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What is the internal resistance of a new D battery with an emf of 1.5 V and a wire of negligible resistance is connected between the terminals of the battery a current of 27 A is produced?
Wiki User
∙ 13y ago
"In short, it is 0.055555555555555555555555555555556Ohms
You can use ohms law to calculate this.
You will use the formula: Resistance Equals Voltage Divided by Current.
It is written: R=V/I
So use the numbers you provided, and the formula above.
1.5 / 27 = 0.055555555555555555555555555555556"
I dispute this answer. emf = I(R+r) is the actual equation u need. From this equation, u will find that V = emf - Ir. Ohm's law is not applicable for this situation. However I do think you will need a voltage to find the internal resistance.
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∙ 13y ago
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How do you calculate the maximum current that can be delivered by a power source of e.m.f. 1.5 V and an internal resistance 0.50 ohms?
This is a terrible way to deal with your power supply, but just going by the givenspecifications:If the output terminals were to be connected (shorted) together, then youwould simply have a 0.5 ohm 'resistor' in a loop with a 1.5 volt source. Thecurrent would beI = E/R = (1.5)/(0.5) = 3 amperes.We have no idea whether the supply is actually capable of delivering that muchcurrent, and in the real world, the internal resistance changes depending on thecurrent being squeezed out of the battery or power supply. But if the supply isrobust and the internal resistance is constant, then 3 amps would be its current"limit" because of that internal 1/2 ohm.To look at it from a slightly different angle ... when the current is 3 amp, thevoltage drop across the internal resistance is (3 x 0.5 ohm) = 1.5 volts, andthere's no remaining voltage available to show up across the output terminals.
Can terminal potential difference be greater than the emf supplied?
All cells have internal resistance. When connected to a load, the resulting load current results in an internal voltage drop across the internal resistance. This voltage drop acts in the opposite sense to the cell's e.m.f., thus causing its terminal voltage to fall below that of the e.m.f. The greater the load current, the greater the difference between the terminal voltage and the e.m.f.
What will happen if you join the two terminals of a cell without connecting them through a switch or a bulb?
A current will flow through whatever you have used to connect the terminals. How much will depend upon the internal resistance of the cell and the resistance of the connecting wire. On cells with a very low internal resistance, such as alkaline cells or lead-acid batteries, the current may be enough to melt the wire. If the connecting wire survives, the cell will be discharged very quickly. If an alkaline battery is discharged too quickly, it will explode. If a lead-acid battery is discharged too quickly, the plates will buckle and the acid may boil.As a general rule, don't try this at home.
What is lost volts?
A battery is rated to supply a certain number of volts. However, it actually supplies less, because they are "lost" as the current has to get out of the battery in the first place.(The battery has internal resistance)The amount of lost volts depends on the current being drawn:The less resistance a circuit has, the more current is drawn, because it's easier to flow.Example:If the circuit has little resistance, it draws a large current and the battery's internal resistance causes more lost volts.If the circuit has high resistance, it draws a small current and there are fewer lost volts.This is why when you short-circuit a battery (give it hardly any resistance to go through) it heats up and may explode. A large current is drawn and all the volts are used by the battery's internal resistance.
How is soul force different from body force?
Soul force is resistance of evil with internal strength. Body force is physical resistance of evil. Gandhi practiced soul force.
Why voltage measured at terminals of battery to which load is connected is always lower than open circuit voltage?
Real-world batteries do not have zero internal resistance. When one connects a load (resistance) to a battery, current begins to flow and the open-circuit potential is divided between the battery's internal resistance and the resistance of the load. Thus, one will measure a lower voltage at the battery terminals when a load is connected, compared to no-load conditions.
A battery has emf of 6 v and an internal resistance of 4 ohm it is connected to a 2.6 ohm resistor a switch when switch is open the potential difference between the terminals of a battery is?
6 volts
Why do you have specific resistance combination inside a voltmeter and an ammeter Explain?
in voltmeter we have internal Resistance and connected in series , to current don't transfer in voltmeter , and we have internal resistance in ammeter and connected in parallel , to most current transfer through the ammeter.
How many amperes does a 6 volt DC battery put out?
That will depend on the sum of the load resistance and the internal resistance of the battery (this is true for all power sources, not just 6 volt batteries). Small compact batteries tend to have higher internal resistance and therefore are more limited in the current they can deliver to a given load than larger batteries.
What is practical voltage source?
in case of ideal voltage source we consider the internal resistance to be zero.but in practical,every battery has some internal resistance then if you connect a load resistance across the terminals of that source,the net potential difference's across the voltage source will be a function of external resistance connects it won't give constant voltage across it's terminals.
Why it is advisable to remove the source and complete the circuit by internal resistance or by a link instead of short circuiting the terminals of the supply?
Because many electronics, like power supplies, have internal energy storage in the form of capacitors. If you were to short the terminals, you would be shorting out the internal capacitors and thus creating a potentially hazardous situation. "Completing the circuit by internal resistance" allows the energy stored in the internal capacitors to bleed off slowly because the internal resistance is larger than a short.
Why does the terminal or load voltage vary when the load resistance is changed?
The behaviour you are describing is, in fact, due to the internal resistance of the voltage source.When a voltage source, such as a battery or generator, is not connected to a load, its potential difference is simply the electromotive force (or 'no-load voltage') of that source. When a load is connected, a load current flows not only through the load itself, but also through the internal resistance of the source. This causes an internal voltage drop across the internal resistance, which acts in the opposite sense (i.e. in accordance with Kirchhoff's Voltage Law), or direction, to the electromotive force, thus reducing the voltage available at the terminals. The greater the load (i.e. the lower the load resistance), the greater the resulting load current, and the greater the internal voltage drop -and the lower the terminal voltage.
A 19 V battery delivers 114 mA of current when connected to a 72 resistor. Determine the internal resistance of the battery.?
Internal resistance is approximately equal to 94.667
What would the current flow be if a short circuit is put across the battery's terminals?
That will depend on the internal resistance of the battery. I = E / R Where I is the current, E is the open circuit battery voltage, and R is the internal resistance of the battery.
A voltmeter is connected to the terminals of a battery that has emf 12.0V and internal resistance 3.00 ohms The battery is not connected to any other external circuitry What is the vmeter's reading?
The voltmeter would read 12 volts. An ammeter connected to to battery would only read 4 amps (12 volts divided by 3 ohms =4)
What is the Thevenin theorem?
The current flowing through the load terminals Rl connected across any two terminals A & B of a linear, bilateral, active network is given by Voc/Rth + RL where VOC is the open circuit voltage across the terminals A & B & RL is the internal resistance of the network as viewed back into the open circuited network from AB deactivating all the independent sources.
Why are there more than two terminals on some ammeters and voltmeters?
A volt meter needs a high internal resistance not to influence the measurement, an ammeter needs a low internal resistance. Sometimes the easiest way to make that happen is to have different connectors.
