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PRACTICUM REPORT BASIC PHYSICS II "KIRCHOFF LAW" COLLECTION DATE: 16th of March 2018 M DATE: 11th of March 2018 M NAME : Utut Muhammad "KIRCHOFF’S LAW" FINAL PROJECT PRACTICUM A. OBJECTIVE 1. Studying Kirchoff's Law 2. Analyzing the series using Kirchoff Law 3. Compares the c...

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PRACTICUM REPORT BASIC PHYSICS II "KIRCHOFF LAW"

COLLECTION DATE: 16th of March 2018 M DATE: 11th of March 2018 M NAME : Utut Muhammad "KIRCHOFF’S LAW" FINAL PROJECT PRACTICUM A. OBJECTIVE 1. Studying Kirchoff's Law 2. Analyzing the series using Kirchoff Law 3. Compares the current value specified by an Ammeter with the results of a series analysis calculation. 4. Identify what happens when the voltage value is raised. B. BASIC THEORY The a number of voltage values in a closed circuit is always zero. The value will always be zero because there is a voltage that has a negative value. Negative voltages are usually found in voltage sources such as batteries or power supplies. This is often used in simple settings. This statement is called Kirchhoff's Law. (Blocer, 2003: 6-7) A direct current (DC) is an electric current flowing in the wire of a string because of the potential difference by 2 poles whose polarity is constant. The opposite is alternating current (AC) caused by the current source with the polarity of the two functions of the time. The source of electric current and back and forth is called the force of electricity or electric power and is also called emf (electromotive force). There is an oddity in the mention of emf because it means an electric motion force, but it is united with joules. To explain the relationship between

electric current, emf, and resistance in a closed circuit in the same direction can be used Kirchoff's Law. (Jati, 2009: 361) On the basis of a simple circuit can be analyzed by using equations and with series or parallel in each resistor by simplifying the circuit into a loop circuit. So the purpose of Kirchoff's rule is to analyze more complex circuits in order to simplify the simplification well so that the two principles of Kirchoff's Law are used. Kirchoff's law is a law to make it easier for us to calculate flows in a circuit that is quite complicated and cannot be solved directly. In other cases, actually, Kirchoff's Law is the application of the Law of Energy Conservation. (Giancoli, 2014: 105) For each closed path in a series, Kirchhoff's Voltage Law states that the number of other algebraic voltages is produced by currents in passive elements which produce voltage, which is sometimes referred to as voltage drop. This law also applies to circuits that are driven by sources. (Nahvi, 2004: 17) The inventor of Kirchoff's Law was a German physicist named Gustav Kirchoff. He succeeded in developing spectroscopy and science tools related to emission spectroscopes. Kirchoff also managed to conclude that a line spectrum would be obtained if the light was passed into a gas. Using a spectroscope, Kirchoff also managed to show that each element if heated would produce a certain emission line pattern. These results can be used to identify cesium elements in 1860 and rubidium in 1861. (Umar, 2008: 126) Ohm's Law and Kirchoff's Law do dominate the theory of circuit analysis. But the three Laws will not be enough when people have to analyze the changes in each parameter in the series. So that in an electrical circuit in the form of two serial loops it contains parameters which consist of a constant source of pressure and a number of resistors that know the value of the resistance and the voltage source that changes. (Sutrisno, 1986) According to Giancoli Douglass, Kirchoff's Law is divided into two parts, namely Kirchoff's I Law and Kirchoff's II Law. Law I Kirchoff or the so-called law of the branch point is based on the conservation of electric

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charge, which reads: "At any point in the branch, the sum of all currents entering a branch, should be equal to all currents leaving the current." 𝐼𝑚𝑎𝑠𝑢𝑘Σ=Σ

𝐼𝑘𝑒𝑙𝑢𝑎𝑟'slaw Kirchoff also called The Law of Knots or Intersections

which are understood as the sum of all currents leading to a node (i.e. the

intersection where three or more sources or branches of current are connected) must be equal to the sum of all currents leaving the node. If we determine the inflow as positive and the outflow as negative, then the number of flows equals zero is a common alternative to the rule. The current always flows from a high potential to a high potential to a low potential through a resistor. When passing through a resistor in the current direction, the potential change is negative because of a potential decrease. The positive terminal of a pure source of electromotive force (emf) is always a terminal with a high totential and does not depend on the direction of the current passing through the source of the electromotive force (emf). (Bueche, 2006: 202) "at each branch point, the sum of all flows entering the branch must be the same as all the currents leaving the branch" (Giancoli, 2014: 105).

