Electrical Energy
To begin the study of electrical energy, try drawing your attention 14. A battery with a voltage V, during the time t as q electron charge flowing through the resistance R. For that batteries do business W equal to the change in potential energy Ep = Vq.Figure 14. Batteries generateenergy on resistance R
because q = I. t, where I is a strong electric current and t the time, so much work is done are:
W = V.I.t
because V = I. R, then much work W is equal to the electrical energy is
Where: W = electrical energy in JouleI = electric current in AmperesR = resistance in OhmsV = potential difference in Voltst = time in seconds (S)q = charge (C)
Example:
Based on the circuit in addition to specifya. The electrical energy generated by the battery for 1 minute.b. Electrical energy is transformed into heat at R = 4 W for 1 minute.
Where: V = electrical energy in JouleR2 = 4
R2 = 2
t = 1 minute = 60 seconds
Asked:a. battery energy generated W = ....b. energy into heat at R1 = 4, W1 = ....
Installation of fuses on the power tool to anticipate the presence of currents that suddenly enlarged which allows electrical devices can be damaged or frayed. With the fuse, if the flow suddenly enlarged then the fuse will break up and the electrical appliance is not damaged. Fuse on the market has a certain value, namely: 3 A, 5 A, 13 A, 15 A. Fuse form given in the figure (17).
(A) (b)Figure 17. a) fuse wire type b) Fuses type of bullets
Next try you notice the following examples!
Example:Incandescent lamps have a 40 watt/220 Volt specifications. How much power is used on the lamp when mounted at a voltage 110 Volt?Given: P1 = 40 WV1 = 220 WV2 = 110 W
Asked: P2Answer: According to equation (6-8)
Example:A vacuum cleaner has a specification of 440 W/220 V. If the value of existing sakering 3 A, 5 A, 13 A and 15 A. Sakering which one to choose?Given:P = 440 W, V = 220 VValue sakering 3 A, 5 A, 13 A and 15 A
Asked: Where Sakering selected?Answer: P = V iI =
I = 2A3A fuse is used is
EXERCISE 4To test your understanding, try the following exercise you are doing this! Work out in advance do not immediately see the answer key!
1. Load 5 ditimbangkan on the element that has a 12 Volt barriers in 1. How much energy is absorbed at the load 5 W for 5 minutes?
2. Lamp 6 is connected to the accumulator 12 Volt lights turned out normal. What is the power on the lights?3. 600 Volt Watt/220 color TV turned on each day on average for 8 hours. What is the electrical energy used by the TV every day?4. Watt/220 400 Volt heating element is used for cooking as much as 1 kg of water temperature of 20 0C to boil at a temperature of 100 0C. If the heat of water 4200 J / kg 0C, how long the water will boil?5. Watt/250 100 Volt incandescent lamp mounted on a 200 Volt voltage. What is the current flowing in the lamp?6. A color TV sakering 1000 W/220 V requires a security if the value of existing sakering is 3 A, 5 A, 13 A, 15 A. What is the value used sakering?7.
In the picture above A, B and C are identical lamps 24 W/12 V. What is the power dissipated in the lamp?
KEY EXERCISE 4
1. Given:= 12V; r = 1
R = 5; t = 300 s
Asked: erergi W at the load R
Answer: I =
W = I 2 Rt = (2) 2 (5) (300)W = 6000 Joule
2. Given: V = 20R = 6
Asked: Power P
Answer: P =
P = 24 Watts
3. Given: P = 600 WV = 220 Vt = 8 h
Asked: Energy W. ...
Answer: P =
W = (600 W) (8 h)W = 4800 WhW = 4.8 Kwh
4. Given: P = 400 WV = 220 Vm = 1 kgc = 4200 J / kg 0CT= 100 - 20 = 80 0C
Asked: t =
Answer: P.t = m.c. T
t =
t =
5. Given: P1 = 100 WV1 = 250 VV2 = 200 V
Asked: Flow I =....
Answer: P =
Large current I =
6. Given: P = 1000
V = 220 VThe fuse rating = 3A, 5A, 13A, 15A
Answer: I =
= 4.55 AFuses used are: 5A
7.
