The two resistors in branch A-B of the circuit in Figure...

Name: Problem 9, Chapter 5: Electricity 1: Devices, Circuits and Materials
Uploaded: 2021-05-04T23:39:48-08:00
Duration: 7 min 28 s
Channel: Luis Rios
Description: The two resistors in branch A-B of the circuit in Figure 5-13 are of equal value. FIGURE CANT COPY Figure 5-13 Series and parallel. What is the value of each resistor if the ammeter indicates 8 amperes?

The two resistors in branch A-B of the circuit in Figure 5-13 are of equal value. FIGURE CANT COPY Figure 5-13 Series and parallel. What is the value of each resistor if the ammeter indicates 8 amperes? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

The two resistors in branch A-B of the circuit in Figure 5-13 are of equal value.
FIGURE CANT COPY
Figure 5-13 Series and parallel.
What is the value of each resistor if the ammeter indicates 8 amperes?
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____________________________________________________________________
____________________________________________________________________

Key Concepts

Circuit Analysis

This broader concept encompasses the methods and techniques used to break down, simplify, and solve electrical circuits. It involves applying Ohm’s Law, Kirchhoff’s Laws, and resistor combination rules to determine unknown parameters, such as individual resistor values from measured currents in specific configurations.

Current Measurement in Circuits

Understanding how current is measured and the role of an ammeter in a circuit is important. An ammeter is connected in series and provides a reading of the total current flowing through a branch, which can be used together with Ohm's Law and resistor combination principles to analyze the circuit.

Ohm's Law

This fundamental principle of electrical circuits states that the voltage across a resistor is proportional to the current flowing through it (V = IR). It is essential in analyzing circuits because it allows you to determine unknown values, such as resistance, when the current and voltage are known.

Series and Parallel Resistor Combinations

Knowledge of how resistors combine when arranged in series or parallel is crucial. In series, resistances add directly; in parallel, the total resistance is found using the reciprocal sum of the individual resistances. This concept helps in simplifying complex circuits into equivalent resistances.

Transcript

00:01 In this problem, we have this circuit here.

00:02 All we're given is that the current of this 13 .8 on resistor is given as 0 .79 amps.

00:15 We can call this current one, so that means 13 .8 can be called resistor one.

00:21 So let's call 17 .2 r2.

00:25 The 8 .45 is r3.

00:29 The one at the bottom is r4.

00:35 And on the left side, 15 -on will be r5, and 12 .5 -on will be r6.

00:44 And we want to find the current for each resistor.

00:48 So to do this, we can use the junction rule, which is the total current of all it equals the sum of all the currents in the circuit.

01:00 But there's two things to notice here.

01:03 First is that from the battery, we'll have the total current coming out of it.

01:11 So we have the total current flowing into r5 and into r6.

01:17 But then when it comes to this place right here, it'll split off.

01:21 It'll split off to this current going down and to another current going to the right.

01:26 The one to the right will flow through r3, and it'll also flow to r3.

01:32 R4.

01:33 So r5 and r6 are in series, so they will have the same current going through them.

01:39 And on the right side, r3 and r4, they are also in series, so they'll have the same current flowing through them as well.

01:48 So now let's look at the current going through r2.

01:53 So we see that r1 and r2 are in parallel to each other, which means that they will both have the same potential difference or the same voltage across across both of them.

02:07 So we can find the voltage from resistor 1 and then i give us the voltage for resistor 2 and then we can find the current flowing through them.

02:17 So again we're looking for the voltage for resistor 1.

02:25 We can use we can do that with alms law which is voltage equals current time resistance.

02:33 So it'll be v1 equals i1 times r1 and we know what i1 is it's given to us as 0 .795 r1 is 13 .8 so the voltage across resistor 1 is 10 .97 volts and this is the same voltage across resistor 2 so we had v2 is also 10 .97 volts so now we can find the current flowing through i2.

03:10 Again from olms law, we have v2 equals i2 times r2, and we can solve for i2, so divide both sides by r2.

03:24 So we get voltage 2 divided by r2.

03:27 And i'll plug it in our values.

03:29 We have the top 10 .97, divided by r2, which is 17 .2.

03:36 So the current flowing through resistor 2 is 0 .64 amps...

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