(Figure 1. Example of the number of currents entering the branch. Http://www.rumusmatematika.me/2016/11/pengertian-dan-hukum Kirchoff-ii.html) Sounds of Kirchoff II's Law namely voltage (difference potential) in a closed circuit is zero, for example, we buy goods and pay according to the

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price (comparable). This is like Kirchoff's Law I. Kirchoff II's Law states that:

(Figure 2. Amount of potential around lines in a circuit must be 0.) Kirchhoff's second law becomes formulated: becomes ΣΔ𝑉= 0 (Giancoli, 2014) Kirchhoff's I Law is charge conservation law which states that the amount of charge flowing does not change. This means that the charge rate (current strength) leading to the branch point is equal to the rate of charge (current strength) leaving the branch point. (Surya, 2009: 95) Kirchhoff's law for voltage states that the amount of voltage in a closed circuit is zero. Polarity or direction of voltage depends on the direction of electric current. The current will flow from a potential or higher voltage point to a potentially lower point. The point that has a lower voltage is a point that has positive polarity, and a point that has a lower voltage is a point with negative polarity. (Zuhal, 2004: 16-17)

C. TOOLS AND MATERIALS N O

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PICTURES

NAME OF TOOLS AND MATERIALS

AMOUNT

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2 Fruit 1

Power Supply

1 Fruit 2

Barriers

3 Fruit 3

Multimeter

4

Connecting Cables to

Sufficient 3 Fruit

5

Resistor

6

Jumper to

taste

Enough 7

Crocodile Capit

D. WORK STEPS Experiment I Change the voltage at E1 NO

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IMAGES

WORK STEP

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1

Prepare tools and materials

2

Make a Kirchoff series with the sketch guide that was shown in Figure

3

Turn on multimeter by turning the selector on a DC current with 10A

4

Turn on the first power supply or V1 and rotate the selector according to the data provided

5

Record the results of the practicum data obtained.

Experiment II Replace the voltage at E2 NO

1

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IMAGE

WORK STEP

Prepare tools and materials

MUHAMMAD

2

Make a Kirchoff series with the sketch guide that was shown in Figure

3

Turn on multimeter by turning the selector on a DC current with 10A

4

Turn on the second power supply or V2 and rotate the selector according to the data provided

E. DATA EXPERIMENT Experiment I Replace the voltage at V1 𝜀1(V) 2V 4V 6V

No 1 2 3

𝜀2(V) 3V 3V 3V

I1 (A) 0 0.03 0.07

I2 (A) 0.18 0.24 0.3

I3 (A) 0.18 0.27 0.36

I1 (A ) 0 -0.03 -0.07

I2 (A) 0.18 0.31 0.43

I3 (A) 0.18 0.25 0, 34

Experiment II Replace the voltage at V2 𝜀1V 2V 2V 2V

No 1 2 3

𝜀2V 3V 6V 9V

F. DISCUSSION In this practice, we have conducted several experiments to prove Kirchoff's Law, where the practitioner uses two power supplies as a voltage source and three resistors with resistance on the first resistor 22 Ω, the second 12 Ω, and the third 15 Ω. Praktikan conducted two experiments

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where experiments with normal polarity but the values on V1 and V2 were changed, but at the time of the lab began, it was found that the amperemeter used to maintain the I3 current was unstable, and turned on he showed the number -0, 01 Ampere. In the first practicum, with a value of 𝜀1 which is changed, that is,

with a value of 2 volts, 4 volts, and 6 volts, while the value of 𝜀2 is fixed.

Based on the experimental data that has been obtained by the practitioner, it is clear that thevalue of kepadachange in1 has an effect on the flowing electric current, where when the voltage 𝜀1 is greater, the value of the

current in each amperemeter will be even greater. In this first Kirchoff Law

practice the value of 𝐼3 smaa with a value of 𝐼1, which when added to the

value of 𝐼2, at a voltage of 4 volts also has thevalue of same𝐼3 when 𝐼1 is added to the value 𝐼2 except for the voltage source using the value of 6 volts,

then the current that is owned is different from the voltage values V1, and V2. In the next practicum, there is a fixed (unchanging) source value at 𝜀1, which is fixed, while thevalues yaitu2change, namely 3 volts, 6 volts,