Given: Lamp A, B and C type 24 W/12VAsked: The power dissipated in the lamp (p)
Answer: Obstacles each lamp
The current flowing
Power on all the lights
You have completed this module activity 4. Furthermore, the following four tasks do it correctly. To see the truth of your work, match your answers with the answer key that existed at the end of the module. If there is any repeat once again that you really understand it.
TASK 4
1. A coil having energized in 1000 constraints of 2 A for 10 minutes. What is the energy used in the component?
2. An electrical appliance has a resistance 25 when energized for 10 minutes to absorb the energy of 60 kilo Joule. How large a current flowing?
3. Barriers 50 are connected to the battery 12 V. what is the power dissipation in resistance?
4. A lamp has a specification 100 W/220 V. What obstacles such lights?
5. Watt/220 350 Volt electric irons used for 4 hours. How many KWh of electric energy in use?
6. A waterfall 100 feet high dam has a discharge flow 50 m3s-1. Waterfalls are used to turn a generator. If the acceleration of gravity 10 ms-2 and the density of water 100 kgm-3 and 80 waterfalls energy back into electrical energy. How much power is generated?
7. A coil water heater 100 Volt heating Watt/220 5 liters of water for 20 minutes of a temperature of 30 0C, water type heat 4200 J / kg 0C. What is the final temperature of water?
8. Three lamps each 36 W/12V, 24W/12 V and 12 W/12 V arranged in parallel and then connected to a 12 Volt battery. What is the power dissipation on all the lights?
9. A winch 220 V 12 A currents require to lift 800 kg with a speed of 9 m / min. Determine the efficiency of the engine if g = 10 m s-2
10. 100 W/200 V light bulbs will be installed at a voltage 250 V. So that the light is normal, how many obstacles that must be diserikan with lights?
1. STRONG FLOW OF ELECTRICITY
Have you ever heard the word strong electric current? Try to remember! Lights in your home caused by electric current in alternating current circuits.
If you connect a small electric lamp and battery with a cable, what happens? Lamp is lit, which is caused by the flow of electricity in direct current circuits.
Electricity generated by electric charge moving in a conductor. Direction of electric current (I) which arise in the opposite direction to the direction of conductor of electron motion.
Electric charge in a certain amount that penetrates a section of a conductor within a certain time unit called a strong electric current. So strong electric current is the amount of electric charge flowing in the conductor wire per unit time.
Fig. 8 Segment of a current carrying conductor wire
If the time t flow of electric charge Q, then the strong electric current I is:
Greater the number of electric charge that moves, the greater the stronger the current.From the above discussion, if you already understand? If not, consider the example problems below.
Example Problem
If an energized electrical conductor wire electric charge of 360 coulombs in 1 minute, we can determine the strong electric current through the conductor wire. The trick as follows:
Given: Q = 360 coulombs
t = 1 minute = 60 seconds
So strong electric current (I) is ...
So strong electric current (I) it is 6 A.
How, easy is not it! If you already understand it, now you try to solve the following problems!. Remember, this exercise is done independently!
Exercise Problem
1. What is a strong electric current?
2. State the unit of electric current strong!
3. Amount of charge that flows through a wire conductor cross section
with a strong electric current 2 amperes for 15 minutes is ....
ANSWER KEY
1. Strong electric current is the amount of electric charge flowing in the conductor wire per unit time.
2. Amperes or coulombs / second.
3. Given: I = 2 Amperes
t = 15 minutes = 900 seconds
Asked: Q =
Answer:
1. I Kirchhoff's law
This discussion is a continuation of material on the module-1 and 2 activities before. The electrical current that you already know even understand it, if flowing like ... the water flow from higher ground to lower ground or electric current is the flow of current from the high potential is called the positive pole via wires (outer circuit) to the low potential is called the negative pole.
In its flow, electrical current is also experiencing the branches. When an electric current through a branching, the electric current is divided at each bifurcation, and the amount depends on the presence or absence of barriers to the branch. If the resistance at the branch is large, the result of electrical current through the branch is also smaller, and vice versa when the branch, then the obstacles are small electrical currents through the branches of a large electrical currents.
Furthermore, the relationship of strong electric currents that enter the branching points / vertices with a number of strong electric current that comes out of the branching points will be investigated by experiments on the sheet you are expected to experiment and try it.
