0
How do you determine the emf and internal resistance of a cell?
Wiki User
∙ 16y ago
You can measure the emf of a cell by using a voltmeter, as this draws current from a cell. You can use the voltage, the emf, and the load resistance to determine the internal resistance of the cell.
Wiki User
∙ 16y ago
Add your answer:
Earn +20 pts
What formula to find the emf and internal resistance for cell when given the current and the voltage?
I don't think you can do that, with the information provided.
Why emf does not depend on resistance?
Because emf is the very source of voltage, either chemical or inductive, an can be meassured at open circuit only so, internal resistance of the supplier is not affecting it.
What is the difference between the emf of a battery and the lost volts when discharging called?
Internal resistance
What are the applications of comparing Emf of two cells with the help of potentiometer?
I think you are talking about it's internal resistance or it's ability to supply a current to a load without significant drop in voltage. Experiment to find EMF of unknown cell The 1 Metre scaled potentiometer is placed across a cell of known EMF and the 'jockey' or slider connected to one terminal of a galvanometer. The second cell is wired with similar polarity to the first cell and connected from the other side of the galvo to the common negative terminal of the two cells. The slider of the potentiometer is progressively moved up from zero until the galvo shows no current. The EMF of the unknown cell is now a direct proportion to the reading in cm read of the metre scale. Example If the known cell has EMF of 3 volts and the galvo balances at 50cm then the unknown cell has EMF of1.5Volts Dry cells used in flashlights have a fairly high internal resistance whereas a Lead acid car cell has a very low internal resistance allowing the starter motor to draw hundreds of Amperes without volt drop.
An old cell with an emf of 1.6 V has an internal resistance of 1.4 How much current will initially flow if its terminals are short-circuited?
I=e/r =1.6/1.4=1.14Amps
Three similarities between pd and emf?
Both emf and pd have the same unit. namely volt Both have the same definition. It is the work performed in moving unit positive charge right from one point to the other against the electric field. Both are got by the product of current and resistance. But emf is the pd across of a cell or source when it is not in action. So emf is slightly greater than the pd in the circuit as there may be a little bit of internal resistance.
A cell of emf E volt and its internal resistance is 'r' ohm. The cell is connected to a load of 'R' ohm. The potential difference across the terminals of the cell is?
The resistance is equally proportionate to "r" in the case that it is above 1. Assuming "r" is greater than 1, the resistance is 4/3 multiplied by omega (the equal proportionate value for mass times ohms). If "r" is less than or equal to 1, there is no resistance.
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.
The Storage battery of a car has an emf of 12 volt if the internal resistance of the battery is 0.4 resistance what is the maximum current that can be drawn from the battery?
You have to imagine the internal resistance as being in parallel with any load you connect. You get the maximum possible current when the load is zero. In this case, just apply Ohm's Law. That is, divide the voltage by the internal resistance.
Is the emf of a cell equal to the total potential drop in the external circuit of the cell?
The EMF of a cell is the voltage across the terminals at zero current. This is the quoted cell voltage but as soon as a current is dawn from the cell, the voltage will drop. It's due to the internal resistance of the cell. In a circuit diagram, a cell is often shown as a voltage source (a perfect source) and a resistor in series to represent the internal resistance. Using Ohms Law, it can be seen that as soon as a current flows, a voltage will be developed across the internal resistance, so reducing the voltage that is seen at the terminals of the cell. The higher the current draw, the higher the voltage drop inside the cell. Normally, the voltage drop is minimal but in most cells, as it loses charge, the internal resistance rises. Eventually it will reach the point where most of the voltage is dropped across the internal resistance, leaving little to drive the intended load. Often, if a battery is removed from a device and measured, the voltage will be measured as equal to or very close to the quoted cell voltage. It is easy to make a judgment that a battery is good when it is almost dead. The only way to confirm the state of the battery is to measure the voltage at the terminals while the load is attached. The results can be very different to the off load voltage. Alkaline cells have a low internal resistance compared to other dry cells. This makes them well suited for high current drain applications. The internal resistance also rises more slowly than most other cells, so they remain useful far longer than zinc-carbon types.
Why emf of battery driving potentiometer is greater than emf of the cell to be measured?
Bcoz the emf which is to be measured is less than emf of driving cell....
How much electromative force is needed to force a current of 45 amperes through a resistance of 35 ohms?
Assuming a perfect machine (doesn't have any internal resistance), an EMF of 45*35 = 1575 volts is required. Actual EMF required by the machine will include the voltage drop internal to the machine, so 1575V constitutes a minimum necessary value.