and 9 volts. From the experimental data that have been obtained by the practitioner, it was obvious that the value of ε2 fickle also gives effect to the current flow, wherein if the value ofthe ε2 the greatervoltage,t he current

value at each ammeters would be even greater. In this second practicum ,the value of 𝐼3 is not the same as the value of 𝐼1, which is summed with 𝐼2, as well as the second experiment with the power source at 𝜀1 which is different

when added 𝐼1 and 𝐼2, which results do not match 𝐼3. This seems occurs because of a practice error in viewing the ammeters, because based on the

calculation of the circuit analysis the value of 𝐼1 should be positive at the 9 volt and 6 volt voltage sources.

Based on the second data from the trial results, if the voltage value gets bigger, then the value of the current will be even greater, and vice versa, if the voltage value gets smaller, then the value of the current flowing will

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be even smaller. This is in accordance with the relation to Ohm's Law, that the voltage value will be directly proportional to the current flowing. The results of the circuit analysis written in the post-practicum assignment have values from the results of calculations with readable values on the ammeters that are close to but there are also appropriate or appropriate values for the first and second practicums, while in this lab we get the same three data if 𝐼1 and 𝐼2 were added to these two experiments,

and the calculated values with the values read on the ammeters must have varied, namely at 𝐼1 where the calculated values have positive and negative

values in calculating the data and data obtained when practicum.

Before doing a practicum, practice beforehand it has been hypothesized in advance that the current value measured in ammeters based on practicum values is greater than the results. The hypothesis is very simple but it is done to answer the pretest only. In the calculation of circuit analysis, this can happen because the calculation of circuit analysis only calculates the resistance value that exists in the resistor, when in fact there are also wire barriers and resistance to the power supply, and this is evident in the results of calculation of circuit analysis. When our labs get new things in our day without accident, suddenly one of the resistors is burned so that the resistor becomes burnt on the color bracelet, this is because the tension source is too high, causing the resistor to burn, and also the negligence of the practitioner. those who do not understand Kirchoff's practicum, in this Kirchoff Law practicum the practitioner can practice this Kirchoff Law until it is finished with Reza's guidance. G. POST PRACTICUM TASKS 1. Explain the shortcomings of series, parallel, and parallel series! Include the solution! Answer: a. The series circuit has the same large current on each obstacle (eg lights) because the amount of charge that passes through obstacle 1

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will also pass the obstacle 2 at the same time interval. Two elements are in a series if only have a main point that is not connected to the current carrying element on a network. b. In series, material series does not need a cost that is quite expensive because it requires a little cable, but when there is an electrical surge all the obstacles (lights) will completely extinguish because all current lines will be cut off. The popularity of series lights is reduced due to the search for broken lights must try to replace a good light bulb on each seat along with the series until a broken light bulb is found. Series lights are safer to use in Christmas tree decorations. Where the series operates with less energy in each lamp and has a lower temperature. c. Parallel circuits, each obstacle (such as a light bulb) has the same resistance. Where based on the design of parallel pairs of light bulbs that are brighter than mounted series, your name is also hotter than series. (Serway, 2010) 2. Calculate the magnitude of 𝐼1,𝐼2, and 𝐼3 that flow by analyzing the circuit that you practice using Kirchhoff's law! Are there differences in

the values of 𝐼1,𝐼2, and 𝐼3 based on the results of your calculations with the values 𝐼1,𝐼2, and 𝐼3 that are read by the ammeters? Explain! Answer:

Kirchhoff's I Law: 𝐼3 =𝐼1+𝐼2 …………… (1)

Loop 1: 𝜀2−𝐼3𝑅2−𝐼1𝑅1 = 0 𝜀2 =𝐼3𝑅2+𝐼1𝑅1 …………… (2)

Loop 2: 𝜀1−𝐼2𝑅3+𝐼1𝑅1 = 0 𝜀1 =𝐼2𝑅3−𝐼1𝑅1 …………… (3) Equation substitution (1 ) to (2): 𝐼3 =𝐼1+𝐼2

𝜀2 = (𝐼1+𝐼2)𝑅2+𝐼1𝑅1

𝜀2 =𝐼1𝑅2+𝐼2𝑅2+𝐼1𝑅1 …… ……… (4) Elimination of equations (4) and (3): 𝜀2 =𝐼1𝑅2+𝐼2𝑅2+𝐼1𝑅1 KIRCHOFF LAW UTUT

