which i will highlight in red for convenience.

00:11 Oops, let me just highlight in red, so r1.

00:20 And for that, we'll first find the effective resistance of the circuit and start finding out how the current branches out.

00:30 So note that resistors in parallel, if two resistors are in parallel, then the total resistance, the reciprocal of the total resistance, is the sums of the reciprocals of the individual resistances.

00:53 So we'll find the resistance of r1 and the resistor that will highlight in green with which it is parallel in, by substituting r1 equals 4 .35 and r2 is equals 3 .75.

01:14 And we can just add these numbers together using a calculator 1 over 3 .75.

01:24 And what we're left with is 72 over 145.

01:30 That's r inverse.

01:32 And we can just reciprocate both sides to find the effective resistance as 2 .01 oms.

01:45 So this entire block that i'm highlighting currently has a resistance.

01:51 Of 2 .01 oms, which is in series with the resistance of 0 .765 oms, so their total resistance will be simply there's some because the resistances in series add up.

02:04 So that's 2 .01 oms plus 4 .35 oms.

02:09 Oh sorry, rather 0 .765 oms.

02:14 0 .765 oms.

02:21 And that gives us.

02:22 2 .775 oms as our resistance for this entire block containing this resistance.

02:42 Now this block is in parallel with 1 .56 so we can find that total resistance by the method we discussed earlier.

02:52 So 1 .56 is in parallel with 2 .75.

02:56 So 1 .56 plus 2 .75 will give us the total one over these.

03:06 We can just add these expressions using a calculator and what we'll end up with is 1 .00 and reciprocating both sides will still end up with one...

Please give Ace some feedback