Tags: Electronic principles , Understanding components , Understanding principles. Can u guys hep me? Mark Donnison 27 June at pm Hi, without seeing how your classmate is arriving at an answer it is impossible to say. Thea 17 June at pm Hi! Mark Donnison 07 June at am Hi!
Justin chow 29 May at am hi mark. Mark Donnison 24 May at am Hi, there is a worked example that shows how to insert your values into the formula and how to take it forwards to an answer. Sanjay 05 December at am Really I will your instructions is a very helpful.
Tom 14 September at am Just for fun: task 3 example 1 should be 6 Ohms Also formula for two res in paralel and for more than 2 is exactly the same — if you use the latter for two, but solve it with vulgar fractions without finding smallest denominator, you will get the same formula. Mark Donnison 17 July at pm Hi, thanks for getting in touch.
Mark Donnison 04 June at pm Hi Ian, it's difficult to visualise your circuit without seeing it but I will give an answer based on what I think you have. Patrick Lummumba Sunu 27 May at pm This is awesome.
Mark Donnison 05 July at am Hi Yonela, yes you need to use a different formula as is shown in the worked examples. Mark Donnison 03 April at am Hi Bill, you would use the series calculation first to find the combined resistances. Tom 12 October at pm Hi i read this reply to colin and didnt quite understand. Manuna Kawkni 16 July at am very important in my training. Edwin mothibeli 26 May at am This is very helpful,thank you for the initiative.
Colin gilly 27 January at pm What is the formula if I know what resistance I want to achieve but don't have the correct value available. Kitronik Newsletter Sign up now to be the first to know about the latest products and resources! Your email. Follow us. This schematic represents a flashlight with two cells voltage sources and a single bulb load resistance in series. A battery is a multiple connection of voltaic cells. The disadvantage of series connections of cells in this manner, though, is that their internal resistances add.
This can sometimes be problematic. For example, if you placed two 6v batteries in your car instead of the typical 12v single battery, you would be adding both the emfs and the internal resistances of each battery. You would therefore end up with the same 12v emf, though the internal resistance would then be doubled, causing issues for you when you want to start your engine.
But, if the cells oppose one another—such as when one is put into an appliance backwards—the total emf is less, since it is the algebraic sum of the individual emfs. When it is reversed, it produces an emf that opposes the other, and results in a difference between the two voltage sources. Battery Charger : This represents two voltage sources connected in series with their emfs in opposition.
Current flows in the direction of the greater emf and is limited by the sum of the internal resistances. Note that each emf is represented by script E in the figure. A battery charger connected to a battery is an example of such a connection. The charger must have a larger emf than the battery to reverse current through it. When two voltage sources with identical emfs are connected in parallel and also connected to a load resistance, the total emf is the same as the individual emfs. But the total internal resistance is reduced, since the internal resistances are in parallel.
Thus, the parallel connection can produce a larger current. Two Identical EMFs : Two voltage sources with identical emfs each labeled by script E connected in parallel produce the same emf but have a smaller total internal resistance than the individual sources.
Parallel combinations are often used to deliver more current. The output, or terminal voltage of a voltage source such as a battery, depends on its electromotive force and its internal resistance. Express the relationship between the electromotive force and terminal voltage in a form of equation. When you forget to turn off your car lights, they slowly dim as the battery runs down. Their gradual dimming implies that battery output voltage decreases as the battery is depleted.
The reason for the decrease in output voltage for depleted or overloaded batteries is that all voltage sources have two fundamental parts—a source of electrical energy and an internal resistance.
All voltage sources create a potential difference and can supply current if connected to a resistance. On a small scale, the potential difference creates an electric field that exerts force on charges, causing current.
We call this potential difference the electromotive force abbreviated emf. Emf is not a force at all; it is a special type of potential difference of a source when no current is flowing. Units of emf are volts. Electromotive force is directly related to the source of potential difference, such as the particular combination of chemicals in a battery.
However, emf differs from the voltage output of the device when current flows. The voltage across the terminals of a battery, for example, is less than the emf when the battery supplies current, and it declines further as the battery is depleted or loaded down. The voltage output of a device is measured across its terminals and is called its terminal voltage V. Terminal voltage is given by the equation:.
Schematic Representation of a Voltage Source : Any voltage source in this case, a carbon-zinc dry cell has an emf related to its source of potential difference, and an internal resistance r related to its construction.
Note that the script E stands for emf. Also shown are the output terminals across which the terminal voltage V is measured. Now we can analyze the circuit. The current provided by the voltage source is. This current runs through resistor and is designated as.
Looking at Figure 6. The resistors and are in series so the currents and are equal to. The potential drops are and. The final analysis is to look at the power supplied by the voltage source and the power dissipated by the resistors. The power dissipated by the resistors is. The total energy is constant in any process. Therefore, the power supplied by the voltage source is. Analyzing the power supplied to the circuit and the power dissipated by the resistors is a good check for the validity of the analysis; they should be equal.
We can consider to be the resistance of wires leading to and a Find the equivalent resistance of the circuit. Then use this result to find the equivalent resistance of the series connection with. The current through is equal to the current from the battery. The voltage across can be found using. To find the equivalent resistance of the circuit, notice that the parallel connection of R 2 R2 and R 3 R3 is in series with R 1 R1 , so the equivalent resistance is.
The total resistance of this combination is intermediate between the pure series and pure parallel values and , respectively. The current through is equal to the current supplied by the battery:.
The voltage across is. The voltage applied to and is less than the voltage supplied by the battery by an amount. When wire resistance is large, it can significantly affect the operation of the devices represented by and. To find the current through , we must first find the voltage applied to it. The voltage across the two resistors in parallel is the same:.
The current is less than the that flowed through when it was connected in parallel to the battery in the previous parallel circuit example. The power dissipated by is given by.
The analysis of complex circuits can often be simplified by reducing the circuit to a voltage source and an equivalent resistance. Even if the entire circuit cannot be reduced to a single voltage source and a single equivalent resistance, portions of the circuit may be reduced, greatly simplifying the analysis. Consider the electrical circuits in your home. Give at least two examples of circuits that must use a combination of series and parallel circuits to operate efficiently.
One implication of this last example is that resistance in wires reduces the current and power delivered to a resistor.
If wire resistance is relatively large, as in a worn or a very long extension cord, then this loss can be significant.
If a large current is drawn, the drop in the wires can also be significant and may become apparent from the heat generated in the cord. For example, when you are rummaging in the refrigerator and the motor comes on, the refrigerator light dims momentarily. Similarly, you can see the passenger compartment light dim when you start the engine of your car although this may be due to resistance inside the battery itself.
What is happening in these high-current situations is illustrated in Figure 6. The device represented by has a very low resistance, so when it is switched on, a large current flows. This increased current causes a larger drop in the wires represented by , reducing the voltage across the light bulb which is , which then dims noticeably.
Two resistors connected in series are connected to two resistors that are connected in parallel. The series-parallel combination is connected to a battery. Each resistor has a resistance of.
The wires connecting the resistors and battery have negligible resistance. A current of runs through resistor.
What is the voltage supplied by the voltage source? Since they are in series, the current through equals the current through. Since , the current through each will be. The total resistance R of two or more resistors connected in series is the sum of the individual resistances of the resistors.
For the circuit above the total resistance R is given by:. Find the total resistance of the circuit above. This is a series circuit and so total resistance is found using the equation:.
Adding resistors in series always increases the total resistance. The current has to pass through each resistor in turn so adding an additional resistor adds to the resistance already encountered.
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