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𝜀1 =𝐼2𝑅3−𝐼1𝑅1 ..... +

𝜀1+𝜀2 =𝐼1𝑅2+𝐼2𝑅2+𝐼2𝑅3 …………… (5)

Change equation (5) to 𝐼1: 𝜀1+𝜀2 =𝐼1𝑅2+𝐼2𝑅2+𝐼2𝑅3 𝐼1 =𝜀1+𝜀2−𝐼2𝑅2−𝐼2𝑅3𝑅2 …………… (6)

Change equation (5) to 𝐼2: 𝜀1+𝜀2−𝐼1𝑅2 =𝐼2𝑅2+𝐼2𝑅3 𝜀1+𝜀2−𝐼1𝑅2 =𝐼2 (𝑅2+𝑅3) 𝜀1 +𝜀2−𝐼1𝑅2𝑅2+𝑅3 =𝐼2

𝐼2 =𝜀1+𝜀2−𝐼1𝑅2𝑅2+𝑅3 …………… (7)

Substitution of equations (7) to (3 ) and get the value 𝐼1: 𝜀1 =𝐼2𝑅3−𝐼1𝑅1

𝜀1 = (𝜀1+𝜀2−𝐼1𝑅2𝑅2+𝑅3)𝑅3−𝐼1𝑅1

𝜀1 =𝜀1𝑅3+𝜀2𝑅3−𝐼1𝑅2𝑅3𝑅2+𝑅3−𝐼1𝑅1 𝜀1

=𝜀1𝑅3+𝜀2𝑅3−𝐼1𝑅2𝑅3−𝐼1𝑅1𝑅2−𝐼1𝑅1𝑅3𝑅2+𝑅3 𝜀1𝑅2+𝜀1𝑅3

=𝜀1𝑅3+𝜀2𝑅3−𝐼1𝑅2𝑅3−𝐼1𝑅1𝑅2−𝐼1𝑅1𝑅3 𝐼1𝑅1𝑅2+𝐼1𝑅2𝑅3+𝐼1𝑅1𝑅3 =𝜀1𝑅3+𝜀2𝑅3−𝜀1𝑅2−𝜀1𝑅3 𝐼1 (𝑅1𝑅2) +𝑅2𝑅3+𝑅1𝑅3)

=𝜀1𝑅3+𝜀2𝑅3−𝜀1𝑅2−𝜀1𝑅3 𝐼1 =𝜀2𝑅3−𝜀1𝑅2𝑅1𝑅2+𝑅2𝑅3+𝑅1𝑅3 …………… (8)

Subs substitute equation (6) to (3) and get the value 𝐼2: 𝜀1 =𝐼2𝑅3−𝐼1𝑅1

𝜀1 =𝐼2𝑅3− (𝜀1+𝜀2−𝐼2𝑅2−𝐼2𝑅3𝑅2)𝑅1 𝜀1 =𝐼2𝑅3− (𝜀1𝑅1+𝜀2𝑅1−𝐼2𝑅1𝑅2−𝐼2𝑅1𝑅3𝑅2) 𝜀1 =𝐼2𝑅2𝑅3 𝜀1𝑅1−𝜀2𝑅1+𝐼2𝑅1𝑅2+𝐼2𝑅1𝑅3𝑅2 𝜀1𝑅2

=𝐼2𝑅2𝑅3−𝜀1𝑅1−𝜀2𝑅1+𝐼2𝑅1𝑅2+𝐼2𝑅1𝑅3 𝐼2𝑅2𝑅3+𝐼2𝑅1𝑅2+𝐼2𝑅1𝑅3 =𝜀1𝑅2+𝜀1𝑅1+𝜀2𝑅1 𝐼2 (𝑅2𝑅3+𝑅1𝑅2+𝑅1𝑅3) =𝜀1𝑅2+𝜀1𝑅1+𝜀2𝑅1 𝐼2 =𝜀1𝑅2+𝜀1𝑅1+𝜀2𝑅1𝑅1𝑅2 +𝑅1𝑅3+𝑅2𝑅3 …………… (9) Substitution of equations (8) and (9) to (1) 𝐼3 =𝐼1+𝐼2