From the experiments will be obtained that designation A1 ampere meter equal to the sum of the appointment of A2 and A3 (see figure 10)This is known as Kirchhoff's law I which reads:
I Kirchhoff's law is actually none other than call it by the law of conservation of electric charge as shown in drawing an analogy to Kirchhoff berikut.Hukum I can be written mathematically as:
Figure 8. Schematic diagram for Figure 10. Circuit forI Kirchhoff's law to investigate strong currents entering
and out of a node
If you have been listening to the description above, and has understood it, please you trying to accomplish / do the following.If you are unfamiliar with either, please try again until you can understand it very well.
Example question: Consider a node A of an electrical circuit as shown in the picture! Strong currents I1 = 10 A, I2 = 5 A direction toward the point A. Strong currents I3 = 8 A way out from point A What is the magnitude and direction of strong currents I4?
Settlement: according to Kirchhoff's law I =
Next = = 10 +5 = 15 amperes
= 8 A at him out of the point A is I4 must be directed out so that:
= I3 + I4 I4 = 8 +
Finally:
I1 + I2 = I3 + I4I4 I5 = 8 +I4 = 15-8 = 7AI4 = 7 amperes at him out of the point A
1. Work Problem The following exercise
1. Consider the electrical circuit below, then the direction of the electric current of the circuit is ....
2. Mention a powerful tool to measure the electrical currents that!3. In the electrical circuit is arranged how ammeters?4. I write the definition of Kirchhoff's law!5. There are five electricity flowing branching, branching electricity flowing into the I1 = 10 amperes, I2 = 5 ampere electrical current carrying while branching out of I3 = 5 amperes, I4 I5 = 7 amperes while the large and its direction must be determined, determine the I5!
KEY EXERCISE 11. Direction of electric current in an electrical circuit that is the direction of flow out of the positive pole through the outer circuit to the negative pole.2. Ammeters3. Arranged in series4. The number of strong electric current into a node equals the number of strong electric current that comes out of these nodes.5. Completion:
Experiment 3Investigate Strong Electric currents at a node* Tools and materials needed:1. bulb 3 pieces each of 1.5 V (L1, L2, L3)2. ammeters 3 pieces (A1, A2, A3)3. 1.5 volt battery 3 pieces4. Power supply DC to 1.5 volts, 3 volts and 4.5 volts5. connecting cable6. connecting the switch (S)
* How the implementation of the experiment:1. The series of tools as shown below:
2. Are all the lights on?
3. If all the lights on, read the figures used by the tool
A1, A2 and A3.4. Record the figures shown by A2 and A3 with the point P
is a branch point sequence.5. Write down your conclusions from the results of this experiment!
Kirchhoff's laws>> Objectives: After studying this activity, you are expected to understand the description of the material according to Kirchoff's law indicators, namely: I describe the sound of Kirchhoff's law correctly, determine the current strength at a branching point where the necessary data available, determine a strong current in one one resistor of a circuit consisting of three resistors in series - parallel and calculate the strong current in a circuit consisting of three resistors in series - parallel and connected with a battery that has a certain barriers when necessary data are available.| Legal I Kirchhoff | Kirchhoff's Second Law | Task 3 | Learning Activity 1 || Learning Activity 2 | Learning Activity 4 | Home |________________________________________
2. Kirchhoff's Second Law
Use Kirchhoff's second law in a closed circuit that is because there is a series that can not be simplified using a combination of series and parallel.
Generally this occurs when two or more emf in the circuit which is connected by a series of such complicated that simplification requires special techniques to be able to explain or operate the circuit. So the second law of Kirchhoff is a solution for these circuits, which reads:
I Kirchhoff's law is actually none other than call it by the law of conservation of electric charge as shown in drawing an analogy to Kirchhoff berikut.Hukum I can be written mathematically as:
Formulated:
Furthermore, there are several steps you through the activities introduced in the first three of this series with one loop (loop is a closed circuit) and then the circuit with two loops or more. Well ... You further please refer to the following:
Circuit with a loopIn figure 12 below shows a simple circuit with a single loop. In such settings, an electric current that flows is the same, ie I (due to the circuit is closed).