𝐼3 = (𝜀2𝑅3−𝜀1𝑅2 /𝑅1𝑅2+𝑅2𝑅3+𝑅1𝑅3) + (𝜀1𝑅2+𝜀1𝑅1𝜀2𝑅1 /𝑅1𝑅2+𝑅1𝑅3+𝑅2𝑅3)

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𝐼3 =𝜀2𝑅3−𝜀1𝑅2+𝜀1𝑅2+𝜀1𝑅1 +𝜀2𝑅1 /𝑅1𝑅2+𝑅2𝑅3+𝑅1𝑅3 𝐼3 =𝜀2𝑅3+𝜀1𝑅1+𝜀2𝑅1 /𝑅1𝑅2+𝑅2𝑅3+𝑅1𝑅3 So the value: 𝐼1 =𝜀2𝑅3−𝜀1𝑅2 /𝑅1𝑅2+𝑅2𝑅3+𝑅1𝑅3 𝐼2 =𝜀1𝑅2+𝜀1𝑅1+𝜀2𝑅1

/𝑅1𝑅2+𝑅1𝑅3+𝑅2𝑅3 𝐼3 =𝜀2𝑅3+𝜀1𝑅1+𝜀2𝑅1 /𝑅1𝑅2+𝑅2𝑅3+𝑅1𝑅3.

a. Experiment I: A. At I1

𝜀2𝑅3 − 𝜀1𝑅2 𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅1𝑅3 (3) (15) − (2) (12) 𝐼1 = (22)(12) + (12)(15) + (22)(15) (45) − (24) 𝐼1 = 264 + 180 + 330 21 𝐼1 = 774 𝐼1 = 0.027132 𝐴 𝐼1 =

At I1

𝐼1 =

At I1 𝐼1 =

At I2

(3) (15) − (4) (12) (22)(12) + (12)(15) + (22)(15) (45) − (48) 𝐼1 = 264 + 180 + 330 −3 𝐼1 = 774 𝐼1 = −0.003876 𝐴

(3) (15) − (6) (12) (22)(12) + (12)(15) + (22)(15) (45) − (72) 𝐼1 = 264 + 180 + 330 −27 𝐼1 = 774 𝐼1 = −0.034884 𝐴

𝜀1𝑅2 + 𝜀1𝑅1 + 𝜀2𝑅1 𝑅1𝑅2 + 𝑅1𝑅3 + 𝑅2𝑅3 (2)(12) + (2)(12) + (3) (22) 𝐼2 = 264 + 180 + 330 𝐼2 =

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24 + 24 + 66 774 114 𝐼2 = 774 𝐼2 = 0.147287 𝐴

𝐼2 = At I2

𝜀1𝑅2 + 𝜀1𝑅1 + 𝜀2𝑅1 𝑅1𝑅2 + 𝑅1𝑅3 + 𝑅2𝑅3 (4)(12) + (4)(12) + (3) (22) 𝐼2 = 264 + 180 + 330 48 + 48 + 66 𝐼2 = 774 162 𝐼2 = 774 𝐼2 = 0.209302 𝐴 𝐼2 =

At I2

𝜀1𝑅2 + 𝜀1𝑅1 + 𝜀2𝑅1 𝑅1𝑅2 + 𝑅1𝑅3 + 𝑅2𝑅3 (6)(12) + (6)(12) + (3) (22) 𝐼2 = 264 + 180 + 330 72 + 72 + 66 𝐼2 = 774 𝐼2 = 0.271318 𝐴 𝐼2 =

On I3

𝜀2𝑅3 + 𝜀1𝑅1 + 𝜀2𝑅1 𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅1𝑅3 (3)(15) + (2)(22) + ( 3) (22) 𝐼3 = 264 + 180 + 330 45 + 44 + 66 𝐼3 = 264 + 180 + 330 155 𝐼3 = 774 𝐼3 = 0.200258 𝐴 𝐼3 =

On I3

𝜀2𝑅3 + 𝜀1𝑅1 + 𝜀2𝑅1 𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅1𝑅3 (3)(15) + (4)(22) + (3) (22) 𝐼3 = 264 + 180 + 330 45 + 88 + 66 𝐼3 = 264 + 180 + 330 199 𝐼3 = 774 𝐼3 = 0.257106 𝐴 𝐼3 =

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At I3 𝜀2𝑅3 + 𝜀1𝑅1 + 𝜀2𝑅1 𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅1𝑅3 (3)(15) + (6)(22) + (3) (22) 𝐼3 = 264 + 180 + 330 45 + 132 + 66 𝐼3 = 264 + 180 + 330 243 𝐼3 = 774 𝐼3 = 0.313953 𝐴 b. Experiment II 𝐼3 =

B.