In solving the problem in the loop note the following!a. Strong flow is positive if the direction of the loop and are negative if the opposite direction of the loop.
b. Emf is positive if the polar positipnya first encountered the opposite emf loop and negative if the negative pole first encountered in the loop.
Suppose we take the direction of the loop in the direction of I, namely abcda
Figure 12. Circuit with a loop
Strong electrical currents I above can be determined using Kirchhoff's Second Law
If the price of e1, e2, r1, r2 & R is known then we can determine the price I was!
Circuit with two loops or moreCircuit that has two or more loops is called a compound circuit. The steps in solving a series of these compounds is as follows:
Figure 13. Circuit with two loops
a. Draw the electrical circuit of the circuit of the compound!b. Set the direction of strong currents for each branch.c. Write the equations of flow for each branch point by using Kirchhoff's Law I!d. Set the loop along with him on every circuit is closed!e. Write down the equations for each loop by using Kirchhoff's Second Law equations!f. Calculate the quantities in question by using the letter e in the above equation!
Well ... pelikkah enough ... But you need not despair, because here are some examples of questions that will help you understand one of the fundamental principles of Electrical Sciences, please you listen ...
Example ProblemAt first the electrical circuit consisting of a loop!Consider the matter closed circuit consisting of one loop in the picture below!
Calculate:a. Strong electrical currents (I) flowing in the circuit of the above!b. Electric voltage between points B and D (VBD)
Completion:> Notice by you! ... ... That is the direction of the loop, the direction of electric current (I) and
closely will the prices of electrical components that are known!a. According to Kirchhoff's second law, in a closed circuit applies the equation:
Thus, a strong electric current (I) that flow is 0.5 amperes.Now we already know the big powerful electrical currents flowing in the circuit wire above!Next we will determine a large voltage between two points!
b. We can calculate the voltage between A and D (VBD) for the trajectory that took BAD or BCD.{Pay attention to road pricing BAD I negative (-)}
So the voltage between point B to point D of VBD is + 20 volts, in a similar way you can determine that a large VDB = - 20 volts, please try.
Now you'll be shown the following example problems for the loop (closed circuit) with 2 (two) loop with some electrical components, please you listen!
Example question:Consider the following electrical circuit drawings:
Asked:a. Strong electrical currents flowing in the circuit (I1, I2, and I3).b. Potential difference between A and B (VAB).
Completion:a. Based on Kirchhoff's Law I, at node A:
Based on Kirchhoff's second law for the loop I or loop CABDC:
Based on Kirchhoff's second law for loop II or loop FEABF:
Furthermore subtitusikan (equating to enter a value) equation (1) and (2) so that equation (2) becomes:
Furthermore eliminasikan (remove) equations 3 and 4 so that:- Equation (3): - 10 + 6 + 2 I2 I3 = 0- Equation (4): - 10-6 I2 I3 + 8 = 0
- Enter subtitusikan) I3 = 2 A into equation (2), so that:- 10 + 6 I1 + 2 (2) = 0 ... ... .. 6 I1 = 6 ... ....I1 = 1 Ampere and I2 = I3 - I1 = 2 - 1 = 1 Ampere.So the electric current in the circuit branch BDCA ie I1 = 1 A.Electric current in the circuit branch BFEA ie I2 = 1A.Electric current in the circuit branch AB is I3 = 2 A.All prices {I1, I2 and I3bertanda positive (+), mean direction pemisalan that we have set the appropriate direction I}.
b. We can calculate the large potential difference between A and B (VAB) to track the path A - B (direct), street and road ACDB AEFB (Nah! ... .. there are three ways to determine the VAB! ....)For the road A-B (direct)
For Road A-C-D-B are:
For the road A-E-F-B are:
So large potential difference between points A and B is VAB = + 4 volts, in a similar way you can determine that the big BBA = - 4 volts? ... ...Have you tried it ... ..
Now you can run a job-3 activity below is that you can measure the understanding of the material-3 activity is well!
Kirchhoff's laws>> Objectives: After studying this activity, you are expected to understand the description of the material according to Kirchoff's law indicators, namely: I describe the sound of Kirchhoff's law correctly, determine the current strength at a branching point where the necessary data available, determine a strong current in one one resistor of a circuit consisting of three resistors in series - parallel and calculate the strong current in a circuit consisting of three resistors in series - parallel and connected with a battery that has a certain barriers when necessary data are available.| Legal I Kirchhoff | Kirchhoff's Second Law | Task 3 | Learning Activity 1 || Learning Activity 2 | Learning Activity 4 | Home |________________________________________
TASK 3Instructions:a. Choose the one answer you think is most appropriate!b. Answer key at the end of the module visible only after you have answered all the questions below.
1. Analogous to the flow direction with the direction of water flowing from:. A. lowlandB. plateauC. highlands to the lowlandsD. lowlands to the highlandsE. from the highlands to the lowlands
2. Analogous to the polar plateau charged battery ....A. negativeB. positiveC. neutralD. Positive - negativeE. negative - positive
3. In its flow, electrical current if a major obstacle then the large electric current that flows will be more ... the flow.A. changingB. certainC. serpentineD. E. Small large
4. In experiments investigating the strong electric current at a node when more and more branches the more electric current is divided into subdivisions, then you ....A. doubtfulB. not agreeC. agreeD. should be tried againE. can not be known
5. The number of strong electric current into a node equals the number of strong electric current that comes out of these nodes. This statement is known as:A. Electrical Current LawB. Ohm's lawC. I Kirchhoff's lawD. Kirchhoff's Second LawE. Electricity Law of Total Flow
6. In a closed circuit, the algebraic sum of electromotive force (e) the potential drop (IR) is equal to zero. This statement is known as:A. Ohm's lawB. I Kirchhoff's lawC. Kirchhoff's Second LawD. Electricity Law of Motion StyleE. Basic Law of Electricity
7. If I1 = 10 A, I2 = 5 A, I3 and I4 = 5A = 12 A, the i5 can be determined by
A. 8 AB. 2 AC. 1 AD. 32 AE. 12 A
8. From question no. 7 above the direction of I5 that is ... from the point A!A. exit - entryB. spinC. just quietlyD. entryE. exit
9. From the circuit in addition to a large potential difference between point D and point A (VDA), namely:A. - 6 VB. 6 VC. 4 VD. - 4 VE. 8 V
10. From the data series in addition to magnitude and direction of flow on resistance 2 W (wire AB), namely:
A. 8 A from A to BB. 6 A from A to BC. 4 A from A to BD. 2 A from A to BE. 1 A from A to B
When you have completed the task of activity-3 well, then cocokkanlah your answers with the answer key of course and should not be confused or wrong address, the other will happen! ... ..
You can calculate the answer that you made, namely:
When you reach the mastery level of 70% let alone more, you can proceed to the next module or activity. Suppose your mastery level below 70% or less than you expected to learn again, continuously so as to achieve mastery level> 70%.Reach, you penguasan levels as high as the star-studded in the sky and do not be pessimistic, because people who are pessimistic are people who fail to live his life!
.: LEARNING ACTIVITY 1ELECTRICAL MEASURING TOOLS>> Objectives: After studying this activity, you are expected to have the ability to distinguish types of electrical measuring instruments, electrical measuring instruments mentioned functions, explains how the measurement of strong currents and describes how the measurement of electrical voltage.| Ampermeter | Voltmeter | Task 1 | 2 Learning activities | Activities learn 3 || Learning activities 4 | Home |________________________________________
Description
When you talk about electricity, it will not regardless of the quantities that exist in the electricity itself. Do you remember what size is it? In the first module class X, you already know that the scale: something that can be measured and expressed in units. Scale that exist in such a strong electric current is called Ampermeter, while the tools for measuring voltage or potential difference between two points is called Volt meter. The meter will we learn in this activity is a digital electrical measuring instruments. Analogue electrical measuring instruments having inaccuracy of about 3% to 4%. This is the usual tools available in science laboratories in schools.1. Ampermeter
Ampermeter is a tool for measuring electric current. The most important part of Ampermeter is the galvanometer. Galvanometer working principle of the force between magnetic field and current carrying coil.
Galvanometer can be used directly for measuring small direct currents strong. The greater the current through the coil the greater the deviation of the galvanometer. The workings of the galvanometer will be discussed further at the time you study the magnetic field in the class XII science majors.