At I1 𝜀2𝑅3 − 𝜀1𝑅2 𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅1𝑅3 (3) (15) − (2) (12) 𝐼1 = (22)(12) + (12)(15) + (22)(15) (45) − (24) 𝐼1 = 264 + 180 + 330 21 𝐼1 = 774 𝐼1 = 0.027132 𝐴 𝐼1 =

On I1

𝜀2𝑅3 − 𝜀1𝑅2 𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅1𝑅3 (6) (15) − (2) (12) 𝐼1 = (22)(12) + (12)(15) + (22)(15) (90) − (24) 𝐼1 = 264 + 180 + 330 66 𝐼1 = 774 𝐼1 = 0.085271 𝐴 𝐼1 =

On I1

𝜀2𝑅3 − 𝜀1𝑅2 𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅1𝑅3 (9) (15) − (2) (12) 𝐼1 = (22)(12) + (12)(15) + (22)(15) (135) − (24) 𝐼1 = 264 + 180 + 330 111 𝐼1 = 774 𝐼1 =

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𝐼1 = 0.143411 𝐴

At I2

𝜀1𝑅2 + 𝜀1𝑅1 + 𝜀2𝑅1 𝑅1𝑅2 + 𝑅1𝑅3 + 𝑅2𝑅3 (2)(12) + (2)(22) + (3) (22) 𝐼2 = 264 + 180 + 330 24 + 44 + 66 𝐼2 = 774 134 𝐼2 = 774 𝐼2 = 0.173127 𝐴 𝐼2 =

At I2

𝜀1𝑅2 + 𝜀1𝑅1 + 𝜀2𝑅1 𝑅1𝑅2 + 𝑅1𝑅3 + 𝑅2𝑅3 (2)(12) + (2)(22) + (6) (22) 𝐼2 = 264 + 180 + 330 24 + 44 + 132 𝐼2 = 774 200 𝐼2 = 774 𝐼2 = 0.258398 𝐴 𝐼2 =

At I2

𝜀1𝑅2 + 𝜀1𝑅1 + 𝜀2𝑅1 𝑅1𝑅2 + 𝑅1𝑅3 + 𝑅2𝑅3 (2)(12) + (2)(22) + (9) (22) 𝐼2 = 264 + 180 + 330 24 + 44 + 198 𝐼2 = 774 266 𝐼2 = 774 𝐼2 = 0.343669 𝐴 𝐼2 =

On I3

𝜀2𝑅3 + 𝜀1𝑅1 + 𝜀2𝑅1 𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅1𝑅3 (3)(15) + (2)(22) + (3) (22) 𝐼3 = 264 + 180 + 330 45 + 44 + 66 𝐼3 = 264 + 180 + 330 155 𝐼3 = 774 𝐼3 = 0.200258 𝐴 𝐼3 =

On I3

KIRCHOFF LAW UTUT

MUHAMMAD

𝜀2𝑅3 + 𝜀1𝑅1 + 𝜀2𝑅1 𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅1𝑅3 (6)(15) + (2)(22) + (6) (22) 𝐼3 = 264 + 180 + 330 90 + 44 + 132 𝐼3 = 264 + 180 + 330 266 𝐼3 = 774 𝐼3 = 0.343669𝐴 𝐼3 =

At I3

𝜀2𝑅3 + 𝜀1𝑅1 + 𝜀2𝑅1 𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅1𝑅3 (9)(15) + (2)(22) + (9) (22) 𝐼3 = 264 + 180 + 330 135 + 44 + 198 𝐼3 = 264 + 180 + 330 377 𝐼3 = 774 𝐼3 = 0.343669𝐴 3. Make a conclusion that you get from the results of the lab and compare 𝐼3 =

the truth with t existing eori! Answer:

✔ Values in Kirchoff's Law calculations are more accurate than the values indicated on the ammeters because the current flowing in imperfectly enters the circuit. ✔ The amount of all flows entering the branching will be the same as the amount of current coming out of the branching. ✔ The greater the voltage flowed in the circuit, the greater the current. ✔ A current that is negative in reverse polarity indicates that the sequence that has been compiled is wrong and reversed. ✔ Kirchoff's Law is...