Ampermeter consists of a galvanometer connected in parallel with the resistor which has low resistance. The goal is to raise the limit measure ampermeter. The measurement results will be readable on a scale that exist in ampermeter.
How do I use Ampermeter?Suppose you will measure the current strength through the circuit in Figure 1. Suppose R is a light, then:
a b
Figure 1. a. drawing a simple circuit with a dc current source.b. actual circuit
You have to put in series with the lamp ampermeter. So that should cut off one end (the lights to be extinguished). Next connect the cable at both ends with ampermeter, such as figure 2.
Figure 2. The series of how to use Ampermeter
Figure 3. Multimeter that can be used as Ampermeter
Be careful when you read the scale used, because you must pay attention to the limit measure is used. Suppose you use a limit measure 1A, the numbers written on a scale from 0 to 10. This means when the needle pointed to the 10 strong ampermeter current flows only 1A. If it shows number 5 means strong currents flowing 0.5 A. In general, the results of observations on ampermeter readings can be written:
What if when you measure the current strength through the needle deviate measuring the maximum limit? This means you measure the current strength is greater than the measuring instrument. You must enlarge the limit measure by sliding the measuring limit if it is still possible. If you do not have to install a shunt resistance in parallel on Ampermeter as in figure 4 below.
Figure 4. The series shunt resistance (RSH) Ampermeter toenlarge the measuring limit.
Large shunt resistance installed in Ampermeter are:
with shunt resistance RSH = unit (pronounced Ohm)
n = Multiples limit measure
I = Limit gauge mounted after the shunt resistance (A)
IA = limit before the tide measuring shunt resistance (A)
RA = Barriers in Ampermeter ()
To better understand the above description to learn about the following example.
1. How strong current flowing in the circuit below?
Unknown: Scale maximum = 10
Measuring limit = 5A
Asked: Results of observation?
Answer: The results of observations =
= 2A
2. Ampermeter have a bottleneck in 4, only capable of measuring up to 5 M A. Ampermeter will be used to measure the magnitude of the electric current reaches 10 A. Determine the magnitude of the shunt resistance should be installed in parallel on Ampermeter.
TASK 3Instructions:a. Choose the one answer you think is most appropriate!b. Answer key at the end of the module visible only after you have answered all the questions below.
1. Analogous to the flow direction with the direction of water flowing from:. A. lowlandB. plateauC. highlands to the lowlandsD. lowlands to the highlandsE. from the highlands to the lowlands
2. Analogous to the polar plateau charged battery ....A. negativeB. positiveC. neutralD. Positive - negativeE. negative - positive
3. In its flow, electrical current if a major obstacle then the large electric current that flows will be more ... the flow.A. changingB. certainC. serpentineD. E. Small large
4. In experiments investigating the strong electric current at a node when more and more branches the more electric current is divided into subdivisions, then you ....A. doubtfulB. not agreeC. agreeD. should be tried againE. can not be known
5. The number of strong electric current into a node equals the number of strong electric current that comes out of these nodes. This statement is known as:A. Electrical Current LawB. Ohm's lawC. I Kirchhoff's lawD. Kirchhoff's Second LawE. Electricity Law of Total Flow
6. In a closed circuit, the algebraic sum of electromotive force (e) the potential drop (IR) is equal to zero. This statement is known as:A. Ohm's lawB. I Kirchhoff's lawC. Kirchhoff's Second LawD. Electricity Law of Motion StyleE. Basic Law of Electricity
7. If I1 = 10 A, I2 = 5 A, I3 and I4 = 5A = 12 A, the i5 can be determined by
A. 8 AB. 2 AC. 1 AD. 32 AE. 12 A
8. From question no. 7 above the direction of I5 that is ... from the point A!A. exit - entryB. spinC. just quietlyD. entryE. exit
9. From the circuit in addition to a large potential difference between point D and point A (VDA), namely:A. - 6 VB. 6 VC. 4 VD. - 4 VE. 8 V
10. From the data series in addition to magnitude and direction of flow on resistance 2 W (wire AB), namely:
A. 8 A from A to BB. 6 A from A to BC. 4 A from A to BD. 2 A from A to BE. 1 A from A to B
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